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Matrices and Determinants — JEE Previous Year Questions

Every Matrices and Determinants question asked in JEE Main and JEE Advanced across the last 186 papers — 342 questions, each with its correct answer. Free to read, no account needed.

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342

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All 342 Matrices and Determinants questions

Most recent papers first.

Q1·MathematicsMultiple correctJEE Advanced 2026
Consider the matrix M=[2−110]M = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}M=[21​−10​]. Let p,q,r,s,a,b,cp, q, r, s, a, b, cp,q,r,s,a,b,c and ddd be integers such that M26=[pqrs]M^{26} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}M26=[pr​qs​] and ∑k=126Mk=[abcd]\sum_{k=1}^{26} M^k = \begin{bmatrix} a & b \\ c & d \end{bmatrix}∑k=126​Mk=[ac​bd​]. Then which of the following statements is (are) TRUE ?
  1. (A)There exists a 2×22 \times 22×2 invertible matrix NNN with real entries such that MN=N[1101]MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}MN=N[10​11​]
  2. (B)The value of aaa is 378
  3. (C)For any two given integers mmm and nnn, there exist unique integers xxx and yyy such that px+qy=mpx + qy = mpx+qy=m and rx+sy=nrx + sy = nrx+sy=n
  4. (D)For each positive real number ttt, the system of linear equations (a+t)x+by=1(a + t)x + by = 1(a+t)x+by=1 and cx+(d+t)y=−1cx + (d + t)y = -1cx+(d+t)y=−1 has a unique solution

Correct answer: (A), (C), (D)

Step-by-step solution →
Q2·MathematicsMultiple correctJEE Advanced 2026
Let R\mathbb{R}R denote the set of all real numbers and let i=−1i = \sqrt{-1}i=−1​. Consider the matrices S=[0−110]S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}S=[01​−10​] and T=[1101]T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}T=[10​11​]. Let a,b,c,da, b, c, da,b,c,d be real numbers such that ST=[abcd]ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}ST=[ac​bd​]. Let H={x+iy:x,y∈RH = \{x + iy : x, y \in \mathbb{R}H={x+iy:x,y∈R and y>0}y > 0\}y>0}. Then which of the following statements is (are) TRUE ?
  1. (A)b+iad+ic=i\dfrac{b + ia}{d + ic} = id+icb+ia​=i
  2. (B)If ω=−1+i32\omega = \dfrac{-1 + i\sqrt{3}}{2}ω=2−1+i3​​, then aω+bcω+d=ω\dfrac{a\omega + b}{c\omega + d} = \omegacω+daω+b​=ω
  3. (C)If mmm is an integer greater than 2 such that (ST)2=(ST)m(ST)^{2} = (ST)^{m}(ST)2=(ST)m, then mmm is an integer multiple of 8
  4. (D)If z∈Hz \in Hz∈H, then az+bcz+d∈H\dfrac{az + b}{cz + d} \in Hcz+daz+b​∈H

Correct answer: (B), (D)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Advanced 2026
Which one of the following matrices can be obtained by performing elementary row transformations on the 3×33 \times 33×3 identity matrix ?
  1. (A)[111111111]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}​111​111​111​​
  2. (B)[111234121]\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}​121​132​141​​
  3. (C)[111234258]\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{bmatrix}​122​135​148​​
  4. (D)[111−112023]\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{bmatrix}​1−10​112​123​​

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Let A=[α12230045]A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}A=​α20​134​205​​ and B=[1000−5α004α−2α]+adj(A)B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \mathrm{adj}(A)B=​100​0−5α4α​00−2α​​+adj(A). If det⁡(B)=66\det(B) = 66det(B)=66, then det⁡(adj(A))\det(\mathrm{adj}(A))det(adj(A)) equals:
  1. (A)289
  2. (B)361
  3. (C)441
  4. (D)529

Correct answer: (C)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
If the system of linear equations: x+y+z=6x + y + z = 6x+y+z=6, x+2y+5z=10x + 2y + 5z = 10x+2y+5z=10, 2x+3y+λz=μ2x + 3y + \lambda z = \mu2x+3y+λz=μ has infinitely many solutions, then the value of λ+μ\lambda + \muλ+μ equals:
  1. (A)12
  2. (B)16
  3. (C)22
  4. (D)28

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
The sum of all possible values of θ ∈ [0, 2π], for which the system of equations : xcos⁡3θ−8y−12z=0x\cos 3\theta - 8y - 12z = 0xcos3θ−8y−12z=0 xcos⁡2θ+3y+3z=0x\cos 2\theta + 3y + 3z = 0xcos2θ+3y+3z=0 x+y+3z=0x + y + 3z = 0x+y+3z=0 has a non-trivial solution, is equal to :
  1. (A)π
  2. (B)2π
  3. (C)3π
  4. (D)4π

Correct answer: (D)

Step-by-step solution →
Q7·MathematicsNumericalJEE Main 2026
Let A = [−11−1101001]\begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}​−110​100​−111​​ satisfy A2^22 + α(adj(adj(A))) + β(adj(A)(adj(adj(A)))) = [2−22−20−100−1]\begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}​2−20​−200​2−1−1​​ for some α, β ∈ ℝ. Then (α − β)2^22 is equal to ______

Correct answer: 4

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let A=[100310931]A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}A=​139​013​001​​ and B=[bij]B = [b_{ij}]B=[bij​], 1≤i,j≤31 \leq i, j \leq 31≤i,j≤3. If B=A99−IB = A^{99} - IB=A99−I, then the value of b31−b21b32\frac{b_{31} - b_{21}}{b_{32}}b32​b31​−b21​​ is :
  1. (A)99
  2. (B)199
  3. (C)149
  4. (D)159

Correct answer: (C)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let AAA be a 3×33 \times 33×3 matrix such that AT[101]=[522]A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix}AT​101​​=​522​​, AT[001]=[311]A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix}AT​001​​=​311​​, A[101]=[344]A \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix}A​101​​=​344​​ and A[001]=[131]A \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix}A​001​​=​131​​. If det⁡(A)=1\det(A) = 1det(A)=1, then det⁡(adj(A2+A))\det(\mathrm{adj}(A^2 + A))det(adj(A2+A)) is equal to:
  1. (A)16
  2. (B)25
  3. (C)49
  4. (D)64

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Consider the system of linear equations in xxx, yyy, zzz: x+2y+tz=0x + 2y + tz = 0x+2y+tz=0, 6x+y+5tz=06x + y + 5tz = 06x+y+5tz=0, 3x+t2y+f(t)z=03x + t^2 y + f(t)z = 03x+t2y+f(t)z=0, where f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a differentiable function. If this system has infinitely many solutions for all t∈Rt \in \mathbb{R}t∈R, then fff
  1. (A)is a constant function
  2. (B)is strictly increasing on R\mathbb{R}R
  3. (C)is strictly decreasing on R\mathbb{R}R
  4. (D)has two critical points

Correct answer: (B)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let M be a 3×33 \times 33×3 matrix such that M(100)=(123)M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}M​100​​=​123​​, M(010)=(012)M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}M​010​​=​012​​ and M(001)=(−111)M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}M​001​​=​−111​​. If M(xyz)=(1711)M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}M​xyz​​=​1711​​, then x+y+zx + y + zx+y+z equals :
  1. (A)4
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (B)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
If f:N→Zf : \mathbf{N} \to \mathbf{Z}f:N→Z is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}f(n)=​n−2n2−3n3​−13(2k+1)3k(2k+1)​−52k+13k(k+2)+1​​, k ∈ N, and ∑n=1kf(n)=98\sum_{n=1}^{k} f(n) = 98∑n=1k​f(n)=98, then k is equal to :
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let A=[112−201135]A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}A=​1−21​103​215​​. Then the sum of all elements of the matrix adj(adj(2(adjA)−1))\mathrm{adj}(\mathrm{adj}(2(\mathrm{adj}A)^{-1}))adj(adj(2(adjA)−1)) is equal to:
  1. (A)333
  2. (B)444
  3. (C)−4-4−4
  4. (D)−3-3−3

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let S={A=[abcd]:a,b,c,d∈{0,1,2,3,4} and A2−4A+3I=0}S = \left\{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4\} \text{ and } A^2 - 4A + 3I = 0 \right\}S={A=[ac​bd​]:a,b,c,d∈{0,1,2,3,4} and A2−4A+3I=0} be a set of 2×22 \times 22×2 matrices. Then the number of matrices in SSS, for which the sum of the diagonal elements is equal to 4, is:
  1. (A)202020
  2. (B)171717
  3. (C)212121
  4. (D)191919

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
If the system of equations: x+y+z=5x + y + z = 5x+y+z=5 x+2y+3z=9x + 2y + 3z = 9x+2y+3z=9 x+3y+λz=μx + 3y + \lambda z = \mux+3y+λz=μ has infinitely many solutions, then the value of λ+μ\lambda + \muλ+μ is:
  1. (A)16
  2. (B)18
  3. (C)19
  4. (D)21

Correct answer: (B)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
Let A=[121α]A=\begin{bmatrix}1 & 2\\ 1 & \alpha\end{bmatrix}A=[11​2α​] and B=[33β2]B=\begin{bmatrix}3 & 3\\ \beta & 2\end{bmatrix}B=[3β​32​]. If A2−4A+I=OA^{2}-4A+I=OA2−4A+I=O and B2−5B−6I=OB^{2}-5B-6I=OB2−5B−6I=O, then among the two statements : (S1): [(B−A)(B+A)]T=[1315710][(B-A)(B+A)]^{T}=\begin{bmatrix}13 & 15\\ 7 & 10\end{bmatrix}[(B−A)(B+A)]T=[137​1510​] and (S2): det⁡(adj⁡(A+B))=−5\det(\operatorname{adj}(A+B))=-5det(adj(A+B))=−5,
  1. (A)only (S1) is correct
  2. (B)only (S2) is correct
  3. (C)both (S1) and (S2) are correct
  4. (D)both (S1) and (S2) are wrong

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsNumericalJEE Main 2026
Consider the matrices A=[2−24−2]A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}A=[24​−2−2​] and B=[3913]B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}B=[31​93​]. If matrices P and Q are such that PA=BPA = BPA=B and AQ=BAQ = BAQ=B, then the absolute value of the sum of the diagonal elements of 2(P+Q)2(P + Q)2(P+Q) is ________.

Correct answer: 34

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
Let α, β ∈ ℝ be such that the system of linear equations x+2y+z=5x+2y+z=5x+2y+z=5 2x+y+αz=52x+y+\alpha z=52x+y+αz=5 8x+4y+βz=188x+4y+\beta z=188x+4y+βz=18 has no solution. Then βα\frac{\beta}{\alpha}αβ​ is equal to :
  1. (A)−4
  2. (B)4
  3. (C)8
  4. (D)−8

Correct answer: (B)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
If the system of equations x+5y+6z=4x + 5y + 6z = 4x+5y+6z=4, 2x+3y+4z=72x + 3y + 4z = 72x+3y+4z=7, x+6y+az=bx + 6y + az = bx+6y+az=b has infinitely many solutions, then the point (a,b)(a, b)(a,b) lies on the line
  1. (A)y−x=3y - x = 3y−x=3
  2. (B)x−y=3x - y = 3x−y=3
  3. (C)x+y=11x + y = 11x+y=11
  4. (D)x+y=12x + y = 12x+y=12

Correct answer: (B)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2026
Let A, B and C be three 2×22\times22×2 matrices with real entries such that B=(I+A)−1B = (I + A)^{-1}B=(I+A)−1 and A+C=IA + C = IA+C=I. If BC=[1−5−12]BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}BC=[1−1​−52​] and CB[x1x2]=[12−6]CB\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}CB[x1​x2​​]=[12−6​], then x1+x2x_1 + x_2x1​+x2​ is
  1. (A)2
  2. (B)0
  3. (C)–2
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q21·MathematicsNumericalJEE Main 2026
Let A = [3−41−1]\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}[31​−4−1​] and B be two matrices such that A100=100B+IA^{100} = 100B + IA100=100B+I. Then the sum of all the elements of B100B^{100}B100 is ______.

Correct answer: 0

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
Let P=[pij]P = [p_{ij}]P=[pij​] and Q=[qij]Q = [q_{ij}]Q=[qij​] be two square matrices of order 3 such that qij=2(i+j−1)pijq_{ij} = 2^{(i + j - 1)} p_{ij}qij​=2(i+j−1)pij​ and det⁡(Q)=210\det(Q) = 2^{10}det(Q)=210. Then the value of det(adj(adj P)) is :
  1. (A)32
  2. (B)16
  3. (C)81
  4. (D)124

Correct answer: (B)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2026
Let f(x)=∫7x10+9x8(1+x2+2x9)2dxf(x) = \int \frac{7x^{10} + 9x^{8}}{(1 + x^{2} + 2x^{9})^{2}}dxf(x)=∫(1+x2+2x9)27x10+9x8​dx, x>0x > 0x>0, lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0limx→0​f(x)=0 and f(1)=14f(1) = \frac{1}{4}f(1)=41​. If A=[00114f′(1)1α241]A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^{2} & 4 & 1 \end{bmatrix}A=​041​α2​0f′(1)4​111​​ and B = adj(adj A) be such that ∣B∣=81|B| = 81∣B∣=81, then α2\alpha^{2}α2 is equal to
  1. (A)2
  2. (B)3
  3. (C)1
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q24·MathematicsNumericalJEE Main 2026
The number of 3 × 2 matrices A, which can be formed using the elements of the set {−2, −1, 0, 1, 2}\{-2,\ -1,\ 0,\ 1,\ 2\}{−2, −1, 0, 1, 2} such that the sum of all the diagonal elements of ATA\mathrm{A}^{\mathrm{T}}\mathrm{A}ATA is 5, is ______

Correct answer: 312

Step-by-step solution →
Q25·MathematicsNumericalJEE Main 2026
Let ∣A∣=6|A| = 6∣A∣=6, where A is a 3×33 \times 33×3 matrix. If ∣adj(3adj(A2⋅adj(2A)))∣=2m⋅3n|\mathrm{adj}(3\mathrm{adj}(A^{2} \cdot \mathrm{adj}(2A)))| = 2^{m} \cdot 3^{n}∣adj(3adj(A2⋅adj(2A)))∣=2m⋅3n, m,n∈Nm, n \in \mathbf{N}m,n∈N, then m+nm + nm+n is equal to __________.

Correct answer: 62

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2026
Let A=[02−3−2013−10]A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}A=​0−23​20−1​−310​​ and B be a matrix such that B(I−A)=I+AB(I - A) = I + AB(I−A)=I+A. Then the sum of the diagonal elements of BTBB^{T}BBTB is equal to______.

Correct answer: 3

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2026
Among the statements : I : If ∣1cos⁡αcos⁡βcos⁡α1cos⁡γcos⁡βcos⁡γ1∣=∣0cos⁡αcos⁡βcos⁡α0cos⁡γcos⁡βcos⁡γ0∣\begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix}​1cosαcosβ​cosα1cosγ​cosβcosγ1​​=​0cosαcosβ​cosα0cosγ​cosβcosγ0​​, then cos⁡2α+cos⁡2β+cos⁡2γ=32\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2}cos2α+cos2β+cos2γ=23​, and II : If ∣x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1∣=px+q\begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q​x2+x2x2+3x−1x2+2x+3​x+13x2x−1​x−23x−32x−1​​=px+q, then p2=196q2p^2 = 196q^2p2=196q2,
  1. (A)both are false
  2. (B)only II is true
  3. (C)both are true
  4. (D)only I is true

Correct answer: (A)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2026
The system of linear equations x+y+z=6x + y + z = 6x+y+z=6 2x+5y+az=362x + 5y + az = 362x+5y+az=36 x+2y+3z=bx + 2y + 3z = bx+2y+3z=b has
  1. (A)unique solution for a=8a = 8a=8 and b=16b = 16b=16
  2. (B)infinitely many solutions for a=8a = 8a=8 and b=14b = 14b=14
  3. (C)infinitely many solutions for a=8a = 8a=8 and b=16b = 16b=16
  4. (D)unique solution for a=8a = 8a=8 and b=14b = 14b=14

Correct answer: (B)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2026
If X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}X=​xyz​​ is a solution of the system of equations AX=BAX = BAX=B, where adj A=[422−5051−23]A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}A=​4−51​20−2​253​​ and B=[402]B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}B=​402​​, then ∣x+y+z∣|x + y + z|∣x+y+z∣ is equal to :
  1. (A)3
  2. (B)32\frac{3}{2}23​
  3. (C)1
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2026
Let n be the number obtained on rolling a fair die. If the probability that the system x−ny+z=6x - ny + z = 6x−ny+z=6 x+(n−2)y+(n+1)z=8x + (n-2)y + (n+1)z = 8x+(n−2)y+(n+1)z=8 (n−1)y+z=1(n-1)y + z = 1(n−1)y+z=1 Has a unique solution is k6\frac{k}{6}6k​, then the sum of k and all possible values of n is :
  1. (A)21
  2. (B)24
  3. (C)20
  4. (D)22

Correct answer: (D)

Step-by-step solution →
Q31·MathematicsNumericalJEE Main 2026
Let A be a 3×33 \times 33×3 matrix such that A+AT=OA + A^{T} = OA+AT=O. If A[1−10]=[332]A\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}3\\3\\2\end{bmatrix}A​1−10​​=​332​​, A2[1−10]=[−319−24]A^{2}\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}-3\\19\\-24\end{bmatrix}A2​1−10​​=​−319−24​​ and det⁡(adj(2adj(A+I)))=(2)α.(3)β.(11)γ\det(\text{adj}(2\text{adj}(A + I))) = (2)^{\alpha}.(3)^{\beta}.(11)^{\gamma}det(adj(2adj(A+I)))=(2)α.(3)β.(11)γ, α\alphaα, β\betaβ, γ\gammaγ are non-negative integers, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to ______

Correct answer: 18

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2026
If A=[2335]A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}A=[23​35​], then the determinant of the matrix (A2025−3A2024+A2023)(A^{2025} - 3A^{2024} + A^{2023})(A2025−3A2024+A2023) is
  1. (A)28
  2. (B)12
  3. (C)24
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsNumericalJEE Main 2026
For some α, β ∈ R, let A=[α212]A=\begin{bmatrix}\alpha & 2\\ 1 & 2\end{bmatrix}A=[α1​22​] and B=[111β]B=\begin{bmatrix}1 & 1\\ 1 & \beta\end{bmatrix}B=[11​1β​] be such that A2−4A+2I=B2−3B+I=OA^{2}-4A+2I=B^{2}-3B+I=OA2−4A+2I=B2−3B+I=O. Then (det⁡(adj⁡(A3−B3)))2(\det(\operatorname{adj}(A^{3}-B^{3})))^{2}(det(adj(A3−B3)))2 is equal to ………

Correct answer: 225

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2026
If the system of equations 3x+y+4z=33x + y + 4z = 33x+y+4z=3 2x+αy−z=−32x + \alpha y - z = -32x+αy−z=−3 X+2y+z=4X + 2y + z = 4X+2y+z=4 has no solution, then the value of α\alphaα is equal to:
  1. (A)19
  2. (B)4
  3. (C)13
  4. (D)23

Correct answer: (A)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2026
For the matrices A=[3−41−1]A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}A=[31​−4−1​] and B=[−2949−1318]B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}B=[−29−13​4918​], if (A15+B)[xy]=[00](A^{15} + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}(A15+B)[xy​]=[00​], then among the following which one is true?
  1. (A)x=5x = 5x=5, y=7y = 7y=7
  2. (B)x=18x = 18x=18, y=11y = 11y=11
  3. (C)x=11x = 11x=11, y=2y = 2y=2
  4. (D)x=16x = 16x=16, y=3y = 3y=3

Correct answer: (C)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Advanced 2025
Consider the matrix P=(200020003)P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}P=​200​020​003​​. Let the transpose of a matrix X be denote by XTX^{T}XT. Then the number of 3×33 \times 33×3 invertible matrices Q with integer entries, such that Q−1=QTQ^{-1} = Q^{T}Q−1=QT and PQ=QPPQ = QPPQ=QP, is
  1. (A)32
  2. (B)8
  3. (C)16
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q37·MathematicsMultiple correctJEE Advanced 2025
Let I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}I=(10​01​) and P=(2003)P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}P=(20​03​). Let Q=(xyz4)Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}Q=(xz​y4​) for some non-zero real numbers x, y and z for which there is a 2 × 2 matrix R with all entries being non-zero real numbers, such that QR = RP. Then which of the following statements is (are) TRUE?
  1. (A)The determinant of Q − 2I is zero
  2. (B)The determinant of Q − 6I is 12
  3. (C)The determinant of Q − 3I is 15
  4. (D)yz = 2

Correct answer: (A), (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Let α\alphaα be a solution of x2+x+1=0x^2+x+1=0x2+x+1=0, and for some aaa and bbb in R\mathbb{R}R, [4ab][11613−1−12−2−14−8]=[000]\begin{bmatrix}4 & a & b\end{bmatrix}\begin{bmatrix}1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\end{bmatrix}[4​a​b​]​1−1−2​16−1−14​132−8​​=[0​0​0​]. If 4α4+mαa+nαb=3\dfrac{4}{\alpha^4}+\dfrac{m}{\alpha^a}+\dfrac{n}{\alpha^b}=3α44​+αam​+αbn​=3, then m+nm+nm+n is equal to ___
  1. (A)333
  2. (B)111111
  3. (C)777
  4. (D)888

Correct answer: (B)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
Let A=[22+p2+p+q46+2p8+3p+2q612+3p20+6p+3q]A=\begin{bmatrix}2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q\end{bmatrix}A=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​. If det⁡(adj(adj(3A)))=2m⋅3n\det\big(\mathrm{adj}(\mathrm{adj}(3A))\big)=2^m\cdot3^ndet(adj(adj(3A)))=2m⋅3n, m,n∈Nm,n\in\mathbb{N}m,n∈N, then m+nm+nm+n is equal to:
  1. (A)222222
  2. (B)242424
  3. (C)262626
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q40·MathematicsIntegerJEE Main 2025
The number of singular matrices of order 2, whose elements are from the set {2,3,6,9}\{2,3,6,9\}{2,3,6,9} is ______.

Correct answer: 36

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Let the system of equations x+5y−z=1x+5y-z=1x+5y−z=1, 4x+3y−3z=74x+3y-3z=74x+3y−3z=7, 24x+y+λz=μ24x+y+\lambda z=\mu24x+y+λz=μ; λ,μ∈R\lambda,\mu\in\mathbb{R}λ,μ∈R, have infinitely many solutions. Then the number of the solutions of this system, if x,y,zx,y,zx,y,z are integers and satisfy 7≤x+y+z≤777\le x+y+z\le 777≤x+y+z≤77, is
  1. (A)3
  2. (B)6
  3. (C)5
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Let AAA be a 3×33\times33×3 matrix such that ∣adj(adj(adj A))∣=81|\text{adj}(\text{adj}(\text{adj}\,A))|=81∣adj(adj(adjA))∣=81. If S={n∈Z:(∣adj(adj A)∣)(n−1)2/2=∣A∣(3n2−5n−4)}S=\left\{n\in\mathbb{Z}:\left(|\text{adj}(\text{adj}\,A)|\right)^{(n-1)^2/2}=|A|^{(3n^2-5n-4)}\right\}S={n∈Z:(∣adj(adjA)∣)(n−1)2/2=∣A∣(3n2−5n−4)}, then ∑n∈S∣A(n2+n)∣\displaystyle\sum_{n\in S}\left|A^{(n^2+n)}\right|n∈S∑​​A(n2+n)​ is equal to
  1. (A)866
  2. (B)750
  3. (C)820
  4. (D)732

Correct answer: (D)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
Let the system of equations 2x+3y+5z=92x+3y+5z=92x+3y+5z=9, 7x+3y−2z=87x+3y-2z=87x+3y−2z=8, 12x+3y−(4+λ)z=16−μ12x+3y-(4+\lambda)z=16-\mu12x+3y−(4+λ)z=16−μ, have infinitely many solutions. Then the radius of the circle centred at (λ,μ)(\lambda,\mu)(λ,μ) and touching the line 4x=3y4x=3y4x=3y is
  1. (A)175\dfrac{17}{5}517​
  2. (B)75\dfrac{7}{5}57​
  3. (C)7
  4. (D)215\dfrac{21}{5}521​

Correct answer: (B)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
Let the matrix A=[100101010]A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}A=​110​001​010​​ satisfy An=An−2+A2−IA^n=A^{n-2}+A^2-IAn=An−2+A2−I for n≥3n\ge3n≥3. Then the sum of all the elements of A50A^{50}A50 is:
  1. (A)53
  2. (B)52
  3. (C)39
  4. (D)44

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsIntegerJEE Main 2025
Let A=[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]A=\begin{bmatrix}\cos\theta & 0 & -\sin\theta\\0 & 1 & 0\\\sin\theta & 0 & \cos\theta\end{bmatrix}A=​cosθ0sinθ​010​−sinθ0cosθ​​. If for some θ∈(0,π)\theta\in(0,\pi)θ∈(0,π), A2=ATA^2=A^TA2=AT, then the sum of the diagonal elements of the matrix (A+I)3+(A−I)3−6A(A+I)^3+(A-I)^3-6A(A+I)3+(A−I)3−6A is equal to ______.

Correct answer: 6

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
Let AAA be a matrix of order 3×33\times 33×3 and ∣A∣=5|A|=5∣A∣=5. If ∣2 adj(3A adj(2A))∣=2α⋅3β⋅5γ\left|2\,\mathrm{adj}\big(3A\,\mathrm{adj}(2A)\big)\right|=2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}​2adj(3Aadj(2A))​=2α⋅3β⋅5γ, α,β,γ∈N\alpha,\beta,\gamma\in\mathbb{N}α,β,γ∈N, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)25
  2. (B)26
  3. (C)27
  4. (D)28

Correct answer: (C)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
If y(x)=∣sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111∣y(x)=\begin{vmatrix}\sin x & \cos x & \sin x+\cos x+1\\27 & 28 & 27\\1 & 1 & 1\end{vmatrix}y(x)=​sinx271​cosx281​sinx+cosx+1271​​, x∈Rx\in\mathbb{R}x∈R, then d2ydx2+y\dfrac{d^2y}{dx^2}+ydx2d2y​+y is equal to:
  1. (A)−1-1−1
  2. (B)28
  3. (C)27
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsIntegerJEE Main 2025
Let III be the identity matrix of order 3×33\times 33×3 and for the matrix A=[λ234567−12]A=\begin{bmatrix}\lambda & 2 & 3\\4 & 5 & 6\\7 & -1 & 2\end{bmatrix}A=​λ47​25−1​362​​, ∣A∣=−1|A|=-1∣A∣=−1. Let BBB be the inverse of the matrix adj(A adj(A2))\mathrm{adj}\big(A\,\mathrm{adj}(A^2)\big)adj(Aadj(A2)). Then ∣(λB+I)∣|(\lambda B+I)|∣(λB+I)∣ is equal to __________.

Correct answer: -38

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
Let A=[α−16β]A=\begin{bmatrix}\alpha & -1\\ 6 & \beta\end{bmatrix}A=[α6​−1β​], α>0\alpha>0α>0, such that det⁡(A)=0\det(A)=0det(A)=0 and α+β=1\alpha+\beta=1α+β=1. If III denotes 2×22\times22×2 identity matrix, then the matrix (1+A)8(1+A)^8(1+A)8 is:
  1. (A)[4−16−1]\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}[46​−1−1​]
  2. (B)[257−64514−127]\begin{bmatrix}257 & -64\\ 514 & -127\end{bmatrix}[257514​−64−127​]
  3. (C)[1025−5112024−1024]\begin{bmatrix}1025 & -511\\ 2024 & -1024\end{bmatrix}[10252024​−511−1024​]
  4. (D)[766−2551530−509]\begin{bmatrix}766 & -255\\ 1530 & -509\end{bmatrix}[7661530​−255−509​]

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2025
Let a∈Ra\in Ra∈R and AAA be a matrix of order 3×33\times33×3 such that det⁡(A)=−4\det(A)=-4det(A)=−4 and A+I=[1a1210a12]A+I=\begin{bmatrix}1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2\end{bmatrix}A+I=​12a​a11​102​​, where III is the identity matrix of order 3×33\times33×3. If det⁡((a+1) adj((a−1)A))\det((a+1)\,\mathrm{adj}((a-1)A))det((a+1)adj((a−1)A)) is 2m 3n2^m\,3^n2m3n, m,n∈{0,1,2,…,20}m,n\in\{0,1,2,\ldots,20\}m,n∈{0,1,2,…,20}, then m+nm+nm+n is equal to:
  1. (A)14
  2. (B)17
  3. (C)15
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2025
If the system of linear equations 3x+y+βz=33x+y+\beta z=33x+y+βz=3; 2x+αy−z=−32x+\alpha y-z=-32x+αy−z=−3; x+2y+z=4x+2y+z=4x+2y+z=4 has infinitely many solutions, then the value of 22β−9α22\beta-9\alpha22β−9α is:
  1. (A)49
  2. (B)31
  3. (C)43
  4. (D)37

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2025
Let AAA be a 3×33\times33×3 real matrix such that A2(A−2I)−4(A−I)=OA^2(A-2I)-4(A-I)=OA2(A−2I)−4(A−I)=O, where III and OOO are the identity and null matrices respectively. If A4=αA2+βA+γIA^4=\alpha A^2+\beta A+\gamma IA4=αA2+βA+γI, where α,β,γ\alpha,\beta,\gammaα,β,γ are real constants, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)12
  2. (B)20
  3. (C)76
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2025
If the system of equations 2x+λy+3z=52x+\lambda y+3z=52x+λy+3z=5, 3x+2y−z=73x+2y-z=73x+2y−z=7, 4x+5y+μz=94x+5y+\mu z=94x+5y+μz=9 has infinitely many solutions, then (λ2+μ2)(\lambda^2+\mu^2)(λ2+μ2) is equal to:
  1. (A)22
  2. (B)18
  3. (C)26
  4. (D)30

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2025
Let α,β\alpha,\betaα,β (α≠β)(\alpha\ne\beta)(α=β) be the values of m, for which the equations x+y+z=1x+y+z=1x+y+z=1; x+2y+4z=mx+2y+4z=mx+2y+4z=m and x+4y+10z=m2x+4y+10z=m^2x+4y+10z=m2 have infinitely many solutions. Then the value of ∑n=110(nα+nβ)\displaystyle\sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)n=1∑10​(nα+nβ) is equal to:
  1. (A)440
  2. (B)3080
  3. (C)3410
  4. (D)560

Correct answer: (A)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2025
Let A=[aij]A=[a_{ij}]A=[aij​] be a matrix of order 3×33\times 33×3, with aij=(2)i+ja_{ij}=(\sqrt{2})^{i+j}aij​=(2​)i+j. If the sum of all the elements in the third row of A2A^2A2 is α+β2\alpha+\beta\sqrt{2}α+β2​, α,β∈Z\alpha,\beta\in Zα,β∈Z, then α+β\alpha+\betaα+β is equal to
  1. (A)280
  2. (B)168
  3. (C)210
  4. (D)224

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2025
Let A=[aij]=[log⁡5128log⁡45log⁡58log⁡425]A=[a_{ij}]=\begin{bmatrix}\log_5 128 & \log_4 5\\ \log_5 8 & \log_4 25\end{bmatrix}A=[aij​]=[log5​128log5​8​log4​5log4​25​]. If AijA_{ij}Aij​ is the cofactor of aija_{ij}aij​, Cij=∑k=12aikAjkC_{ij}=\displaystyle\sum_{k=1}^{2}a_{ik}A_{jk}Cij​=k=1∑2​aik​Ajk​, 1≤i,j≤21\le i, j\le 21≤i,j≤2, and C=[Cij]C=[C_{ij}]C=[Cij​], then 8∣C∣8|C|8∣C∣ is equal to:
  1. (A)262
  2. (B)288
  3. (C)242
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q57·MathematicsIntegerJEE Main 2025
Let S={m∈Z:Am2+Am=3I−A−6}S=\{m\in Z: A^{m^2}+A^{m}=3I-A^{-6}\}S={m∈Z:Am2+Am=3I−A−6}, where A=[2−110]A=\begin{bmatrix}2 & -1\\ 1 & 0\end{bmatrix}A=[21​−10​]. Then n(S)n(S)n(S) is equal to ______.

Correct answer: 2

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2025
Let M and m respectively be the maximum and the minimum values of f(x)=∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣f(x)=\begin{vmatrix}1+\sin^2 x & \cos^2 x & 4\sin 4x\\ \sin^2 x & 1+\cos^2 x & 4\sin 4x\\ \sin^2 x & \cos^2 x & 1+4\sin 4x\end{vmatrix}f(x)=​1+sin2xsin2xsin2x​cos2x1+cos2xcos2x​4sin4x4sin4x1+4sin4x​​, x∈Rx\in Rx∈R. Then M4−m4M^4-m^4M4−m4 is equal to:
  1. (A)1280
  2. (B)1295
  3. (C)1040
  4. (D)1215

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsIntegerJEE Main 2025
Let M denote the set of all real matrices of order 3×33\times33×3 and let S={−3,−2,−1,1,2}S=\{-3, -2, -1, 1, 2\}S={−3,−2,−1,1,2}. Let S1={A=[aij]∈M:A=ATS_1=\{A=[a_{ij}]\in M: A=A^TS1​={A=[aij​]∈M:A=AT and aij∈S,∀i,j}a_{ij}\in S, \forall i, j\}aij​∈S,∀i,j}, S2={A=[aij]∈M:A=−ATS_2=\{A=[a_{ij}]\in M: A=-A^TS2​={A=[aij​]∈M:A=−AT and aij∈S,∀i,j}a_{ij}\in S, \forall i, j\}aij​∈S,∀i,j}, S3={A=[aij]∈M:a11+a22+a33=0S_3=\{A=[a_{ij}]\in M: a_{11}+a_{22}+a_{33}=0S3​={A=[aij​]∈M:a11​+a22​+a33​=0 and aij∈S,∀i,j}a_{ij}\in S, \forall i, j\}aij​∈S,∀i,j}. If n(S1∪S2∪S3)=125αn(S_1\cup S_2\cup S_3)=125\alphan(S1​∪S2​∪S3​)=125α, then α\alphaα equals.

Correct answer: 1613

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2025
Let A=[12−201]A=\begin{bmatrix}\frac{1}{\sqrt{2}}&-2\\0&1\end{bmatrix}A=[2​1​0​−21​] and P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]P=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}P=[cosθsinθ​−sinθcosθ​], θ>0\theta>0θ>0. If B=PAPTB=PAP^TB=PAPT, C=PTB10PC=P^TB^{10}PC=PTB10P and the sum of the diagonal elements of C is mn\frac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is:
  1. (A)65
  2. (B)127
  3. (C)258
  4. (D)2049

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2025
If the system of equations x+2y−3z=2, 2x+λy+5z=5, 14x+3y+μz=33x+2y-3z=2,\ 2x+\lambda y+5z=5,\ 14x+3y+\mu z=33x+2y−3z=2, 2x+λy+5z=5, 14x+3y+μz=33 has infinitely many solutions, then λ+μ\lambda+\muλ+μ is equal to:
  1. (A)13
  2. (B)10
  3. (C)11
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2025
For some a,ba,ba,b, let f(x)=∣a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx∣f(x)=\begin{vmatrix} a+\dfrac{\sin x}{x} & 1 & b \\ a & 1+\dfrac{\sin x}{x} & b \\ a & 1 & b+\dfrac{\sin x}{x}\end{vmatrix}f(x)=​a+xsinx​aa​11+xsinx​1​bbb+xsinx​​​, x≠0x\neq 0x=0, lim⁡x→0f(x)=λ+μa+νb\displaystyle\lim_{x\to 0}f(x)=\lambda+\mu a+\nu bx→0lim​f(x)=λ+μa+νb. Then (λ+μ+ν)2(\lambda+\mu+\nu)^2(λ+μ+ν)2 is equal to:
  1. (A)25
  2. (B)9
  3. (C)36
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsIntegerJEE Main 2025
Let A be a 3×33\times33×3 matrix such that XTAX=OX^T AX=OXTAX=O for all nonzero 3×13\times13×1 matrices X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}X=​xyz​​. If A[111]=[14−5]A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}A​111​​=​14−5​​, A[121]=[04−8]A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}A​121​​=​04−8​​, and det⁡(adj⁡(2(A+I)))=2α3β5γ\det(\operatorname{adj}(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}det(adj(2(A+I)))=2α3β5γ, α,β,γ∈N\alpha,\beta,\gamma\in\mathbb{N}α,β,γ∈N, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is _______

Correct answer: 44

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2025
If the system of equations 2x−y+z=42x-y+z=42x−y+z=4, 5x+λy+3z=125x+\lambda y+3z=125x+λy+3z=12, 100x−47y+μz=212100x-47y+\mu z=212100x−47y+μz=212, has infinitely many solutions, then μ−2λ\mu-2\lambdaμ−2λ is equal to
  1. (A)565656
  2. (B)595959
  3. (C)555555
  4. (D)575757

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2025
If the system of equations (λ−1)x+(λ−4)y+λz=5(\lambda-1)x+(\lambda-4)y+\lambda z=5(λ−1)x+(λ−4)y+λz=5 λx+(λ−1)y+(λ−4)z=7\lambda x+(\lambda-1)y+(\lambda-4)z=7λx+(λ−1)y+(λ−4)z=7 (λ+1)x+(λ+2)y−(λ+2)z=9(\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9(λ+1)x+(λ+2)y−(λ+2)z=9 has infinitely many solutions, then λ2+λ\lambda^2+\lambdaλ2+λ is equal to
  1. (A)101010
  2. (B)121212
  3. (C)666
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2025
If AAA, BBB and (adj⁡(A−1)+adj⁡(B−1))(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))(adj(A−1)+adj(B−1)) are non-singular matrices of same order, then the inverse of A(adj⁡(A−1)+adj⁡(B−1))−1BA\big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\big)^{-1}BA(adj(A−1)+adj(B−1))−1B, is equal to
  1. (A)AB−1+A−1BAB^{-1}+A^{-1}BAB−1+A−1B
  2. (B)adj⁡(B−1)+adj⁡(A−1)\operatorname{adj}(B^{-1})+\operatorname{adj}(A^{-1})adj(B−1)+adj(A−1)
  3. (C)1∣AB∣(adj⁡(B)+adj⁡(A))\dfrac{1}{|AB|}\big(\operatorname{adj}(B)+\operatorname{adj}(A)\big)∣AB∣1​(adj(B)+adj(A))
  4. (D)AB−1∣A∣+BA−1∣B∣\dfrac{AB^{-1}}{|A|}+\dfrac{BA^{-1}}{|B|}∣A∣AB−1​+∣B∣BA−1​

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2025
The system of equations x+y+z=6, x+2y+5z=9, x+5y+λz=μ,x+y+z=6,\ x+2y+5z=9,\ x+5y+\lambda z=\mu,x+y+z=6, x+2y+5z=9, x+5y+λz=μ, has no solution if :
  1. (A)λ=17, μ≠18\lambda=17,\ \mu\neq 18λ=17, μ=18
  2. (B)λ≠17, μ=18\lambda\neq 17,\ \mu=18λ=17, μ=18
  3. (C)λ=15, μ≠18\lambda=15,\ \mu\neq 18λ=15, μ=18
  4. (D)λ=17, μ=18\lambda=17,\ \mu=18λ=17, μ=18

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2025
Let A=[aij]A=[a_{ij}]A=[aij​] be a 3×33\times 33×3 matrix such that A[010]=[001], A[413]=[110]A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix},\ A\begin{bmatrix}4\\1\\3\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix}A​010​​=​001​​, A​413​​=​110​​ and A[212]=[100]A\begin{bmatrix}2\\1\\2\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}A​212​​=​100​​, then a23a_{23}a23​ equals :
  1. (A)−1-1−1
  2. (B)000
  3. (C)222
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsIntegerJEE Main 2025
Let AAA be a 3×33\times33×3 matrix with det⁡A=−2\det A=-2detA=−2. If det⁡(3 adj(−6 adj(3A)))=2m⋅3n\det\big(3\,\mathrm{adj}(-6\,\mathrm{adj}(3A))\big)=2^{m}\cdot 3^{n}det(3adj(−6adj(3A)))=2m⋅3n with m>nm>nm>n, find 4m+2n4m+2n4m+2n (per the paper).

Correct answer: 34

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2025
For a 3×33\times33×3 matrix MMM, let trace (M)(M)(M) denote the sum of all the diagonal elements of MMM. Let AAA be a 3×33\times33×3 matrix such that ∣A∣=12|A|=\dfrac{1}{2}∣A∣=21​ and trace (A)=3(A)=3(A)=3. If B=adj⁡(adj⁡(2A))B=\operatorname{adj}(\operatorname{adj}(2A))B=adj(adj(2A)), then the value of ∣B∣+trace⁡(B)|B|+\operatorname{trace}(B)∣B∣+trace(B) equals:
  1. (A)56
  2. (B)132
  3. (C)174
  4. (D)280

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2025
If the system of linear equations: x+y+2z=6x+y+2z=6x+y+2z=6, 2x+3y+az=a+12x+3y+az=a+12x+3y+az=a+1, −x−3y+bz=2b-x-3y+bz=2b−x−3y+bz=2b, where a,b∈Ra,b\in\mathbb{R}a,b∈R, has infinitely many solutions, then 7a+3b7a+3b7a+3b is equal to:
  1. (A)9
  2. (B)12
  3. (C)16
  4. (D)22

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsIntegerJEE Advanced 2024
Let S={A(01c1ad1be):a,b,c,d,e∈{0,1} and ∣A∣∈{−1,1}}S = \left\{A\begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\}\right\}S=⎩⎨⎧​A​011​1ab​cde​​:a,b,c,d,e∈{0,1} and ∣A∣∈{−1,1}⎭⎬⎫​, where ∣A∣|A|∣A∣ denotes the determinant of AAA. Then the number of elements in SSS is ______ .

Correct answer: 16

Step-by-step solution →
Q73·MathematicsSingle correctJEE Advanced 2024
Let α\alphaα and β\betaβ be the distinct roots of the equation x2+x−1=0x^{2} + x - 1 = 0x2+x−1=0. Consider the set T={1,α,β}T = \{1, \alpha, \beta\}T={1,α,β}. For a 3×33 \times 33×3 matrix M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​, define Ri=ai1+ai2+ai3R_{i} = a_{i1} + a_{i2} + a_{i3}Ri​=ai1​+ai2​+ai3​ and Cj=a1j+a2j+a3jC_{j} = a_{1j} + a_{2j} + a_{3j}Cj​=a1j​+a2j​+a3j​ for i=1,2,3i = 1, 2, 3i=1,2,3 and j=1,2,3j = 1, 2, 3j=1,2,3. Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.The number of matrices M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ with all entries in TTT such that Ri=Cj=0R_{i} = C_{j} = 0Ri​=Cj​=0 for all i,ji, ji,j, is1.1
Q.The number of symmetric matrices M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ with all entries in TTT such that Cj=0C_{j} = 0Cj​=0 for all jjj, is2.12
R.Let M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ be a skew symmetric matrix such that aij∈Ta_{ij} \in Taij​∈T for i>ji > ji>j. Then the number of elements in the set {(xyz):x,y,z∈R,M(xyz)=(a120−a23)}\left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in R, M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix}\right\}⎩⎨⎧​​xyz​​:x,y,z∈R,M​xyz​​=​a12​0−a23​​​⎭⎬⎫​ is3.infinite
S.Let M=(aij)3×3M = (a_{ij})_{3\times3}M=(aij​)3×3​ be a matrix with all entries in TTT such that Ri=0R_{i} = 0Ri​=0 for all iii. Then the absolute value of determinant of MMM is4.6
5.0
  1. (A)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (5) (S) →\to→ (1)
  2. (B)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (1) (S) →\to→ (5)
  3. (C)(P) →\to→ (2) (Q) →\to→ (4) (R) →\to→ (3) (S) →\to→ (5)
  4. (D)(P) →\to→ (1) (Q) →\to→ (5) (R) →\to→ (3) (S) →\to→ (4)

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsNumericalJEE Main 2024
Let AAA be a non-singular matrix of order 333. If det⁡(3 adj(2 adj((det⁡A)A)))=3−13⋅2−10\det(3\,\mathrm{adj}(2\,\mathrm{adj}((\det A)A))) = 3^{-13} \cdot 2^{-10}det(3adj(2adj((detA)A)))=3−13⋅2−10 and det⁡(3 adj(2A))=2m⋅3n\det(3\,\mathrm{adj}(2A)) = 2^m \cdot 3^ndet(3adj(2A))=2m⋅3n, then ∣3m+2n∣|3m + 2n|∣3m+2n∣ is equal to ________.

Correct answer: 14

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
Let B=[1315]B=\begin{bmatrix}1&3\\1&5\end{bmatrix}B=[11​35​] and A be a 2×22\times 22×2 matrix such that AB−1=A−1AB^{-1}=A^{-1}AB−1=A−1. If BCB−1=ABCB^{-1}=ABCB−1=A and C4+αC2+βI=OC^{4}+\alpha C^{2}+\beta I=OC4+αC2+βI=O, then 2β−α2\beta-\alpha2β−α is equal to:
  1. (A)161616
  2. (B)222
  3. (C)888
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2024
Let λ,μ∈R\lambda, \mu \in \mathbb{R}λ,μ∈R. If the system of equations 3x+5y+λz=33x + 5y + \lambda z = 33x+5y+λz=3, 7x+11y−9z=27x + 11y - 9z = 27x+11y−9z=2, 97x+155y−189z=μ97x + 155y - 189z = \mu97x+155y−189z=μ has infinitely many solutions, then μ+2λ\mu + 2\lambdaμ+2λ is equal to:
  1. (A)25
  2. (B)24
  3. (C)27
  4. (D)22

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsNumericalJEE Main 2024
Consider the matrices A=[2−53m]A=\begin{bmatrix}2&-5\\3&m\end{bmatrix}A=[23​−5m​], B=[20m]B=\begin{bmatrix}20\\m\end{bmatrix}B=[20m​] and X=[xy]X=\begin{bmatrix}x\\y\end{bmatrix}X=[xy​]. Let the set of all mmm, for which the system of equations AX=BAX=BAX=B has a negative solution (i.e., x<0x<0x<0 and y<0y<0y<0), be the interval (a,b)(a, b)(a,b). Then 8∫ab∣A∣ dm8\displaystyle\int_{a}^{b}|A|\,dm8∫ab​∣A∣dm is equal to _______.

Correct answer: 450

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2024
Let A=[2a013105b]A=\begin{bmatrix}2 & a & 0\\ 1 & 3 & 1\\ 0 & 5 & b\end{bmatrix}A=​210​a35​01b​​. If A3=4A2−A−21IA^3=4A^2-A-21IA3=4A2−A−21I, where III is the identity matrix of order 3×33\times33×3, then 2a+3b2a+3b2a+3b is equal to:
  1. (A)−10-10−10
  2. (B)−13-13−13
  3. (C)−9-9−9
  4. (D)−12-12−12

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2024
Let A=[2−111]A=\begin{bmatrix}2 & -1\\ 1 & 1\end{bmatrix}A=[21​−11​]. If the sum of the diagonal elements of A13A^{13}A13 is 3n3^n3n, then nnn is equal to ___

Correct answer: 7

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2024
If α≠a\alpha\ne aα=a, β≠b\beta\ne bβ=b, γ≠c\gamma\ne cγ=c and ∣αbcaβcabγ∣=0\begin{vmatrix}\alpha&b&c\\a&\beta&c\\a&b&\gamma\end{vmatrix}=0​αaa​bβb​ccγ​​=0, then aα−a+bβ−b+γγ−c\dfrac{a}{\alpha-a}+\dfrac{b}{\beta-b}+\dfrac{\gamma}{\gamma-c}α−aa​+β−bb​+γ−cγ​ is equal to
  1. (A)222
  2. (B)333
  3. (C)000
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
If the system of equations x+4y−z=λx+4y-z=\lambdax+4y−z=λ, 7x+9y+μz=−37x+9y+\mu z=-37x+9y+μz=−3, 5x+y+2z=−15x+y+2z=-15x+y+2z=−1 has infinitely many solutions, then (2μ+3λ)(2\mu+3\lambda)(2μ+3λ) is equal to
  1. (A)222
  2. (B)−3-3−3
  3. (C)333
  4. (D)−2-2−2

Correct answer: (B)

Step-by-step solution →
Q82·MathematicsNumericalJEE Main 2024
Let αβγ=45\alpha\beta\gamma=45αβγ=45; α,β,γ∈R\alpha,\beta,\gamma\in\mathbb{R}α,β,γ∈R. If x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0)x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0)x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0) for some x,y,z∈Rx,y,z\in\mathbb{R}x,y,z∈R, xyz≠0xyz\neq 0xyz=0, then 6α+4β+γ6\alpha+4\beta+\gamma6α+4β+γ is equal to _______.

Correct answer: 55

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2024
If AAA is a square matrix of order 333 such that det⁡(A)=3\det(A)=3det(A)=3 and det⁡(adj⁡(−4 adj⁡(−3 adj⁡(3 adj⁡((2A)−1)))))=2m3n\det\left(\operatorname{adj}\left(-4\,\operatorname{adj}\left(-3\,\operatorname{adj}\left(3\,\operatorname{adj}\left((2A)^{-1}\right)\right)\right)\right)\right)=2^{m}3^{n}det(adj(−4adj(−3adj(3adj((2A)−1)))))=2m3n, then m+2nm+2nm+2n is equal to:
  1. (A)333
  2. (B)222
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2024
If the system of equations 2x+7y+λz=32x+7y+\lambda z=32x+7y+λz=3, 3x+2y+5z=43x+2y+5z=43x+2y+5z=4, x+μy+32z=−1x+\mu y+32z=-1x+μy+32z=−1 has infinitely many solutions, then (λ−μ)(\lambda-\mu)(λ−μ) is equal to ______.

Correct answer: 38

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2024
For α,β∈R\alpha,\beta\in\mathbb{R}α,β∈R and a natural number nnn, let Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣A_{r}=\begin{vmatrix} r & 1 & \tfrac{n^{2}}{2}+\alpha \\ 2r & 2 & n^{2}-\beta \\ 3r-2 & 3 & \tfrac{n(3n-1)}{2}\end{vmatrix}Ar​=​r2r3r−2​123​2n2​+αn2−β2n(3n−1)​​​. Then 2A10−A82A_{10}-A_{8}2A10​−A8​ is
  1. (A)4α+2β4\alpha+2\beta4α+2β
  2. (B)2α+4β2\alpha+4\beta2α+4β
  3. (C)2β2\beta2β
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2024
Let A and B be two square matrices of order 3 such that ∣A∣=3|A| = 3∣A∣=3 and ∣B∣=2|B| = 2∣B∣=2. Then ∣ATA(adj(2A))−1(adj(4B))(adj(AB))−1AAT∣\left|A^T A(\text{adj}(2A))^{-1} (\text{adj}(4B))(\text{adj}(AB))^{-1} A A^T\right|​ATA(adj(2A))−1(adj(4B))(adj(AB))−1AAT​ is equal to:
  1. (A)646464
  2. (B)818181
  3. (C)323232
  4. (D)108108108

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2024
If the system of equations 11x+y+λz=−511x + y + \lambda z = -511x+y+λz=−5, 2x+3y+5z=32x + 3y + 5z = 32x+3y+5z=3, 8x−19y−39z=μ8x - 19y - 39z = \mu8x−19y−39z=μ has infinitely many solutions, then λ4−μ\lambda^4 - \muλ4−μ is equal to:
  1. (A)494949
  2. (B)454545
  3. (C)474747
  4. (D)515151

Correct answer: (C)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2024
The values of m,nm,nm,n, for which the system of equations x+y+z=4x+y+z=4x+y+z=4, 2x+5y+5z=172x+5y+5z=172x+5y+5z=17, x+2y+mz=nx+2y+mz=nx+2y+mz=n has infinitely many solutions, satisfy the equation:
  1. (A)m2+n2−m−n=46m^2+n^2-m-n=46m2+n2−m−n=46
  2. (B)m2+n2+m+n=64m^2+n^2+m+n=64m2+n2+m+n=64
  3. (C)m2+n2+mn=68m^2+n^2+mn=68m2+n2+mn=68
  4. (D)m2+n2−mn=39m^2+n^2-mn=39m2+n2−mn=39

Correct answer: (D)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2024
Let αβ≠0\alpha\beta\ne 0αβ=0 and A=[βα3ααβ−βα2α]A=\begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{bmatrix}A=​βα−β​ααα​3β2α​​. If B=[3α−93α−α7−2α−2α5−2β]B=\begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix}B=​3α−α−2α​−975​3α−2α−2β​​ is the matrix of cofactors of the elements of AAA, then det⁡(AB)\det(AB)det(AB) is equal to:
  1. (A)343
  2. (B)125
  3. (C)64
  4. (D)216

Correct answer: (D)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2024
Let α∈(0,∞)\alpha\in(0,\infty)α∈(0,∞) and A=[12α101012]A=\begin{bmatrix}1 & 2 & \alpha\\ 1 & 0 & 1\\ 0 & 1 & 2\end{bmatrix}A=​110​201​α12​​. If det⁡(adj(2A−AT)⋅adj(A−2AT))=28\det\big(\mathrm{adj}(2A-A^T)\cdot\mathrm{adj}(A-2A^T)\big)=2^8det(adj(2A−AT)⋅adj(A−2AT))=28, then ∣A∣2|A|^2∣A∣2 is equal to:
  1. (A)111
  2. (B)494949
  3. (C)161616
  4. (D)444

Correct answer: (C)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2024
Let A=[1201]A=\begin{bmatrix}1&2\\0&1\end{bmatrix}A=[10​21​] and B=I+adj⁡(A)+(adj⁡A)2+…+(adj⁡A)10B=I+\operatorname{adj}(A)+(\operatorname{adj}A)^2+\ldots+(\operatorname{adj}A)^{10}B=I+adj(A)+(adjA)2+…+(adjA)10. Then, the sum of all the elements of the matrix B is
  1. (A)−110-110−110
  2. (B)222222
  3. (C)−88-88−88
  4. (D)−124-124−124

Correct answer: (C)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2024
Let the system of linear equations x+(2sin⁡α)y+(2cos⁡α)z=0x+(\sqrt2\sin\alpha)y+(\sqrt2\cos\alpha)z=0x+(2​sinα)y+(2​cosα)z=0, x+(cos⁡α)y+(sin⁡α)z=0x+(\cos\alpha)y+(\sin\alpha)z=0x+(cosα)y+(sinα)z=0, x+(sin⁡α)y−(cos⁡α)z=0x+(\sin\alpha)y-(\cos\alpha)z=0x+(sinα)y−(cosα)z=0 has a non-trivial solution, then α∈(0,π2)\alpha\in\left(0,\dfrac\pi2\right)α∈(0,2π​) is equal to:
  1. (A)3π4\dfrac{3\pi}{4}43π​
  2. (B)7π24\dfrac{7\pi}{24}247π​
  3. (C)5π24\dfrac{5\pi}{24}245π​
  4. (D)11π24\dfrac{11\pi}{24}2411π​

Correct answer: (C)

Step-by-step solution →
Q93·MathematicsNumericalJEE Main 2024
Let A be a 2×22\times 22×2 symmetric matrix such that A[11]=[37]A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}A[11​]=[37​] and the determinant of A be 1. If A−1=αA+βIA^{-1}=\alpha A+\beta IA−1=αA+βI, where I is an identity matrix of order 2×22\times 22×2, then α+β\alpha+\betaα+β equals

Correct answer: 5

Step-by-step solution →
Q94·MathematicsNumericalJEE Main 2024
Let AAA be a 3×33\times33×3 matrix of non-negative real elements such that A[111]=3[111]A\begin{bmatrix}1\\1\\1\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}A​111​​=3​111​​. Then the maximum value of det⁡(A)\det(A)det(A) is ___

Correct answer: 27

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2024
Let A=I2−2MMTA=I_2-2MM^TA=I2​−2MMT, where MMM is real matrix of order 2×12\times 12×1 such that the relation MTM=I1M^T M=I_1MTM=I1​ holds. If λ\lambdaλ is a real number such that the relation AX=λXAX=\lambda XAX=λX holds for some non-zero real matrix XXX of order 2×12\times 12×1, then the sum of squares of all possible values of λ\lambdaλ is equal to __________.

Correct answer: 2

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2024
If the system of equations 2x+3y−z=52x+3y-z=52x+3y−z=5, x+αy+3z=−4x+\alpha y+3z=-4x+αy+3z=−4, 3x−y+βz=73x-y+\beta z=73x−y+βz=7 has infinitely many solutions, then 13αβ13\alpha\beta13αβ is equal to:
  1. (A)111011101110
  2. (B)112011201120
  3. (C)121012101210
  4. (D)122012201220

Correct answer: (B)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2024
Let the system of equations x+2y+3z=5x+2y+3z=5x+2y+3z=5, 2x+3y+z=92x+3y+z=92x+3y+z=9, 4x+3y+λz=μ4x+3y+\lambda z=\mu4x+3y+λz=μ have infinite number of solutions. Then λ+2μ\lambda+2\muλ+2μ is equal to:
  1. (A)28
  2. (B)17
  3. (C)22
  4. (D)15

Correct answer: (B)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2024
If A=[21−12]A=\begin{bmatrix}\sqrt2 & 1\\ -1 & \sqrt2\end{bmatrix}A=[2​−1​12​​], B=[1011]B=\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}B=[11​01​], C=ABATC=ABA^TC=ABAT and X=ATC2AX=A^T C^2 AX=ATC2A, then det⁡X\det XdetX is equal to:
  1. (A)243243243
  2. (B)729729729
  3. (C)272727
  4. (D)891891891

Correct answer: (B)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2024
If the system of linear equations x−2y+z=−4x-2y+z=-4x−2y+z=−4; 2x+αy+3z=52x+\alpha y+3z=52x+αy+3z=5; 3x−y+βz=33x-y+\beta z=33x−y+βz=3 has infinitely many solutions, then 12α+13β12\alpha+13\beta12α+13β is equal to
  1. (A)606060
  2. (B)646464
  3. (C)545454
  4. (D)585858

Correct answer: (D)

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2024
Let A be a 3×33\times33×3 real matrix such that A(101)=2(101)A\begin{pmatrix}1\\0\\1\end{pmatrix}=2\begin{pmatrix}1\\0\\1\end{pmatrix}A​101​​=2​101​​, A(−101)=4(−101)A\begin{pmatrix}-1\\0\\1\end{pmatrix}=4\begin{pmatrix}-1\\0\\1\end{pmatrix}A​−101​​=4​−101​​, A(010)=2(010)A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}A​010​​=2​010​​. Then, the system (A−3I)(xyz)=(123)(A-3I)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}(A−3I)​xyz​​=​123​​ has
  1. (A)unique solution
  2. (B)exactly two solutions
  3. (C)no solution
  4. (D)infinitely many solutions

Correct answer: (A)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2024
If f(x)=∣x32x2+11+3x3x2+22xx3+6x3−x4x2−2∣f(x)=\begin{vmatrix}x^3 & 2x^2+1 & 1+3x\\ 3x^2+2 & 2x & x^3+6\\ x^3-x & 4 & x^2-2\end{vmatrix}f(x)=​x33x2+2x3−x​2x2+12x4​1+3xx3+6x2−2​​ for all x∈Rx\in Rx∈R, then 2f(0)+f′(0)2f(0)+f'(0)2f(0)+f′(0) is equal to
  1. (A)484848
  2. (B)242424
  3. (C)424242
  4. (D)181818

Correct answer: (C)

Step-by-step solution →
Q102·MathematicsNumericalJEE Main 2024
Let A be a 3×33\times33×3 matrix and det⁡(A)=2\det(A)=2det(A)=2. If n=det⁡(adj(adj(…(adjA))))n=\det\left(\text{adj}\left(\text{adj}\left(\ldots\left(\text{adj}A\right)\right)\right)\right)n=det(adj(adj(…(adjA)))) (adj applied 2024 times). Then the remainder when nnn is divided by 9 is equal to ______.

Correct answer: 7

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2024
Consider the system of linear equations x+y+z=5x+y+z=5x+y+z=5, x+2y+λ2z=9x+2y+\lambda^2 z=9x+2y+λ2z=9, x+3y+λz=μx+3y+\lambda z=\mux+3y+λz=μ, where λ,μ∈R\lambda,\mu\in Rλ,μ∈R. Then, which of the following statement is NOT correct?
  1. (A)System has infinite number of solutions if λ=1\lambda=1λ=1 and μ=13\mu=13μ=13
  2. (B)System is inconsistent if λ=1\lambda=1λ=1 and μ≠13\mu\ne 13μ=13
  3. (C)System is consistent if λ≠1\lambda\ne 1λ=1 and μ=13\mu=13μ=13
  4. (D)System has unique solution if λ≠1\lambda\ne 1λ=1 and μ≠13\mu\ne 13μ=13

Correct answer: (D)

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2024
If f(x)=∣2cos⁡2x2sin⁡2x3+sin⁡22x3+2cos⁡2x2sin⁡2xsin⁡22x2cos⁡2x3+2sin⁡2xsin⁡22x∣f(x)=\begin{vmatrix}2\cos^2x & 2\sin^2x & 3+\sin^2 2x\\ 3+2\cos^2x & 2\sin^2x & \sin^2 2x\\ 2\cos^2x & 3+2\sin^2x & \sin^2 2x\end{vmatrix}f(x)=​2cos2x3+2cos2x2cos2x​2sin2x2sin2x3+2sin2x​3+sin22xsin22xsin22x​​, then 13f′(0)\dfrac13 f'(0)31​f′(0) is equal to:
  1. (A)000
  2. (B)111
  3. (C)222
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2024
Consider the system of linear equations x+y+z=4μx+y+z=4\mux+y+z=4μ, x+2y+2λz=10μx+2y+2\lambda z=10\mux+2y+2λz=10μ, x+3y+4λ2z=μ2+15x+3y+4\lambda^2 z=\mu^2+15x+3y+4λ2z=μ2+15, where λ,μ∈R\lambda,\mu\in\mathbb{R}λ,μ∈R. Which one of the following statements is NOT correct?
  1. (A)The system has unique solution if λ≠12\lambda\ne\dfrac12λ=21​ and μ≠1,15\mu\ne1,15μ=1,15
  2. (B)The system is inconsistent if λ=12\lambda=\dfrac12λ=21​ and μ≠1\mu\ne1μ=1
  3. (C)The system has infinite number of solutions if λ=12\lambda=\dfrac12λ=21​ and μ=15\mu=15μ=15
  4. (D)The system is consistent if λ≠12\lambda\ne\dfrac12λ=21​

Correct answer: (B)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2024
Let R=[x000y000z]R=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}R=​x00​0y0​00z​​ be a non-zero 3×33\times 33×3 matrix, where xsin⁡θ=ysin⁡(θ+2π3)=zsin⁡(θ+4π3)≠0x\sin\theta=y\sin\left(\theta+\dfrac{2\pi}{3}\right)=z\sin\left(\theta+\dfrac{4\pi}{3}\right)\ne 0xsinθ=ysin(θ+32π​)=zsin(θ+34π​)=0, θ∈(0,2π)\theta\in(0,2\pi)θ∈(0,2π). For a square matrix MMM, let trace(M)(M)(M) denote the sum of all the diagonal entries of MMM. Then, among the statements: (I) Trace(R)=0(R)=0(R)=0 (II) If trace(adj(adj(R)))=0(\text{adj}(\text{adj}(R)))=0(adj(adj(R)))=0, then RRR has exactly one non-zero entry.
  1. (A)Both (I) and (II) are true
  2. (B)Neither (I) nor (II) is true
  3. (C)Only (II) is true
  4. (D)Only (I) is true

Correct answer: (B)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2024
Let A=[2126211332]A=\begin{bmatrix}2 & 1 & 2\\ 6 & 2 & 11\\ 3 & 3 & 2\end{bmatrix}A=​263​123​2112​​ and P=[120502715]P=\begin{bmatrix}1 & 2 & 0\\ 5 & 0 & 2\\ 7 & 1 & 5\end{bmatrix}P=​157​201​025​​. The sum of the prime factors of ∣P−1AP−2I∣|P^{-1}AP-2I|∣P−1AP−2I∣ is equal to:
  1. (A)26
  2. (B)27
  3. (C)66
  4. (D)23

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsNumericalJEE Main 2024
Let for any three distinct consecutive terms a,b,ca,b,ca,b,c of an A.P, the lines ax+by+c=0ax+by+c=0ax+by+c=0 be concurrent at the point PPP and Q (α,β)Q\,(\alpha,\beta)Q(α,β) be a point such that the system of equations x+y+z=6x+y+z=6x+y+z=6, 2x+5y+αz=β2x+5y+\alpha z=\beta2x+5y+αz=β and x+2y+3z=4x+2y+3z=4x+2y+3z=4, has infinitely many solutions. Then (PQ)2(PQ)^2(PQ)2 is equal to ___.

Correct answer: 113

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2024
Let A=[1000αβ0βα]A=\begin{bmatrix}1 & 0 & 0\\ 0 & \alpha & \beta\\ 0 & \beta & \alpha\end{bmatrix}A=​100​0αβ​0βα​​ and ∣2A∣3=221|2A|^3=2^{21}∣2A∣3=221 where α,β∈Z\alpha,\beta\in Zα,β∈Z. Then a value of α\alphaα is
  1. (A)333
  2. (B)555
  3. (C)171717
  4. (D)999

Correct answer: (B)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2024
Let A be a square matrix such that AAT=IAA^T=IAAT=I. Then 12A[(A+AT)2+(A−AT)2]\dfrac{1}{2}A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]21​A[(A+AT)2+(A−AT)2] is equal to
  1. (A)A2+IA^2+IA2+I
  2. (B)A3+IA^3+IA3+I
  3. (C)A2+ATA^2+A^TA2+AT
  4. (D)A3+ATA^3+A^TA3+AT

Correct answer: (D)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2024
Consider the matrix f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}f(x)=​cosxsinx0​−sinxcosx0​001​​. Given below are two statements: Statement I: f(−x)f(-x)f(−x) is the inverse of the matrix f(x)f(x)f(x). Statement II: f(x)⋅f(y)=f(x+y)f(x)\cdot f(y)=f(x+y)f(x)⋅f(y)=f(x+y). In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are true

Correct answer: (D)

Step-by-step solution →
Q112·MathematicsNumericalJEE Main 2024
Let A=[201110101]A=\begin{bmatrix}2 & 0 & 1\\ 1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}A=​211​010​101​​, B=[B1,B2,B3]B=[B_1,B_2,B_3]B=[B1​,B2​,B3​], where B1,B2,B3B_1,B_2,B_3B1​,B2​,B3​ are column matrices, and AB1=[100]AB_1=\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}AB1​=​100​​, AB2=[230]AB_2=\begin{bmatrix}2\\ 3\\ 0\end{bmatrix}AB2​=​230​​, AB3=[321]AB_3=\begin{bmatrix}3\\ 2\\ 1\end{bmatrix}AB3​=​321​​. If α=∣B∣\alpha=|B|α=∣B∣ and β\betaβ is the sum of all the diagonal elements of BBB, then α3+β3\alpha^3+\beta^3α3+β3 is equal to __________.

Correct answer: 28

Step-by-step solution →
Q113·MathematicsNumericalJEE Main 2024
Let A be a 2×22\times 22×2 real matrix and I be the identity matrix of order 2. If the roots of the equation ∣A−xI∣=0|A-xI|=0∣A−xI∣=0 be −1-1−1 and 333, then the sum of the diagonal elements of the matrix A2A^2A2 is __________.

Correct answer: 10

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2024
The values of α\alphaα, for which ∣132α+32113α+132α+33α+10∣=0\begin{vmatrix} 1 & \dfrac32 & \alpha+\dfrac32 \\ 1 & \dfrac13 & \alpha+\dfrac13 \\ 2\alpha+3 & 3\alpha+1 & 0\end{vmatrix}=0​112α+3​23​31​3α+1​α+23​α+31​0​​=0, lie in the interval
  1. (A)(−2,1)(-2,1)(−2,1)
  2. (B)(−3,0)(-3,0)(−3,0)
  3. (C)(−32,32)\left(-\dfrac32,\dfrac32\right)(−23​,23​)
  4. (D)(0,3)(0,3)(0,3)

Correct answer: (B)

Step-by-step solution →
Q115·MathematicsMultiple correctJEE Advanced 2023
Let M = (aij)(a_{ij})(aij​), i,j∈{1,2,3}i, j \in \{1, 2, 3\}i,j∈{1,2,3}, be the 3×33 \times 33×3 matrix such that aij=1a_{ij} = 1aij​=1 if j+1j + 1j+1 is divisible by iii, otherwise aij=0a_{ij} = 0aij​=0. Then which of the following statements is(are) true?
  1. (A)MMM is invertible
  2. (B)There exists a nonzero column matrix (a1a2a3)\begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \end{pmatrix}​a1​a2​a3​​​ and such that M(a1a2a3)=(−a1−a2−a3)M \begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \end{pmatrix} = \begin{pmatrix} -a_{1} \\ -a_{2} \\ -a_{3} \end{pmatrix}M​a1​a2​a3​​​=​−a1​−a2​−a3​​​
  3. (C)The set {X∈R3:MX=0}≠{0}\left\{ X \in \mathbb{R}^{3} : MX = \mathbf{0} \right\} \neq \{\mathbf{0}\}{X∈R3:MX=0}={0}, where 0=(000)\mathbf{0} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}0=​000​​
  4. (D)The matrix (M−2I)(M - 2I)(M−2I) is invertible, where III is the 3×33 \times 33×3 identity matrix

Correct answer: (B), (C)

Step-by-step solution →
Q116·MathematicsIntegerJEE Advanced 2023
Let R={(a3bc2d050):a,b,c,d∈{0,3,5,7,11,13,17,19}}R = \left\{ \begin{pmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{pmatrix} : a, b, c, d \in \{0, 3, 5, 7, 11, 13, 17, 19\} \right\}R=⎩⎨⎧​​ac0​325​bd0​​:a,b,c,d∈{0,3,5,7,11,13,17,19}⎭⎬⎫​. Then the number of invertible matrices in R is

Correct answer: 3780

Step-by-step solution →
Q117·MathematicsSingle correctJEE Advanced 2023
Let α\alphaα, β\betaβ and γ\gammaγ be real numbers. Consider the following system of linear equations x+2y+z=7x + 2y + z = 7x+2y+z=7 x+αz=11x + \alpha z = 11x+αz=11 2x−3y+βz=γ2x - 3y + \beta z = \gamma2x−3y+βz=γ Match each entry in List-I to the correct entries in List-II. The correct option is:
List – IList – II
P.If β=12(7α−3)\beta = \frac{1}{2}(7\alpha - 3)β=21​(7α−3) and γ=28\gamma = 28γ=28, then the system has1.A unique solution
Q.If β=12(7α−3)\beta = \frac{1}{2}(7\alpha - 3)β=21​(7α−3) and γ≠28\gamma \neq 28γ=28, then the system has2.No solution
R.If β≠12(7α−3)\beta \neq \frac{1}{2}(7\alpha - 3)β=21​(7α−3) where α=1\alpha = 1α=1 and γ≠28\gamma \neq 28γ=28, then the system has3.Infinitely many solution
S.If β≠12(7α−3)\beta \neq \frac{1}{2}(7\alpha - 3)β=21​(7α−3) where α=1\alpha = 1α=1 and γ=28\gamma = 28γ=28, then the system has4.x=11x = 11x=11, y=−2y = -2y=−2 and z=0z = 0z=0 as a solution
5.x=−15x = -15x=−15, y=4y = 4y=4 and z=0z = 0z=0 as a solution
  1. (A)(P) → (3) (Q) → (2) (R) → (1) (S) → (4)
  2. (B)(P) → (3) (Q) → (2) (R) → (5) (S) → (4)
  3. (C)(P) → (2) (Q) → (1) (R) → (4) (S) → (5)
  4. (D)(P) → (2) (Q) → (1) (R) → (1) (S) → (3)

Correct answer: (A)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2023
Let the system of linear equations −x+2y−9z=7-x+2y-9z=7−x+2y−9z=7, −x+3y+7z=9-x+3y+7z=9−x+3y+7z=9, −2x+y+5z=8-2x+y+5z=8−2x+y+5z=8, −3x+y+13z=λ-3x+y+13z=\lambda−3x+y+13z=λ has a unique solution x=α,y=β,z=γx=\alpha, y=\beta, z=\gammax=α,y=β,z=γ. Then the distance of the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) from the plane 2x−2y+z=λ2x-2y+z=\lambda2x−2y+z=λ is
  1. (A)9
  2. (B)11
  3. (C)13
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2023
Let the determinant of a square matrix A of order m be m−nm-nm−n, where m and n satisfy 4m+n=224m+n=224m+n=22 and 17m+4n=9317m+4n=9317m+4n=93. If det⁡(n adj(adj(mA)))=3a5b6c\det(n\,\mathrm{adj}(\mathrm{adj}(mA)))=3^a 5^b 6^cdet(nadj(adj(mA)))=3a5b6c, then a+b+ca+b+ca+b+c is equal to:
  1. (A)96
  2. (B)101
  3. (C)109
  4. (D)84

Correct answer: (A)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
For the system of linear equations 2x+4y+2az=b, x+2y+3z=4, 2x−5y+2z=82x+4y+2az=b,\,x+2y+3z=4,\,2x-5y+2z=82x+4y+2az=b,x+2y+3z=4,2x−5y+2z=8, which of the following is NOT correct?
  1. (A)It has infinitely many solutions if a=3, b=6a=3,\,b=6a=3,b=6
  2. (B)It has unique solution if a=b=6a=b=6a=b=6
  3. (C)It has unique solution if a=b=8a=b=8a=b=8
  4. (D)It has infinitely many solution if a=3, b=8a=3,\,b=8a=3,b=8

Correct answer: (A)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2023
If the system of equations 2x+y−z=52x + y - z = 52x+y−z=5, 2x−5y+λz=μ2x - 5y + \lambda z = \mu2x−5y+λz=μ, x+2y−5z=7x + 2y - 5z = 7x+2y−5z=7 has infinitely many solutions, then (λ+μ)2+(λ−μ)2(\lambda + \mu)^2 + (\lambda - \mu)^2(λ+μ)2+(λ−μ)2 is equal to
  1. (A)916
  2. (B)912
  3. (C)920
  4. (D)904

Correct answer: (A)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2023
Let for A=[123α31112]A = \begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}A=​1α1​231​312​​, ∣A∣=2|A| = 2∣A∣=2. If ∣2 adj (2 adj (2A))∣=32n\left|2\,\text{adj}\,(2\,\text{adj}\,(2A))\right| = 32^n∣2adj(2adj(2A))∣=32n, then 3n+α3n + \alpha3n+α is equal to
  1. (A)10
  2. (B)9
  3. (C)12
  4. (D)11

Correct answer: (D)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2023
Let B=[13α123αα4], α>2B=\begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix},\,\alpha>2B=​11α​32α​α34​​,α>2 be the adjoint of a matrix AAA and ∣A∣=2|A|=2∣A∣=2, then [α −2α α][α−2αα][\alpha\,-2\alpha\,\alpha]\begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}[α−2αα]​α−2αα​​ is equal to:
  1. (A)161616
  2. (B)323232
  3. (C)−16-16−16
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2023
The number of symmetric matrices of order 333, with all the entries from the set {0,1,2,3,4,5,6,7,8,9}\{0,1,2,3,4,5,6,7,8,9\}{0,1,2,3,4,5,6,7,8,9} is:
  1. (A)6106^{10}610
  2. (B)9109^{10}910
  3. (C)10310^{3}103
  4. (D)10610^{6}106

Correct answer: (D)

Step-by-step solution →
Q125·MathematicsNumericalJEE Main 2023
Let Dk=∣12k2k−1nn2+n+2n2nn2+nn2+n+2∣D_k=\begin{vmatrix}1&2k&2k-1\\n&n^2+n+2&n^2\\n&n^2+n&n^2+n+2\end{vmatrix}Dk​=​1nn​2kn2+n+2n2+n​2k−1n2n2+n+2​​. If ∑k=1nDk=96\sum_{k=1}^{n}D_k=96∑k=1n​Dk​=96, then n is equal to

Correct answer: 6

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2023
Let A=[115101]A=\begin{bmatrix}1&\frac{1}{51}\\0&1\end{bmatrix}A=[10​511​1​]. If B=[12−1−1]A[−1−211]B=\begin{bmatrix}1&2\\-1&-1\end{bmatrix}A\begin{bmatrix}-1&-2\\1&1\end{bmatrix}B=[1−1​2−1​]A[−11​−21​], then the sum of all the elements of the matrix ∑n=150Bn\sum_{n=1}^{50}B^n∑n=150​Bn is equal to
  1. (A)100
  2. (B)50
  3. (C)75
  4. (D)125

Correct answer: (A)

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2023
If ∣x+1xxxx+λxxxx+λ2∣=98(103x+81)\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix}=\dfrac{9}{8}(103x+81)​x+1xx​xx+λx​xxx+λ2​​=89​(103x+81), then λ,λ3\lambda,\dfrac{\lambda}{3}λ,3λ​ are the roots of the equation
  1. (A)4λ2+24λ+27=04\lambda^2+24\lambda+27=04λ2+24λ+27=0
  2. (B)4λ2−24λ+27=04\lambda^2-24\lambda+27=04λ2−24λ+27=0
  3. (C)4λ2+24λ−27=04\lambda^2+24\lambda-27=04λ2+24λ−27=0
  4. (D)4λ2−24λ−27=04\lambda^2-24\lambda-27=04λ2−24λ−27=0

Correct answer: (B)

Step-by-step solution →
Q128·MathematicsNumericalJEE Main 2023
Let A=[012a031c0]A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}A=​0a1​10c​230​​, where a,c∈Ra, c \in \mathbb{R}a,c∈R. If A3=AA^3 = AA3=A and the positive value of aaa belongs to the interval (n−1,n](n - 1, n](n−1,n], where n∈Nn \in \mathbb{N}n∈N, then nnn is equal to _______ .

Correct answer: 2

Step-by-step solution →
Q129·MathematicsSingle correctJEE Main 2023
Let AAA be a 2×22 \times 22×2 matrix with real entries such that A′=αA+IA' = \alpha A + IA′=αA+I, where α∈R−{−1,1}\alpha \in \mathbb{R} - \{-1, 1\}α∈R−{−1,1}. If det⁡(A2−A)=4\det(A^2 - A) = 4det(A2−A)=4, then the sum of all possible values of α\alphaα is equal to
  1. (A)000
  2. (B)32\frac{3}{2}23​
  3. (C)52\frac{5}{2}25​
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2023
If the system of linear equations 7x+11y+αz=137x+11y+\alpha z=137x+11y+αz=13, 5x+4y+7z=β5x+4y+7z=\beta5x+4y+7z=β, 175x+194y+57z=361175x+194y+57z=361175x+194y+57z=361 has infinitely many solutions, then α+β+2\alpha+\beta+2α+β+2 is equal to
  1. (A)444
  2. (B)333
  3. (C)555
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2023
If A=15! 6! 7![5!6!7!6!7!8!7!8!9!]A=\frac{1}{5!\,6!\,7!}\begin{bmatrix}5! & 6! & 7!\\ 6! & 7! & 8!\\ 7! & 8! & 9!\end{bmatrix}A=5!6!7!1​​5!6!7!​6!7!8!​7!8!9!​​, then ∣adj(adj(2A))∣\left|\mathrm{adj}\left(\mathrm{adj}(2A)\right)\right|∣adj(adj(2A))∣ is equal to:
  1. (A)282^828
  2. (B)2122^{12}212
  3. (C)2202^{20}220
  4. (D)2162^{16}216

Correct answer: (D)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2023
If A is a 3×33\times33×3 matrix and ∣A∣=2|A|=2∣A∣=2, then ∣3 adj(∣3A∣A2)∣\left|3\,adj\left(\left|3A\right|A^2\right)\right|​3adj(∣3A∣A2)​ is equal to
  1. (A)311⋅6103^{11}\cdot6^{10}311⋅610
  2. (B)312⋅6103^{12}\cdot6^{10}312⋅610
  3. (C)310⋅6113^{10}\cdot6^{11}310⋅611
  4. (D)312⋅6113^{12}\cdot6^{11}312⋅611

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsNumericalJEE Main 2023
Let S be the set of values of λ\lambdaλ for which the system of equations 6λx−3y+3z=4λ26\lambda x-3y+3z=4\lambda^26λx−3y+3z=4λ2, 2x+6λy+4z=12x+6\lambda y+4z=12x+6λy+4z=1, 3x+2y+3λz=λ3x+2y+3\lambda z=\lambda3x+2y+3λz=λ has no solution. Then 12∑λ∈S∣λ∣12\sum_{\lambda\in S}|\lambda|12∑λ∈S​∣λ∣ is equal to _______ .

Correct answer: 24

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2023
For the system of linear equations 2x−y+3z=52x-y+3z=52x−y+3z=5, 3x+2y−z=73x+2y-z=73x+2y−z=7, 4x+5y+αz=β4x+5y+\alpha z=\beta4x+5y+αz=β, which of the following is NOT correct?
  1. (A)The system has infinitely many solutions for α=−5\alpha=-5α=−5 and β=9\beta=9β=9
  2. (B)The system has a unique solution for α≠−5\alpha\neq-5α=−5 and β=8\beta=8β=8
  3. (C)The system has infinitely many solutions for α=−6\alpha=-6α=−6 and β=9\beta=9β=9
  4. (D)The system is inconsistent for α=−5\alpha=-5α=−5 and β=8\beta=8β=8

Correct answer: (C)

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2023
Let A=[21012−10−12]A=\begin{bmatrix}2 & 1 & 0\\ 1 & 2 & -1\\ 0 & -1 & 2\end{bmatrix}A=​210​12−1​0−12​​. If ∣ adj(adj(adj 2A)) ∣=(16)n|\,\mathrm{adj}(\mathrm{adj}(\mathrm{adj}\,2A))\,|=(16)^{n}∣adj(adj(adj2A))∣=(16)n, then nnn is equal to
  1. (A)10
  2. (B)9
  3. (C)12
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2023
Let P=[3212−1232]P=\begin{bmatrix}\frac{\sqrt{3}}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}P=[23​​−21​​21​23​​​], A=[1101]A=\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}A=[10​11​] and Q=PAPTQ=PAP^{T}Q=PAPT. If PTQ2007P=[abcd]P^{T}Q^{2007}P=\begin{bmatrix}a & b\\ c & d\end{bmatrix}PTQ2007P=[ac​bd​], then 2a+b−3c−4d2a+b-3c-4d2a+b−3c−4d equal to
  1. (A)2007
  2. (B)2005
  3. (C)2006
  4. (D)2004

Correct answer: (B)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2023
Let S be the set of all values of θ∈[−π,π]\theta\in[-\pi,\pi]θ∈[−π,π] for which the system of linear equations x+y+3 z=0x+y+\sqrt{3}\,z=0x+y+3​z=0, −x+(tan⁡θ)y+7 z=0-x+(\tan\theta)y+\sqrt{7}\,z=0−x+(tanθ)y+7​z=0, x+y+(tan⁡θ)z=0x+y+(\tan\theta)z=0x+y+(tanθ)z=0 has non-trivial solution. Then 120π∑θ∈Sθ\dfrac{120}{\pi}\sum_{\theta\in S}\thetaπ120​∑θ∈S​θ is equal to
  1. (A)40
  2. (B)10
  3. (C)20
  4. (D)30

Correct answer: (C)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2023
If A=[15λ10]A=\begin{bmatrix}1&5\\\lambda&10\end{bmatrix}A=[1λ​510​], A−1=αA+βIA^{-1}=\alpha A+\beta IA−1=αA+βI and α+β=−2\alpha+\beta=-2α+β=−2, then 4α2+β2+λ24\alpha^{2}+\beta^{2}+\lambda^{2}4α2+β2+λ2 is equal to
  1. (A)12
  2. (B)10
  3. (C)19
  4. (D)14

Correct answer: (D)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2023
If the system of equations x+y+az=b, 2x+5y+2z=6, x+2y+3z=3x+y+az=b,\,2x+5y+2z=6,\,x+2y+3z=3x+y+az=b,2x+5y+2z=6,x+2y+3z=3 has infinitely many solutions, then 2a+3b2a+3b2a+3b is equal to:
  1. (A)23
  2. (B)28
  3. (C)25
  4. (D)20

Correct answer: (A)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2023
Let PPP be a square matrix such that P2=I−PP^2=I-PP2=I−P. For α,β,γ,δ∈N\alpha,\beta,\gamma,\delta\in \mathbb{N}α,β,γ,δ∈N, if Pα+Pβ=γI−29PP^\alpha+P^\beta=\gamma I-29PPα+Pβ=γI−29P and Pα−Pβ=δI−13PP^\alpha-P^\beta=\delta I-13PPα−Pβ=δI−13P, then α+β+γ−δ\alpha+\beta+\gamma-\deltaα+β+γ−δ is equal to
  1. (A)18
  2. (B)40
  3. (C)24
  4. (D)22

Correct answer: (C)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2023
Let A=[pqrs]A=\begin{bmatrix} p & q \\ r & s \end{bmatrix}A=[pr​qs​] be such that A2=bIA^{2}=bIA2=bI, where b∈R∖{0}, Ib\in\mathbb{R}\setminus\{0\},\,Ib∈R∖{0},I is the identity matrix and all entries aija_{ij}aij​ of AAA are non-zero. If ∣A∣=a, a≠0|A|=a,\,a\ne 0∣A∣=a,a=0 then 3a2+4b23a^{2}+4b^{2}3a2+4b2 equals:
  1. (A)777
  2. (B)555
  3. (C)888
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2023
For the system of equations x+y+z=6x+y+z=6x+y+z=6, x+2y+αz=10x+2y+\alpha z=10x+2y+αz=10, x+3y+5z=βx+3y+5z=\betax+3y+5z=β, which one of the following is NOT true?
  1. (A)System has a unique solution for α=3,β≠14\alpha=3,\beta\neq 14α=3,β=14.
  2. (B)System has no solution for α=3,β=24\alpha=3,\beta=24α=3,β=24.
  3. (C)System has a unique solution for α=−3,β=14\alpha=-3,\beta=14α=−3,β=14.
  4. (D)System has infinitely many solutions for α=3,β=14\alpha=3,\beta=14α=3,β=14.

Correct answer: (A)

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2023
Let SSS denote the set of all real values of λ\lambdaλ such that the system of equations λx+y+z=1\lambda x+y+z=1λx+y+z=1, x+λy+z=1x+\lambda y+z=1x+λy+z=1, x+y+λz=1x+y+\lambda z=1x+y+λz=1 is inconsistent, then ∑λ∈S(∣λ∣2+∣λ∣)\sum_{\lambda\in S}\left(|\lambda|^2+|\lambda|\right)∑λ∈S​(∣λ∣2+∣λ∣) is equal to
  1. (A)444
  2. (B)121212
  3. (C)666
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2023
For the system of linear equations αx+y+z=1\alpha x+y+z=1αx+y+z=1, x+αy+z=1x+\alpha y+z=1x+αy+z=1, x+y+αz=βx+y+\alpha z=\betax+y+αz=β, which one of the following statements is NOT correct?
  1. (A)It has infinitely many solutions if α=2\alpha=2α=2 and β=−1\beta=-1β=−1
  2. (B)It has no solution if α=−2\alpha=-2α=−2 and β=1\beta=1β=1
  3. (C)x+y+z=34x+y+z=\dfrac34x+y+z=43​ if α=2\alpha=2α=2 and β=1\beta=1β=1
  4. (D)It has infinitely many solutions if α=1\alpha=1α=1 and β=1\beta=1β=1

Correct answer: (A)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2023
Let f(x)=∣1+sin⁡2xcos⁡2xsin⁡2xsin⁡2x1+cos⁡2xsin⁡2xsin⁡2xcos⁡2x1+sin⁡2x∣f(x)=\begin{vmatrix}1+\sin^2 x & \cos^2 x & \sin 2x\\ \sin^2 x & 1+\cos^2 x & \sin 2x\\ \sin^2 x & \cos^2 x & 1+\sin 2x\end{vmatrix}f(x)=​1+sin2xsin2xsin2x​cos2x1+cos2xcos2x​sin2xsin2x1+sin2x​​, x∈[π6,π3]x\in\left[\frac{\pi}{6},\frac{\pi}{3}\right]x∈[6π​,3π​]. If α\alphaα and β\betaβ respectively are the maximum and the minimum values of fff, then
  1. (A)α2+β2=92\alpha^2+\beta^2=\frac{9}{2}α2+β2=29​
  2. (B)β2−2α=194\beta^2-2\sqrt{\alpha}=\frac{19}{4}β2−2α​=419​
  3. (C)α2−β2=43\alpha^2-\beta^2=4\sqrt{3}α2−β2=43​
  4. (D)β2+2α=194\beta^2+2\sqrt{\alpha}=\frac{19}{4}β2+2α​=419​

Correct answer: (B)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2023
If A=12[13−31]A=\dfrac12\begin{bmatrix}1&\sqrt3\\-\sqrt3&1\end{bmatrix}A=21​[1−3​​3​1​], then:
  1. (A)A30+A25+A=IA^{30}+A^{25}+A=IA30+A25+A=I
  2. (B)A30=A25A^{30}=A^{25}A30=A25
  3. (C)A30+A25−A=IA^{30}+A^{25}-A=IA30+A25−A=I
  4. (D)A30−A25=2IA^{30}-A^{25}=2IA30−A25=2I

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsNumericalJEE Main 2023
Let AAA be a n×nn\times nn×n matrix such that ∣A∣=2|A|=2∣A∣=2. If the determinant of the matrix Adj(2⋅Adj(2A−1))\mathrm{Adj}\left(2\cdot\mathrm{Adj}(2A^{-1})\right)Adj(2⋅Adj(2A−1)) is 2842^{84}284, then nnn is equal to

Correct answer: 84

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2023
For the system of linear equations x+y+z=6x+y+z=6x+y+z=6, αx+βy+7z=3\alpha x+\beta y+7z=3αx+βy+7z=3, x+2y+3z=14x+2y+3z=14x+2y+3z=14 which of the following is NOT true ?
  1. (A)If α=β\alpha=\betaα=β and α≠7\alpha\neq 7α=7, then the system has a unique solution
  2. (B)If α=β=7\alpha=\beta=7α=β=7, then the system has no solution
  3. (C)For every point (α,β)≠(7,7)(\alpha,\beta)\neq(7,7)(α,β)=(7,7) on the line x−2y+7=0x-2y+7=0x−2y+7=0, the system has infinitely many solutions
  4. (D)There is a unique point (α,β)(\alpha,\beta)(α,β) on the line x+2y+18=0x+2y+18=0x+2y+18=0 for which the system has infinitely many solutions

Correct answer: (C)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2023
Let A=(10004−1012−3)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}A=​100​0412​0−1−3​​. Then the sum of the diagonal elements of the matrix (A+I)11(A + I)^{11}(A+I)11 is equal to
  1. (A)614461446144
  2. (B)205020502050
  3. (C)409740974097
  4. (D)409440944094

Correct answer: (C)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2023
If a point P(α,β,γ)P(\alpha,\beta,\gamma)P(α,β,γ) satisfying (α β γ)(2108938848)=(0 0 0)(\alpha\ \beta\ \gamma)\begin{pmatrix}2&10&8\\9&3&8\\8&4&8\end{pmatrix}=(0\ 0\ 0)(α β γ)​298​1034​888​​=(0 0 0) lies on the plane 2x+4y+3z=52x+4y+3z=52x+4y+3z=5, then 6α+9β+7γ6\alpha+9\beta+7\gamma6α+9β+7γ is equal to:
  1. (A)−1-1−1
  2. (B)115\dfrac{11}{5}511​
  3. (C)54\dfrac5445​
  4. (D)111111

Correct answer: (D)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2023
If PPP is a 3×33 \times 33×3 real matrix such that PT=aP+(a−1)IP^T=aP + (a - 1)IPT=aP+(a−1)I, where a>1a > 1a>1, then:
  1. (A)∣Adj P∣=12|\mathrm{Adj}\,P|=\dfrac{1}{2}∣AdjP∣=21​
  2. (B)∣Adj P∣=1|\mathrm{Adj}\,P|=1∣AdjP∣=1
  3. (C)PPP is a singular matrix
  4. (D)∣Adj P∣>1|\mathrm{Adj}\,P| > 1∣AdjP∣>1

Correct answer: (B)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2023
Let A=(mnpq)A=\begin{pmatrix}m&n\\p&q\end{pmatrix}A=(mp​nq​), d=∣A∣≠0d=|A|\neq0d=∣A∣=0 and ∣A−d(Adj A)∣=0|A-d(\mathrm{Adj}\,A)|=0∣A−d(AdjA)∣=0. Then
  1. (A)1+d2=m2+q21+d^2=m^2+q^21+d2=m2+q2
  2. (B)1+d2=(m+q)21+d^2=(m+q)^21+d2=(m+q)2
  3. (C)(1+d)2=m2+q2(1+d)^2=m^2+q^2(1+d)2=m2+q2
  4. (D)(1+d)2=(m+q)2(1+d)^2=(m+q)^2(1+d)2=(m+q)2

Correct answer: (D)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2023
Let the system of linear equations x+y+kz=2x+y+kz=2x+y+kz=2, 2x+3y−z=12x+3y-z=12x+3y−z=1, 3x+4y+2z=k3x+4y+2z=k3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(k+1)x+(2k-1)y=7(k+1)x+(2k−1)y=7, (2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10 has:
  1. (A)infinitely many solutions
  2. (B)unique solution satisfying x−y=1x-y=1x−y=1
  3. (C)unique solution satisfying x+y=1x+y=1x+y=1
  4. (D)no solution

Correct answer: (C)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2023
For α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R, suppose the system of linear equations x−y+z=1x - y + z=1x−y+z=1, 2x+2y+αz=82x + 2y + \alpha z=82x+2y+αz=8, 3x−y+4z=β3x - y + 4z=\beta3x−y+4z=β has infinitely many solutions. Then α\alphaα and β\betaβ are the roots of:
  1. (A)x2+14x+24=0x^2 + 14x + 24=0x2+14x+24=0
  2. (B)x2+18x+56=0x^2 + 18x + 56=0x2+18x+56=0
  3. (C)x2−18x+56=0x^2 - 18x + 56=0x2−18x+56=0
  4. (D)x2−10x+16=0x^2 - 10x + 16=0x2−10x+16=0

Correct answer: (C)

Step-by-step solution →
Q155·MathematicsNumericalJEE Main 2023
Let AAA be a symmetric matrix such that ∣A∣=2|A|=2∣A∣=2 and [21332]A=[12αβ]\begin{bmatrix} 2 & 1 \\ 3 & \tfrac32 \end{bmatrix}A=\begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}[23​123​​]A=[1α​2β​]. If the sum of the diagonal elements of AAA is sss, then βsα2\dfrac{\beta s}{\alpha^2}α2βs​ is equal to _____.

Correct answer: 5

Step-by-step solution →
Q156·MathematicsSingle correctJEE Main 2023
Let α\alphaα and β\betaβ be real numbers. Consider a 3×33 \times 33×3 matrix AAA such that A2=3A+αIA^2 = 3A + \alpha IA2=3A+αI. If A4=21A+βIA^4 = 21A + \beta IA4=21A+βI, then
  1. (A)β=−8\beta = -8β=−8
  2. (B)β=8\beta = 8β=8
  3. (C)α=4\alpha = 4α=4
  4. (D)α=1\alpha = 1α=1

Correct answer: (A)

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2023
Consider the following system of equations αx+2y+z=1\alpha x + 2y + z = 1αx+2y+z=1, 2αx+3y+z=12\alpha x + 3y + z = 12αx+3y+z=1, 3x+αy+2z=β3x + \alpha y + 2z = \beta3x+αy+2z=β for some β∈R\beta \in \mathbb{R}β∈R. Then which of the following is NOT correct.
  1. (A)It has a solution if α=−1\alpha = -1α=−1 and β≠2\beta \neq 2β=2
  2. (B)It has a solution for all α≠−1\alpha \neq -1α=−1 and β=2\beta = 2β=2
  3. (C)It has no solution for α=3\alpha = 3α=3 and for all β≠2\beta \neq 2β=2
  4. (D)It has no solution for α=−1\alpha = -1α=−1 and for all β∈R\beta \in \mathbb{R}β∈R

Correct answer: (D)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2023
The set of all values of t∈Rt\in\mathbb{R}t∈R, for which the matrix [ete−t(sin⁡t−2cos⁡t)e−t(−2sin⁡t−cos⁡t)ete−t(2sin⁡t+cos⁡t)e−t(sin⁡t−2cos⁡t)ete−tcos⁡te−tsin⁡t]\begin{bmatrix} e^t & e^{-t}(\sin t-2\cos t) & e^{-t}(-2\sin t-\cos t) \\ e^t & e^{-t}(2\sin t+\cos t) & e^{-t}(\sin t-2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}​etetet​e−t(sint−2cost)e−t(2sint+cost)e−tcost​e−t(−2sint−cost)e−t(sint−2cost)e−tsint​​ is invertible, is:
  1. (A)R\mathbb{R}R
  2. (B){kπ+π4, k∈Z}\{k\pi+\dfrac{\pi}{4},\,k\in\mathbb{Z}\}{kπ+4π​,k∈Z}
  3. (C){kπ, k∈Z}\{k\pi,\,k\in\mathbb{Z}\}{kπ,k∈Z}
  4. (D){(2k+1)π2, k∈Z}\{(2k+1)\dfrac{\pi}{2},\,k\in\mathbb{Z}\}{(2k+1)2π​,k∈Z}

Correct answer: (A)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2023
Let A=[110310−310110]A = \begin{bmatrix} \dfrac{1}{\sqrt{10}} & \dfrac{3}{\sqrt{10}} \\ \dfrac{-3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}} \end{bmatrix}A=​10​1​10​−3​​10​3​10​1​​​ and B=[1−i01]B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}B=[10​−i1​], where i=−1i = \sqrt{-1}i=−1​. If M=ATBAM = A^{T}BAM=ATBA, then the inverse of the matrix AM2023ATAM^{2023}A^{T}AM2023AT is
  1. (A)[10−2023i1]\begin{bmatrix} 1 & 0 \\ -2023i & 1 \end{bmatrix}[1−2023i​01​]
  2. (B)[1−2023i01]\begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}[10​−2023i1​]
  3. (C)[102023i1]\begin{bmatrix} 1 & 0 \\ 2023i & 1 \end{bmatrix}[12023i​01​]
  4. (D)[12023i01]\begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}[10​2023i1​]

Correct answer: (D)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2023
Let x,y,z>1x,y,z>1x,y,z>1 and A=[1log⁡xylog⁡xzlog⁡yx2log⁡yzlog⁡zxlog⁡zy3]A=\begin{bmatrix}1 & \log_x y & \log_x z\\ \log_y x & 2 & \log_y z\\ \log_z x & \log_z y & 3\end{bmatrix}A=​1logy​xlogz​x​logx​y2logz​y​logx​zlogy​z3​​. Then ∣adj(adj A2)∣|\mathrm{adj}(\mathrm{adj}\,A^2)|∣adj(adjA2)∣ is equal to:
  1. (A)282^828
  2. (B)494^949
  3. (C)646^464
  4. (D)242^424

Correct answer: (A)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2023
Let A,B,CA, B, CA,B,C be 3×33\times 33×3 matrices such that AAA is symmetric and BBB and CCC are skew-symmetric. Consider the statements (S1) A13B26−B26A13A^{13}B^{26}-B^{26}A^{13}A13B26−B26A13 is symmetric (S2) A26C13−C13A26A^{26}C^{13}-C^{13}A^{26}A26C13−C13A26 is symmetric Then,
  1. (A)Only S2 is true
  2. (B)Both S1 and S2 are false
  3. (C)Only S1 is true
  4. (D)Both S1 and S2 are true

Correct answer: (A)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2023
Let S1S_1S1​ and S2S_2S2​ be respectively the sets of all a∈R−{0}a\in\mathbb{R}-\{0\}a∈R−{0} for which the system of linear equations ax+2ay−3az=1ax+2ay-3az=1ax+2ay−3az=1; (2a+1)x+(2a+3)y+(a+1)z=2(2a+1)x+(2a+3)y+(a+1)z=2(2a+1)x+(2a+3)y+(a+1)z=2; (3a+5)x+(a+5)y+(a+2)z=3(3a+5)x+(a+5)y+(a+2)z=3(3a+5)x+(a+5)y+(a+2)z=3 has unique solution and infinitely many solutions. Then:
  1. (A)S1S_1S1​ is an infinite set and n(S2)=2n(S_2)=2n(S2​)=2
  2. (B)S2S_2S2​ is an infinite set and n(S1)=2n(S_1)=2n(S1​)=2
  3. (C)S1=ΦS_1=\PhiS1​=Φ and S2=R−{0}S_2=\mathbb{R}-\{0\}S2​=R−{0}
  4. (D)S1=R−{0}S_1=\mathbb{R}-\{0\}S1​=R−{0} and S2=ΦS_2=\PhiS2​=Φ

Correct answer: (D)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2023
If the system of equations x+2y+3z=3x+2y+3z=3x+2y+3z=3, 4x+3y−4z=44x+3y-4z=44x+3y−4z=4, 8x+4y+λz=9+μ8x+4y+\lambda z=9+\mu8x+4y+λz=9+μ has infinitely many solutions, then the ordered pair (λ,μ)(\lambda,\mu)(λ,μ) is equal to:
  1. (A)(−725,215)\left(-\dfrac{72}{5},\dfrac{21}{5}\right)(−572​,521​)
  2. (B)(−725,−215)\left(-\dfrac{72}{5},-\dfrac{21}{5}\right)(−572​,−521​)
  3. (C)(725,−215)\left(\dfrac{72}{5},-\dfrac{21}{5}\right)(572​,−521​)
  4. (D)(725,215)\left(\dfrac{72}{5},\dfrac{21}{5}\right)(572​,521​)

Correct answer: (C)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2023
Let α\alphaα be a root of the equation (a−c)x2+(b−a)x+(c−b)=0(a-c)x^2+(b-a)x+(c-b)=0(a−c)x2+(b−a)x+(c−b)=0 where a,b,ca,b,ca,b,c are distinct real numbers such that the matrix [α2α1111abc]\begin{bmatrix}\alpha^2 & \alpha & 1\\ 1 & 1 & 1\\ a & b & c\end{bmatrix}​α21a​α1b​11c​​ is singular. Then, the value of (a−c)2(b−a)(c−b)+(b−a)2(a−c)(c−b)+(c−b)2(a−c)(b−a)\dfrac{(a-c)^2}{(b-a)(c-b)}+\dfrac{(b-a)^2}{(a-c)(c-b)}+\dfrac{(c-b)^2}{(a-c)(b-a)}(b−a)(c−b)(a−c)2​+(a−c)(c−b)(b−a)2​+(a−c)(b−a)(c−b)2​ is
  1. (A)121212
  2. (B)999
  3. (C)333
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2023
Let A be a 3×33\times33×3 matrix such that ∣adj(adj(adj A))∣=124|\mathrm{adj}(\mathrm{adj}(\mathrm{adj}\,A))|=12^4∣adj(adj(adjA))∣=124. Then ∣A−1 adj A∣|A^{-1}\,\mathrm{adj}\,A|∣A−1adjA∣ is equal to
  1. (A)6\sqrt66​
  2. (B)232\sqrt323​
  3. (C)121212
  4. (D)111

Correct answer: (B)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2023
If AAA and BBB are two non-zero n×nn\times nn×n matrices such that A2+B=A2BA^2+B=A^2 BA2+B=A2B, then
  1. (A)A2=IA^2=IA2=I or B=IB=IB=I
  2. (B)A2B=IA^2 B=IA2B=I
  3. (C)AB=IAB=IAB=I
  4. (D)A2B=BA2A^2 B=BA^2A2B=BA2

Correct answer: (D)

Step-by-step solution →
Q167·MathematicsMultiple correctJEE Advanced 2022
Let ∣M∣|M|∣M∣ denote the determinant of a square matrix MMM. Let g:[0,π2]→Rg : \left[0, \frac{\pi}{2}\right] \rightarrow Rg:[0,2π​]→R be the function defined by g(θ)=f(θ)−1+f(π2−θ)−1g(\theta) = \sqrt{f(\theta) - 1} + \sqrt{f\left(\frac{\pi}{2} - \theta\right) - 1}g(θ)=f(θ)−1​+f(2π​−θ)−1​ where f(θ)=12∣1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1∣+∣sin⁡πcos⁡(θ+π4)tan⁡(θ−π4)sin⁡(θ−π4)−cos⁡π2log⁡e(4π)cot⁡(θ+π4)log⁡e(π4)tan⁡π∣f(\theta) = \frac{1}{2}\begin{vmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{vmatrix} + \begin{vmatrix} \sin\pi & \cos\left(\theta + \frac{\pi}{4}\right) & \tan\left(\theta - \frac{\pi}{4}\right) \\ \sin\left(\theta - \frac{\pi}{4}\right) & -\cos\frac{\pi}{2} & \log_{e}\left(\frac{4}{\pi}\right) \\ \cot\left(\theta + \frac{\pi}{4}\right) & \log_{e}\left(\frac{\pi}{4}\right) & \tan\pi \end{vmatrix}f(θ)=21​​1−sinθ−1​sinθ1−sinθ​1sinθ1​​+​sinπsin(θ−4π​)cot(θ+4π​)​cos(θ+4π​)−cos2π​loge​(4π​)​tan(θ−4π​)loge​(π4​)tanπ​​. Let p(x)p(x)p(x) be a quadratic polynomial whose roots are the maximum and minimum values of the function g(θ)g(\theta)g(θ) and p(2)=2−2p(2) = 2 - \sqrt{2}p(2)=2−2​. Then, which of the following is/are TRUE?
  1. (A)p(3+24)<0p\left(\frac{3 + \sqrt{2}}{4}\right) < 0p(43+2​​)<0
  2. (B)p(1+324)>0p\left(\frac{1 + 3\sqrt{2}}{4}\right) > 0p(41+32​​)>0
  3. (C)p(52−14)>0p\left(\frac{5\sqrt{2} - 1}{4}\right) > 0p(452​−1​)>0
  4. (D)p(5−24)<0p\left(\frac{5 - \sqrt{2}}{4}\right) < 0p(45−2​​)<0

Correct answer: (A), (C)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Advanced 2022
If M = (5232−32−12)\begin{pmatrix} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{pmatrix}(25​−23​​23​−21​​) , then which of the following matrices is equal to M2022^{2022}2022?
  1. (A)(30343033−3033−3032)\begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}(3034−3033​3033−3032​)
  2. (B)(3034−30333033−3032)\begin{pmatrix} 3034 & -3033 \\ 3033 & -3032 \end{pmatrix}(30343033​−3033−3032​)
  3. (C)(30333032−3032−3031)\begin{pmatrix} 3033 & 3032 \\ -3032 & -3031 \end{pmatrix}(3033−3032​3032−3031​)
  4. (D)(30323031−3031−3030)\begin{pmatrix} 3032 & 3031 \\ -3031 & -3030 \end{pmatrix}(3032−3031​3031−3030​)

Correct answer: (A)

Step-by-step solution →
Q169·MathematicsIntegerJEE Advanced 2022
Let β be real number. Consider the matrix A = (β0121−231−2)\begin{pmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{pmatrix}​β23​011​1−2−2​​. If A7−(β−1)A6−βA5A^{7} - \left(\beta - 1\right)A^{6} - \beta A^{5}A7−(β−1)A6−βA5 is a singular matrix, then the value of 9β is ________.

Correct answer: 3

Step-by-step solution →
Q170·MathematicsSingle correctJEE Advanced 2022
Let ppp, qqq, rrr be nonzero real numbers that are, respectively, the 10th10^{\text{th}}10th, 100th100^{\text{th}}100th and 1000th1000^{\text{th}}1000th terms of a harmonic progression. Consider the system of linear equation x+y+z=1x + y + z = 1x+y+z=1 10x+100y+1000z=010x + 100y + 1000z = 010x+100y+1000z=0 qrx+pry+pqz=0qrx + pry + pqz = 0qrx+pry+pqz=0 The correct option is:
List-IList-II
I.If qr=10\frac{q}{r} = 10rq​=10, then the system of linear equations hasP.x=0x = 0x=0, y=109y = \frac{10}{9}y=910​, z=−19z = -\frac{1}{9}z=−91​
II.If pr≠100\frac{p}{r} \neq 100rp​=100, then the system of linear equations hasQ.x=109x = \frac{10}{9}x=910​, y=−19y = -\frac{1}{9}y=−91​, z=0z = 0z=0 as a solution
III.If pq≠10\frac{p}{q} \neq 10qp​=10, then the system of linear equation hasR.infinitely many solutions
IV.If pq=10\frac{p}{q} = 10qp​=10, then the system of linear equations hasS.no solution
T.at least one solution
  1. (A)(I) → (T); (II) → (R); (III) → (S); (IV) → (T)
  2. (B)(I) → (Q); (II) → (S); (III) → (S); (IV) → (R)
  3. (C)(I) → (Q); (II) → (R); (III) → (P); (IV) → (R)
  4. (D)(I) → (T); (II) → (S); (III) → (P); (IV) → (T)

Correct answer: (B)

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2022
If the system of equations x+y+z=6x + y + z = 6x+y+z=6 2x+5y+αz=β2x + 5y + \alpha z = \beta2x+5y+αz=β x+2y+3z=14x + 2y + 3z = 14x+2y+3z=14 has infinitely many solutions, then α+β\alpha + \betaα+β is equal to :
  1. (A)888
  2. (B)363636
  3. (C)444444
  4. (D)484848

Correct answer: (C)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2022
Let A and B be two 3×33 \times 33×3 non-zero real matrices such that AB is a zero matrix. Then
  1. (A)The system of linear equations AX=0AX = 0AX=0 has a unique solution
  2. (B)The system of linear equations AX=0AX = 0AX=0 has infinitely many solutions
  3. (C)B is an invertible matrix
  4. (D)adj (A) is an invertible matrix

Correct answer: (B)

Step-by-step solution →
Q173·MathematicsNumericalJEE Main 2022
Let p and p + 2 be prime numbers and let Δ=∣p!(p+1)!(p+2)!(p+1)!(p+2)!(p+3)!(p+2)!(p+3)!(p+4)!∣\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}Δ=​p!(p+1)!(p+2)!​(p+1)!(p+2)!(p+3)!​(p+2)!(p+3)!(p+4)!​​ Then the sum of the maximum values of α\alphaα and β\betaβ, such that pαp^{\alpha}pα and (p+2)β(p+2)^{\beta}(p+2)β divide Δ\DeltaΔ, is __________.

Correct answer: 4

Step-by-step solution →
Q174·MathematicsNumericalJEE Main 2022
Let x=[111]x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}x=​111​​ and A=[−12301600−1]A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}A=​−100​210​36−1​​. For k∈Nk \in \mathbb{N}k∈N, if X′AkX=33X' A^k X = 33X′AkX=33, then k is equal to:

Correct answer: 10

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2022
Which of the following matrices can NOT be obtained from the matrix [−121−1]\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}[−11​2−1​] by a single elementary row operation?
  1. (A)[011−1]\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}[01​1−1​]
  2. (B)[1−1−12]\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}[1−1​−12​]
  3. (C)[−12−27]\begin{bmatrix} -1 & 2 \\ -2 & 7 \end{bmatrix}[−1−2​27​]
  4. (D)[−12−13]\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}[−1−1​23​]

Correct answer: (C)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2022
Let A and B be any two 3×33 \times 33×3 symmetric and skew symmetric matrices respectively. Then which of the following is <b>NOT</b> true?
  1. (A)A4−B4A^{4}-B^{4}A4−B4 is a symmetric matrix
  2. (B)AB−BAAB-BAAB−BA is a symmetric matrix
  3. (C)B5−A5B^{5}-A^{5}B5−A5 is a skew-symmetric matrix
  4. (D)AB+BAAB+BAAB+BA is a skew-symmetric matrix

Correct answer: (C)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2022
Let the matrix A=[010001100]A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}A=​001​100​010​​ and the matrix B0=A49+2A98B_{0} = A^{49} + 2A^{98}B0​=A49+2A98. If Bn=Adj(Bn−1)B_{n} = Adj(B_{n-1})Bn​=Adj(Bn−1​) for all n≥1n \geq 1n≥1, then det⁡(B4)\det(B_{4})det(B4​) is equal to :
  1. (A)3283^{28}328
  2. (B)3303^{30}330
  3. (C)3323^{32}332
  4. (D)3363^{36}336

Correct answer: (C)

Step-by-step solution →
Q178·MathematicsNumericalJEE Main 2022
Let A=[1−12α]A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}A=[12​−1α​] and B=[β110]B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}B=[β1​10​], α,β∈R\alpha, \beta \in Rα,β∈R. Let α1\alpha_{1}α1​ be the value of α\alphaα which satisfies (A+B)2=A2+[2222](A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}(A+B)2=A2+[22​22​] and α2\alpha_{2}α2​ be the value of α\alphaα which satisfies (A+B)2=B2(A + B)^{2} = B^{2}(A+B)2=B2. Then ∣α1−α2∣\left| \alpha_{1} - \alpha_{2} \right|∣α1​−α2​∣ is equal to _______.

Correct answer: 2

Step-by-step solution →
Q179·MathematicsNumericalJEE Main 2022
Consider a matrix A=[αβγα2β2γ2β+γγ+αα+β]A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}A=​αα2β+γ​ββ2γ+α​γγ2α+β​​, where α,β,γ\alpha, \beta, \gammaα,β,γ are three distinct natural numbers. If det⁡(adj⁡(adj⁡(adj⁡(adj⁡A))))(α−β)16(β−γ)16(γ−α)16=232×316\frac{\det\left(\operatorname{adj}\left(\operatorname{adj}\left(\operatorname{adj}\left(\operatorname{adj}A\right)\right)\right)\right)}{\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}} = 2^{32} \times 3^{16}(α−β)16(β−γ)16(γ−α)16det(adj(adj(adj(adjA))))​=232×316, then the number of such 3 – tuples (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) is ________.

Correct answer: 42

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2022
Let S be the set containing all 3×33\times 33×3 matrices with entries from {−1, 0, 1}\{-1,\,0,\,1\}{−1,0,1}. The total number of matrices A∈SA\in SA∈S such that the sum of all the diagonal elements of ATAA^{T}AATA is 6 is __________.

Correct answer: 5376

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2022
Let A=(4−2αβ)A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}A=(4α​−2β​). If A2+γA+18I=OA^{2}+\gamma A + 18I = OA2+γA+18I=O, then det (A) is equal to ______.
  1. (A)−18-18−18
  2. (B)181818
  3. (C)−50-50−50
  4. (D)505050

Correct answer: (B)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2022
Let A=(12−2−5)A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix}A=(1−2​2−5​). Let α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R be such that αA2+βA=2I\alpha A^2 + \beta A = 2IαA2+βA=2I. Then α+β\alpha + \betaα+β is equal to -
  1. (A)-10
  2. (B)-6
  3. (C)6
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q183·MathematicsNumericalJEE Main 2022
The number of matrices A=[abcd]A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​], where a,b,c,d∈{−1,0,1,2,3,……,10}a,b,c,d\in\{-1,0,1,2,3,\ldots\ldots,10\}a,b,c,d∈{−1,0,1,2,3,……,10}, such that A=A−1A=A^{-1}A=A−1, is______.

Correct answer: 50

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2022
Let A=[111]A=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}A=​111​​ and B=[92−102112122132−142−152162172]B=\begin{bmatrix} 9^{2} & -10^{2} & 11^{2} \\ 12^{2} & 13^{2} & -14^{2} \\ -15^{2} & 16^{2} & 17^{2} \end{bmatrix}B=​92122−152​−102132162​112−142172​​, then the value of A′BAA'BAA′BA is:
  1. (A)1224
  2. (B)1042
  3. (C)540
  4. (D)539

Correct answer: (D)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2022
Let A be a 2 × 2 matrix with det (A) = –1 and det ((A + I) (Adj (A) + I)) = 4. Then the sum of the diagonal elements of A can be :
  1. (A)–1
  2. (B)2
  3. (C)1
  4. (D)−2-\sqrt{2}−2​

Correct answer: (B)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2022
If the system of linear equations. 8x + y + 4z = –2 x + y + z = 0 λx – 3y = μ has infinitely many solutions, then the distance of the point (λ,μ,−12)\left(\lambda,\mu,-\dfrac{1}{2}\right)(λ,μ,−21​) from the plane 8x + y + 4z + 2 = 0 is :
  1. (A)353\sqrt{5}35​
  2. (B)4
  3. (C)269\dfrac{26}{9}926​
  4. (D)103\dfrac{10}{3}310​

Correct answer: (D)

Step-by-step solution →
Q187·MathematicsNumericalJEE Main 2022
Let A=[2−1−110−11−10]A = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}A=​211​−10−1​−1−10​​ and B=A−IB = A - IB=A−I. If ω=3i−12\omega = \frac{\sqrt{3}i - 1}{2}ω=23​i−1​, then the number of elements in the set {n∈{1,2,…,100}:An+(ωB)n=A+B}\{n \in \{1, 2, \ldots, 100\} : A^{n} + (\omega B)^{n} = A + B\}{n∈{1,2,…,100}:An+(ωB)n=A+B} is equal to ________.

Correct answer: 17

Step-by-step solution →
Q188·MathematicsSingle correctJEE Main 2022
The number of θ∈(0,4π)\theta \in (0, 4\pi)θ∈(0,4π) for which the system of linear equations 3(sin⁡3θ)x−y+z=23 (\sin 3\theta) x - y + z = 23(sin3θ)x−y+z=2 3(cos⁡2θ)x+4y+3z=33 (\cos 2\theta) x + 4y + 3z = 33(cos2θ)x+4y+3z=3 6x+7y+7z=96x + 7y + 7z = 96x+7y+7z=9 has no solution is :
  1. (A)6
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q189·MathematicsNumericalJEE Main 2022
Let M=[0−αα0]M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}M=[0α​−α0​], where α\alphaα is a non-zero real number an N=∑k=149M2kN = \sum_{k=1}^{49} M^{2k}N=∑k=149​M2k. If (I−M2)N=−2I(I - M^2)N = -2I(I−M2)N=−2I, then the positive integral value of α\alphaα is ____.

Correct answer: 1

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2022
Let A=[aij]A = [a_{ij}]A=[aij​] be a square matrix of order 3 such that aij=2 j−ia_{ij} = 2^{\,j-i}aij​=2j−i, for all i, j = 1, 2, 3. Then, the matrix A2+A3+…+A10A^{2} + A^{3} + \ldots + A^{10}A2+A3+…+A10 is equal to :
  1. (A)(310−32)A\left(\frac{3^{10} - 3}{2}\right)A(2310−3​)A
  2. (B)(310−12)A\left(\frac{3^{10} - 1}{2}\right)A(2310−1​)A
  3. (C)(310+12)A\left(\frac{3^{10} + 1}{2}\right)A(2310+1​)A
  4. (D)(310+32)A\left(\frac{3^{10} + 3}{2}\right)A(2310+3​)A

Correct answer: (A)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2022
Let A=(2−102)A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}A=(20​−12​). If B=I−5C1 (adjA)+5C2 (adjA)2−...−5C5 (adjA)5B = I - {}^{5}C_{1}\,(\mathrm{adj}A) + {}^{5}C_{2}\,(\mathrm{adj}A)^{2} - ... - {}^{5}C_{5}\,(\mathrm{adj}A)^{5}B=I−5C1​(adjA)+5C2​(adjA)2−...−5C5​(adjA)5, then the sum of all elements of the matrix B is:
  1. (A)−5-5−5
  2. (B)−6-6−6
  3. (C)−7-7−7
  4. (D)−8-8−8

Correct answer: (C)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2022
If the system of linear equations 2x+y−z=72x + y - z = 72x+y−z=7 x−3y+2z=1x - 3y + 2z = 1x−3y+2z=1 x+4y+δz=kx + 4y + \delta z = kx+4y+δz=k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to:
  1. (A)−3
  2. (B)3
  3. (C)6
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsNumericalJEE Main 2022
Let A=(1+i1−i0)A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}A=(1+i−i​10​) where i=−1i = \sqrt{-1}i=−1​. Then, the number of elements in the set {n∈{1,2,…,100}:An=A}\{n \in \{1, 2, \ldots, 100\} : A^n = A\}{n∈{1,2,…,100}:An=A} is

Correct answer: 25

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2022
Let A be a matrix of order 3 × 3 and det (A) = 2. Then det (det (A) adj (5 adj (A3\mathrm{A}^{3}A3))) is equal to ____.
  1. (A)512×106512 \times 10^{6}512×106
  2. (B)256×106256 \times 10^{6}256×106
  3. (C)1024×1061024 \times 10^{6}1024×106
  4. (D)256×1011256 \times 10^{11}256×1011

Correct answer: (A)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2022
If the system of linear equations 2x+3y−z=−22x + 3y - z = -22x+3y−z=−2 x+y+z=4x + y + z = 4x+y+z=4 x−y+∣λ∣z=4λ−4x - y + |\lambda| z = 4\lambda - 4x−y+∣λ∣z=4λ−4 where λ∈R,\lambda \in \mathbb{R},λ∈R, has no solution, then
  1. (A)λ=7\lambda = 7λ=7
  2. (B)λ=−7\lambda = -7λ=−7
  3. (C)λ=8\lambda = 8λ=8
  4. (D)λ2=1\lambda^{2} = 1λ2=1

Correct answer: (B)

Step-by-step solution →
Q196·MathematicsNumericalJEE Main 2022
If the system of linear equations 2x−3y=γ+52x - 3y = \gamma + 52x−3y=γ+5, αx+5y=β+1\alpha x + 5y = \beta + 1αx+5y=β+1, where α,β,γ∈R\alpha, \beta, \gamma \in \mathbf{R}α,β,γ∈R has infinitely many solutions, then the value of ∣9α+3β+5γ∣\left|9\alpha + 3\beta + 5\gamma\right|∣9α+3β+5γ∣ is equal to

Correct answer: 58

Step-by-step solution →
Q197·MathematicsNumericalJEE Main 2022
The positive value of the determinant of the matrix A, whose Adj(Adj(A))=(1428−14−14142828−1414)Adj\left(Adj(A)\right)=\begin{pmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{pmatrix}Adj(Adj(A))=​14−1428​2814−14​−142814​​, is __________.

Correct answer: 14

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2022
Let the system of linear equations x+2y+z=2x + 2y + z = 2x+2y+z=2, αx+3y−z=α\alpha x + 3y - z = \alphaαx+3y−z=α, −αx+y+2z=−α-\alpha x + y + 2z = -\alpha−αx+y+2z=−α be inconsistent. Then α\alphaα is equal to :
  1. (A)52\frac{5}{2}25​
  2. (B)−52-\frac{5}{2}−25​
  3. (C)72\frac{7}{2}27​
  4. (D)−72-\frac{7}{2}−27​

Correct answer: (D)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2022
Let f(x)=∣a−10axa−1ax2axa∣f(x)=\begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^{2} & ax & a \end{vmatrix}f(x)=​aaxax2​−1aax​0−1a​​, a∈Ra \in \mathbb{R}a∈R. Then the sum of which the squares of all the values of a for 2f′(10)−f′(5)+100=02f'(10)-f'(5)+100=02f′(10)−f′(5)+100=0 is :
  1. (A)117117117
  2. (B)106106106
  3. (C)125125125
  4. (D)136136136

Correct answer: (C)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2022
Let A and B be two 3×33\times33×3 matrices such that AB=IAB=IAB=I and ∣A∣=18|A|=\dfrac{1}{8}∣A∣=81​ then ∣adj(B adj(2A))∣\left|\mathrm{adj}(B\,\mathrm{adj}(2A))\right|∣adj(Badj(2A))∣ is equal to
  1. (A)161616
  2. (B)323232
  3. (C)646464
  4. (D)128128128

Correct answer: (C)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2022
Let A be a 3×33 \times 33×3 invertible matrix. If ∣adj(24A)∣=adj(3adj(2A))∣|adj (24A)| = adj(3adj(2A))|∣adj(24A)∣=adj(3adj(2A))∣, then ∣A∣2|A|^{2}∣A∣2 is equal to :
  1. (A)666^{6}66
  2. (B)2122^{12}212
  3. (C)262^{6}26
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q202·MathematicsSingle correctJEE Main 2022
If the system of equations αx + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = β, has infinitely many solutions, then the ordered pair (α, β) is equal to :
  1. (A)(1,–3)
  2. (B)(–1, 3)
  3. (C)(1, 3)
  4. (D)(–1, –3)

Correct answer: (C)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2022
The ordered pair (a, b), for which the system of linear equations 3x−2y+z=b3x - 2y + z = b3x−2y+z=b 5x−8y+9z=35x - 8y + 9z = 35x−8y+9z=3 2x+y+az=−12x + y + az = -12x+y+az=−1 has no solution, is :
  1. (A)(3,13)\left(3, \frac{1}{3}\right)(3,31​)
  2. (B)(−3,13)\left(-3, \frac{1}{3}\right)(−3,31​)
  3. (C)(−3,−13)\left(-3, -\frac{1}{3}\right)(−3,−31​)
  4. (D)(3,−13)\left(3, -\frac{1}{3}\right)(3,−31​)

Correct answer: (C)

Step-by-step solution →
Q204·MathematicsNumericalJEE Main 2022
Let X=[010001000]X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}X=​000​100​010​​, Y=αI+βX+γX2Y=\alpha I+\beta X+\gamma X^{2}Y=αI+βX+γX2 and Z=α2I−αβX+(β2−αγ)X2Z=\alpha^{2}I-\alpha\beta X+\left(\beta^{2}-\alpha\gamma\right)X^{2}Z=α2I−αβX+(β2−αγ)X2, α,β,γ∈R\alpha,\beta,\gamma\in\mathbb{R}α,β,γ∈R. If Y−1=[15−2515015−250015]Y^{-1}=\begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}Y−1=​51​00​5−2​51​0​51​5−2​51​​​, then (α−β+γ)2(\alpha-\beta+\gamma)^{2}(α−β+γ)2 is equal to ______ .

Correct answer: 100

Step-by-step solution →
Q205·MathematicsSingle correctJEE Main 2022
Let A=[0−220]A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}A=[02​−20​]. If M and N are two matrices given by M=∑k=110A2kM = \sum_{k=1}^{10} A^{2k}M=∑k=110​A2k and N=∑k=110A2k−1N = \sum_{k=1}^{10} A^{2k-1}N=∑k=110​A2k−1 then MN2MN^2MN2 is
  1. (A)a non-identity symmetric matrix
  2. (B)a skew-symmetric matrix
  3. (C)neither symmetric nor skew-symmetric matrix
  4. (D)an identify matrix

Correct answer: (A)

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2022
The system of equations −kx+3y−14z=25-kx + 3y - 14z = 25−kx+3y−14z=25 −15x+4y−kz=3-15x + 4y - kz = 3−15x+4y−kz=3 −4x+y+3z=4-4x + y + 3z = 4−4x+y+3z=4 is consistent for all k in the set
  1. (A)RRR
  2. (B)R−{−11,13}R - \{-11,13\}R−{−11,13}
  3. (C)R−{13}R - \{13\}R−{13}
  4. (D)R−{−11,11}R - \{-11,11\}R−{−11,11}

Correct answer: (D)

Step-by-step solution →
Q207·MathematicsNumericalJEE Main 2022
Let A=(2−21−1)A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}A=(21​−2−1​) and B=(−12−12)B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}B=(−1−1​22​). Then the number of elements in the set {(n,m):n,m∈{1,2,.....,10}\{(n, m) : n, m \in \{1, 2, ....., 10\}{(n,m):n,m∈{1,2,.....,10} and nAn+mBm=I}nA^{n} + mB^{m} = I\}nAn+mBm=I} is ______

Correct answer: 1

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2022
Let A be a 3×33 \times 33×3 real matrix such that A(110)=(110)A\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}A​110​​=​110​​; A(101)=(−101)A\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}A​101​​=​−101​​ and A(001)=(112)A\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}A​001​​=​112​​. If X=(x1,x2,x3)TX = (x_1, x_2, x_3)^TX=(x1​,x2​,x3​)T and I is an identity matrix of order 3, then the system (A−2I)X=(411)(A - 2I)X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}(A−2I)X=​411​​ has
  1. (A)no solution
  2. (B)infinitely many solutions
  3. (C)unique solution
  4. (D)exactly two solutions

Correct answer: (B)

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2022
Let the system of linear equations x+y+αz=2x + y + \alpha z = 2x+y+αz=2 3x+y+z=43x + y + z = 43x+y+z=4 x+2z=1x + 2z = 1x+2z=1 have a unique solution (x∗,y∗,z∗)(x^*, y^*, z^*)(x∗,y∗,z∗). If (α,x∗)(\alpha, x^*)(α,x∗), (y∗,α)(y^*, \alpha)(y∗,α) and (x∗,−y∗)(x^*, -y^*)(x∗,−y∗) are collinear points, then the sum of absolute values of all possible values of α\alphaα is :
  1. (A)4
  2. (B)3
  3. (C)2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q210·MathematicsSingle correctJEE Main 2022
Let S={n:1≤n≤50 and n is odd}S=\{\sqrt{n}:1\le n\le 50\ \text{and n is odd}\}S={n​:1≤n≤50 and n is odd}. Let a∈Sa\in Sa∈S and A=[10a−110−a01]A=\begin{bmatrix}1&0&a\\-1&1&0\\-a&0&1\end{bmatrix}A=​1−1−a​010​a01​​. If ∑a∈Sdet⁡(adjA)=100λ\sum_{a\in S}\det(\mathrm{adj}A)=100\lambda∑a∈S​det(adjA)=100λ, then λ\lambdaλ is equal to
  1. (A)218
  2. (B)221
  3. (C)663
  4. (D)1717

Correct answer: (B)

Step-by-step solution →
Q211·MathematicsNumericalJEE Main 2022
Let S={(−1a0b);a,b∈{1,2,3,…100}}S = \left\{\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix}; a, b \in \{1, 2, 3, \ldots 100\}\right\}S={(−10​ab​);a,b∈{1,2,3,…100}} and let Tn={A∈S:An(n+1)=I}T_n = \{A \in S : A^{n(n+1)} = I\}Tn​={A∈S:An(n+1)=I}. Then the number of elements in ⋂n=1100Tn\bigcap_{n=1}^{100} T_n⋂n=1100​Tn​ is ______.

Correct answer: 100

Step-by-step solution →
Q212·MathematicsSingle correctJEE Main 2022
The number of values of α\alphaα for which the system of equations : x+y+z=αx + y + z = \alphax+y+z=α αx+2αy+3z=−1\alpha x + 2\alpha y + 3z = -1αx+2αy+3z=−1 x+3αy+5z=4x + 3\alpha y + 5z = 4x+3αy+5z=4 is inconsistent, is
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q213·MathematicsMultiple correctJEE Advanced 2021
For any 3×33 \times 33×3 matrix M, let ∣M∣\left|M\right|∣M∣ denote the determinant of M. Let I be the 3×33 \times 33×3 identity matrix. Let E and F be two 3×33 \times 33×3 matrices such that (I−EF)(I - EF)(I−EF) is invertible. If G=(I−EF)−1G = (I - EF)^{-1}G=(I−EF)−1, then which of the following statements is (are) TRUE ?
  1. (A)∣FE∣=∣I−FE∣∣FGE∣\left|FE\right| = \left|I - FE\right|\left|FGE\right|∣FE∣=∣I−FE∣∣FGE∣
  2. (B)(I−FE)(I+FGE)=I(I - FE)(I + FGE) = I(I−FE)(I+FGE)=I
  3. (C)EFG=GEFEFG = GEFEFG=GEF
  4. (D)(I−FE)(I−FGE)=I\left(I - FE\right)\left(I - FGE\right) = I(I−FE)(I−FGE)=I

Correct answer: (A), (B), (C)

Step-by-step solution →
Q214·MathematicsMultiple correctJEE Advanced 2021
For any 3×33 \times 33×3 matrix M, let ∣M∣\left|M\right|∣M∣ denote the determinant of M. Let E=[12323481318]E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix}E=​128​2313​3418​​, P=[100001010]P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}P=​100​001​010​​ and F=[13281813243]F = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix}F=​182​3184​2133​​ If Q is a nonsingular matrix of order 3×33 \times 33×3, then which of the following statements is (are) TRUE ?
  1. (A)F=PEPF = PEPF=PEP and P2=[100010001]P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}P2=​100​010​001​​
  2. (B)∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣\left|EQ + PFQ^{-1}\right| = \left|EQ\right| + \left|PFQ^{-1}\right|​EQ+PFQ−1​=∣EQ∣+​PFQ−1​
  3. (C)∣(EF)3∣>∣EF∣2\left|(EF)^3\right| > \left|EF\right|^2​(EF)3​>∣EF∣2
  4. (D)Sum of the diagonal entries of P−1EP+FP^{-1}EP + FP−1EP+F is equal to the sum of diagonal entries of E+P−1FPE + P^{-1}FPE+P−1FP

Correct answer: (A), (B), (D)

Step-by-step solution →
Q215·MathematicsNumericalJEE Advanced 2021
Let α\alphaα, β\betaβ and γ\gammaγ be real numbers such that the system of linear equations x+2y+3z=αx + 2y + 3z = \alphax+2y+3z=α 4x+5y+6z=β4x + 5y + 6z = \beta4x+5y+6z=β 7x+8y+9z=γ−17x + 8y + 9z = \gamma - 17x+8y+9z=γ−1 is consistent. Let ∣M∣|M|∣M∣ represent the determinant of the matrix M=[α2γβ10−101]M = \begin{bmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}M=​αβ−1​210​γ01​​ Let P be the plane containing all those (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P. The value of ∣M∣\left|M\right|∣M∣ is ________.

Correct answer: 1.00

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2021
Let Jn,m=∫012xnxm−1dxJ_{n,m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1}dxJn,m​=∫021​​xm−1xn​dx, ∀ n > m and n, m ∈ N . Consider a matrix A=[aij]3×3A = [a_{ij}]_{3 \times 3}A=[aij​]3×3​ where aij={J6+i,3−Ji+3,3,i≤j0,i>ja_{ij} = \begin{cases} J_{6+i,3} - J_{i+3,3}, & i \le j \\ 0, & i > j \end{cases}aij​={J6+i,3​−Ji+3,3​,0,​i≤ji>j​. Then ∣adjA−1∣\left|\mathrm{adj}A^{-1}\right|​adjA−1​ is :
  1. (A)(15)2×242(15)^{2} \times 2^{42}(15)2×242
  2. (B)(15)2×234(15)^{2} \times 2^{34}(15)2×234
  3. (C)(105)2×238(105)^{2} \times 2^{38}(105)2×238
  4. (D)(105)2×236(105)^{2} \times 2^{36}(105)2×236

Correct answer: (C)

Step-by-step solution →
Q217·MathematicsSingle correctJEE Main 2021
Consider the system of linear equations − x + y + 2z = 0 3x − ay + 5z =1 2x − 2y − az = 7 Let S1S_{1}S1​ be the set of all a ∈ ℝ for which the system is inconsistent and S2S_{2}S2​ be the set of all a ∈ ℝ for which the system has infinitely many solutions. If n(S1)n(S_{1})n(S1​) and n(S2)n(S_{2})n(S2​) denote the number of elements in S1S_{1}S1​ and S2S_{2}S2​ respectively, then
  1. (A)n(S1)=2n(S_{1}) = 2n(S1​)=2, n(S2)=2n(S_{2}) = 2n(S2​)=2
  2. (B)n(S1)=1n(S_{1}) = 1n(S1​)=1, n(S2)=0n(S_{2}) = 0n(S2​)=0
  3. (C)n(S1)=2n(S_{1}) = 2n(S1​)=2, n(S2)=0n(S_{2}) = 0n(S2​)=0
  4. (D)n(S1)=0n(S_{1}) = 0n(S1​)=0, n(S2)=2n(S_{2}) = 2n(S2​)=2

Correct answer: (C)

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2021
If the following system of linear equations 2x+y+z=52x + y + z = 52x+y+z=5 x−y+z=3x - y + z = 3x−y+z=3 x+y+az=bx + y + az = bx+y+az=b has no solution, then :
  1. (A)a=−13, b≠73a = -\frac{1}{3},\ b \ne \frac{7}{3}a=−31​, b=37​
  2. (B)a≠13, b=73a \ne \frac{1}{3},\ b = \frac{7}{3}a=31​, b=37​
  3. (C)a≠−13, b=73a \ne -\frac{1}{3},\ b = \frac{7}{3}a=−31​, b=37​
  4. (D)a=13, b≠73a = \frac{1}{3},\ b \ne \frac{7}{3}a=31​, b=37​

Correct answer: (D)

Step-by-step solution →
Q219·MathematicsSingle correctJEE Main 2021
If α+β+γ=2π\alpha + \beta + \gamma = 2\piα+β+γ=2π, then the system of equations x+(cos⁡γ)y+(cos⁡β)z=0x + (\cos \gamma)y + (\cos \beta)z = 0x+(cosγ)y+(cosβ)z=0 (cos⁡γ)x+y+(cos⁡α)z=0(\cos \gamma)x + y + (\cos \alpha)z = 0(cosγ)x+y+(cosα)z=0 (cos⁡β)x+(cos⁡α)y+z=0(\cos \beta)x + (\cos \alpha)y + z = 0(cosβ)x+(cosα)y+z=0 has :
  1. (A)no solution
  2. (B)infinitely many solution
  3. (C)exactly two solutions
  4. (D)a unique solution

Correct answer: (B)

Step-by-step solution →
Q220·MathematicsNumericalJEE Main 2021
The number of elements in the set {A=(ab0d):a,b,d∈{−1,0,1} and (I−A)3=I−A3}\left\{ A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I - A)^{3} = I - A^{3} \right\}{A=(a0​bd​):a,b,d∈{−1,0,1} and (I−A)3=I−A3}, where I is 2×22 \times 22×2 identity matrix, is :

Correct answer: 8

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2021
Let [λ] be the greatest integer less than or equal to λ. The set of all values of λ for which the system of linear equations x + y + z = 4, 3x + 2y+ 5z = 3, 9x + 4y + (28+ [λ])z = [λ] has a solution is:
  1. (A)ℝ
  2. (B)(−∞,−9)∪(−9,∞)(-\infty, -9) \cup (-9, \infty)(−∞,−9)∪(−9,∞)
  3. (C)[−9, −8)
  4. (D)(−∞,−9)∪[−8,∞)(-\infty, -9) \cup [-8, \infty)(−∞,−9)∪[−8,∞)

Correct answer: (A)

Step-by-step solution →
Q222·MathematicsSingle correctJEE Main 2021
Let A=([x+1][x+2][x+3][x][x+3][x+3][x][x+2][x+4])A = \begin{pmatrix} [x+1] & [x+2] & [x+3] \\ [x] & [x+3] & [x+3] \\ [x] & [x+2] & [x+4] \end{pmatrix}A=​[x+1][x][x]​[x+2][x+3][x+2]​[x+3][x+3][x+4]​​, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval:
  1. (A)[68, 69)
  2. (B)[62, 63)
  3. (C)[65, 66)
  4. (D)[60, 61)

Correct answer: (B)

Step-by-step solution →
Q223·MathematicsNumericalJEE Main 2021
If the system of linear equations 2x+y−z=32x + y - z = 32x+y−z=3 x−y−z=αx - y - z = \alphax−y−z=α 3x+3y+βz=33x + 3y + \beta z = 33x+3y+βz=3 has infinitely many solution, then α+β−αβ\alpha + \beta - \alpha\betaα+β−αβ is equal to __________.

Correct answer: 5

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2021
If the matrix A=(02K−1)A = \begin{pmatrix} 0 & 2 \\ K & -1 \end{pmatrix}A=(0K​2−1​) satisfies A(A3+3I)=2IA(A^3 + 3I) = 2IA(A3+3I)=2I, then the value of K is :
  1. (A)12\frac{1}{2}21​
  2. (B)−12-\frac{1}{2}−21​
  3. (C)−1-1−1
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q225·MathematicsSingle correctJEE Main 2021
Let A=(100011100)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix}A=​101​010​010​​. Then A2025−A2020A^{2025} - A^{2020}A2025−A2020 is equal to :
  1. (A)A6−AA^{6} - AA6−A
  2. (B)A5A^{5}A5
  3. (C)A5−AA^{5} - AA5−A
  4. (D)A6A^{6}A6

Correct answer: (A)

Step-by-step solution →
Q226·MathematicsNumericalJEE Main 2021
Let A be a 3×33 \times 33×3 real matrix. If det⁡(2Adj(2 Adj(Adj(2A))))=241\det(2\mathrm{Adj}(2\ \mathrm{Adj}(\mathrm{Adj}(2A)))) = 2^{41}det(2Adj(2 Adj(Adj(2A))))=241, then the value of det⁡(A2)\det(A^{2})det(A2) equal ________.

Correct answer: 4

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Q227·MathematicsSingle correctJEE Main 2021
Two fair dice are thrown. The numbers on them are taken as λ\lambdaλ and μ\muμ, and a system of linear equations x+y+z=5x + y + z = 5x+y+z=5 x+2y+3z=μx + 2y + 3z = \mux+2y+3z=μ x+3y+λz=1x + 3y + \lambda z = 1x+3y+λz=1 is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :
  1. (A)p=16p = \frac{1}{6}p=61​ and q=136q = \frac{1}{36}q=361​
  2. (B)p=56p = \frac{5}{6}p=65​ and q=536q = \frac{5}{36}q=365​
  3. (C)p=56p = \frac{5}{6}p=65​ and q=136q = \frac{1}{36}q=361​
  4. (D)p=16p = \frac{1}{6}p=61​ and q=536q = \frac{5}{36}q=365​

Correct answer: (B)

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Q228·MathematicsSingle correctJEE Main 2021
Let θ∈(0,π2)\theta \in \left(0, \frac{\pi}{2}\right)θ∈(0,2π​). If the system of linear equations (1+cos⁡2θ)x+sin⁡2θ y+4sin⁡3θ z=0(1 + \cos^2\theta)x + \sin^2\theta\, y + 4\sin 3\theta\, z = 0(1+cos2θ)x+sin2θy+4sin3θz=0 cos⁡2θ x+(1+sin⁡2θ) y+4sin⁡3θ z=0\cos^2\theta\, x + (1 + \sin^2\theta)\, y + 4\sin 3\theta\, z = 0cos2θx+(1+sin2θ)y+4sin3θz=0 cos⁡2θ x+sin⁡2θ y+(1+4sin⁡3θ) z=0\cos^2\theta\, x + \sin^2\theta\, y + (1 + 4\sin 3\theta)\, z = 0cos2θx+sin2θy+(1+4sin3θ)z=0 has a non-trivial solution, then the value of θ\thetaθ is :
  1. (A)4π9\frac{4\pi}{9}94π​
  2. (B)7π18\frac{7\pi}{18}187π​
  3. (C)π18\frac{\pi}{18}18π​
  4. (D)5π18\frac{5\pi}{18}185π​

Correct answer: (B)

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Q229·MathematicsNumericalJEE Main 2021
Let f(x)=∣sin⁡2x−2+cos⁡2xcos⁡2x2+sin⁡2xcos⁡2xcos⁡2xsin⁡2xcos⁡2x1+cos⁡2x∣f\left(x\right)=\begin{vmatrix}\sin^2 x & -2+\cos^2 x & \cos 2x \\ 2+\sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1+\cos 2x\end{vmatrix}f(x)=​sin2x2+sin2xsin2x​−2+cos2xcos2xcos2x​cos2xcos2x1+cos2x​​, x∈[0,π]x \in \left[0,\pi\right]x∈[0,π]. Then the maximum value of f(x)f\left(x\right)f(x) is equal to ____.

Correct answer: 6

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Q230·MathematicsSingle correctJEE Main 2021
Let A and B be two 3 x 3 real matrices such that (A2−B2)(A^2 - B^2)(A2−B2) is invertible matrix. If A5=B5A^5 = B^5A5=B5 and A3B2=A2B3A^3B^2 = A^2B^3A3B2=A2B3, then the value of the determinant of the matrix A3+B3A^3 + B^3A3+B3 is equal to :
  1. (A)2
  2. (B)1
  3. (C)0
  4. (D)4

Correct answer: (C)

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Q231·MathematicsNumericalJEE Main 2021
For real numbers α\alphaα and β\betaβ, consider the following system of linear equations: x+y−z=2x+y-z=2x+y−z=2, x+2y+αz=1x+2y+\alpha z=1x+2y+αz=1, 2x−y+z=β2x-y+z=\beta2x−y+z=β. If the system has infinite solutions, then α+β\alpha+\betaα+β is equal to ____.

Correct answer: 5

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Q232·MathematicsSingle correctJEE Main 2021
Let A=[12−14]A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}A=[1−1​24​]. If A−1=αI+βAA^{-1} = \alpha I + \beta AA−1=αI+βA, α,β∈R\alpha, \beta \in Rα,β∈R. I is a 2 x 2 identify matrix, then 4(α−β)4\left(\alpha - \beta\right)4(α−β) is equal to :
  1. (A)2
  2. (B)5
  3. (C)4
  4. (D)83\frac{8}{3}38​

Correct answer: (C)

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Q233·MathematicsSingle correctJEE Main 2021
If P=[10121]P=\begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix}P=[121​​01​], then P50P^{50}P50 is :
  1. (A)[10251]\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}[125​01​]
  2. (B)[12501]\begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix}[10​251​]
  3. (C)[10501]\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}[150​01​]
  4. (D)[15001]\begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}[10​501​]

Correct answer: (A)

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Q234·MathematicsNumericalJEE Main 2021
Let M={A=(abcd):a,b,c,d∈{±3,±2,±1,0}}M = \left\{ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a,b,c,d \in \left\{ \pm 3, \pm 2, \pm 1, 0 \right\} \right\}M={A=(ac​bd​):a,b,c,d∈{±3,±2,±1,0}}. Define f:M→Zf : M \rightarrow Zf:M→Z, as f(A)=det⁡(A)f\left( A \right) = \det\left( A \right)f(A)=det(A), for all A∈MA \in MA∈M, where ZZZ is set of all integers. Then the number of A∈MA \in MA∈M such that f(A)=15f\left( A \right) = 15f(A)=15 is equal to...........

Correct answer: 16

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Q235·MathematicsSingle correctJEE Main 2021
The number of distinct real roots of ∣sin⁡xcos⁡xcos⁡xcos⁡xsin⁡xcos⁡xcos⁡xcos⁡xsin⁡x∣=0\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0​sinxcosxcosx​cosxsinxcosx​cosxcosxsinx​​=0 in the interval −π4≤x≤π4-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}−4π​≤x≤4π​ is:
  1. (A)2
  2. (B)4
  3. (C)1
  4. (D)3

Correct answer: (C)

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Q236·MathematicsSingle correctJEE Main 2021
The values of a and b, for which the system of equations 2x + 3y + 6z = 8 x + 2y + az = 5 3x + 5y + 9z = b Has no solution, are :
  1. (A)a≠3a \neq 3a=3, b = 3
  2. (B)a≠3a \neq 3a=3, b ≠\neq= 13
  3. (C)a = 3, b ≠\neq= 13
  4. (D)a = 3, b = 13

Correct answer: (C)

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Q237·MathematicsSingle correctJEE Main 2021
The values of λ\lambdaλ and μ\muμ such that the system of equations x+y+z=6x + y + z = 6x+y+z=6, 3x+5y+5z=263x + 5y + 5z = 263x+5y+5z=26, x+2y+λz=μx + 2y + \lambda z = \mux+2y+λz=μ has no solution, are :
  1. (A)λ=2,μ≠10\lambda = 2, \mu \neq 10λ=2,μ=10
  2. (B)λ=3,μ≠10\lambda = 3, \mu \neq 10λ=3,μ=10
  3. (C)λ=3,μ=5\lambda = 3, \mu = 5λ=3,μ=5
  4. (D)λ≠2,μ=10\lambda \neq 2, \mu = 10λ=2,μ=10

Correct answer: (A)

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Q238·MathematicsSingle correctJEE Main 2021
Let A=∣aij∣A = \left|a_{ij}\right|A=∣aij​∣ be a real matrix of order 3 x 3, such that a11+ai2+ai3=1a_{11} + a_{i2} + a_{i3} = 1a11​+ai2​+ai3​=1, for i=1,2,3i = 1, 2, 3i=1,2,3. Then, the sum of all the entries of the matrix A3A^3A3 is equal to :
  1. (A)999
  2. (B)333
  3. (C)111
  4. (D)222

Correct answer: (B)

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Q239·MathematicsNumericalJEE Main 2021
Let A=(010100001)A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}A=​010​100​001​​. Then the number of 3 x 3 matrices B with entries from the set {1,2,3,4,5}\{1,2,3,4,5\}{1,2,3,4,5} and satisfying AB = BA is...........

Correct answer: 3125

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Q240·MathematicsSingle correctJEE Main 2021
The value of k∈Rk\in Rk∈R, for which the following system of linear equations 3x−y+4z=3,3x-y+4z=3,3x−y+4z=3, x+2y−3z=−2,x+2y-3z=-2,x+2y−3z=−2, 6x+5y+kz=−3,6x+5y+kz=-3,6x+5y+kz=−3, has infinitely many solutions, is :
  1. (A)333
  2. (B)−3-3−3
  3. (C)−5-5−5
  4. (D)555

Correct answer: (C)

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Q241·MathematicsNumericalJEE Main 2021
Let A=(1−1001−1001)A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}A=​100​−110​0−11​​ and B=7A20−20A7+2IB = 7A^{20} - 20A^{7} + 2IB=7A20−20A7+2I , where III is an identity matrix of order 3 x 3. If B=[bij]B = \left[ b_{ij} \right]B=[bij​], then b13b_{13}b13​ is equal to..........

Correct answer: 910

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Q242·MathematicsSingle correctJEE Main 2021
Let A=[23a0]A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}A=[2a​30​], a∈Ra \in Ra∈R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det⁡(Q)=9\det(Q) = 9det(Q)=9, then the modulus of the sum of all possible values of determinant of P is equal to :
  1. (A)18
  2. (B)36
  3. (C)24
  4. (D)45

Correct answer: (B)

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Q243·MathematicsNumericalJEE Main 2021
Let a, b, c, d be in arithmetic progression with common difference λ\lambdaλ. If ∣x+a−cx+bx+ax−1x+cx+bx−b+dx+dx+c∣=2\begin{vmatrix} x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c \end{vmatrix} = 2​x+a−cx−1x−b+d​x+bx+cx+d​x+ax+bx+c​​=2, then the value of λ2\lambda^2λ2 is equal to.........

Correct answer: 1

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Q244·MathematicsNumericalJEE Main 2021
Let A={aij}A=\{a_{ij}\}A={aij​} be a 3 x 3 matrix, where aij={(−1)j−iif i<j,2if i=j,(−1)i+jif i>j,a_{ij}=\begin{cases}(-1)^{j-i} & \text{if } i<j, \\ 2 & \text{if } i=j, \\ (-1)^{i+j} & \text{if } i>j,\end{cases}aij​=⎩⎨⎧​(−1)j−i2(−1)i+j​if i<j,if i=j,if i>j,​ then det⁡(3 Adj(2A−1))\det\left(3\,\text{Adj}\left(2A^{-1}\right)\right)det(3Adj(2A−1)) is equal to…….

Correct answer: 108

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Q245·MathematicsSingle correctJEE Main 2021
The solutions of the equation ∣1+sin⁡2xsin⁡2xsin⁡2xcos⁡2x1+cos⁡2xcos⁡2x4sin⁡2x4sin⁡2x1+4sin⁡2x∣=0,(0<x<π)\begin{vmatrix} 1+\sin^{2}x & \sin^{2}x & \sin^{2}x \\ \cos^{2}x & 1+\cos^{2}x & \cos^{2}x \\ 4\sin 2x & 4\sin 2x & 1+4\sin 2x \end{vmatrix} = 0,\left(0 < x < \pi\right)​1+sin2xcos2x4sin2x​sin2x1+cos2x4sin2x​sin2xcos2x1+4sin2x​​=0,(0<x<π), are
  1. (A)π12,π6\dfrac{\pi}{12}, \dfrac{\pi}{6}12π​,6π​
  2. (B)π6,5π6\dfrac{\pi}{6}, \dfrac{5\pi}{6}6π​,65π​
  3. (C)5π12,7π12\dfrac{5\pi}{12}, \dfrac{7\pi}{12}125π​,127π​
  4. (D)7π12,11π12\dfrac{7\pi}{12}, \dfrac{11\pi}{12}127π​,1211π​

Correct answer: (D)

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Q246·MathematicsSingle correctJEE Main 2021
Let A+2B=[1206−33−531]A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}A+2B=​16−5​2−33​031​​ and 2A−B=[2−152−16012]2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}2A−B=​220​−1−11​562​​. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) −-− Tr(B) has value equal to
  1. (A)111
  2. (B)222
  3. (C)000
  4. (D)333

Correct answer: (B)

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Q247·MathematicsSingle correctJEE Main 2021
Let the system of linear equations 4x + λy + 2z = 0 2x − y + z = 0 μx + 2y + 3z = 0, λ, μ ∈ R. has a non-trivial solution. Then which of the following is true ?
  1. (A)μ = 6, λ ∈ R
  2. (B)λ = 2, μ ∈ R
  3. (C)λ = 3, μ ∈ R
  4. (D)μ = −6, λ ∈ R

Correct answer: (A)

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Q248·MathematicsNumericalJEE Main 2021
Let I be an identity matrix of order 2 × 2 and P=[2−15−3]P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}P=[25​−1−3​]. Then the value of n ∈ N for which Pn=5I−8PP^{n} = 5I - 8PPn=5I−8P is equal to ________ .

Correct answer: 6

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Q249·MathematicsSingle correctJEE Main 2021
Let α,β,γ\alpha, \beta, \gammaα,β,γ be the real roots of the equation, x3+ax2+bx+c=0x^{3} + ax^{2} + bx + c = 0x3+ax2+bx+c=0, (a,b,c∈R(a, b, c \in R(a,b,c∈R and a,b≠0)a, b \neq 0)a,b=0). If the system of equations (in, u, v, w) given by αu+βv+γw=0\alpha u + \beta v + \gamma w = 0αu+βv+γw=0, βu+γv+αw=0\beta u + \gamma v + \alpha w = 0βu+γv+αw=0; γu+αv+βw=0\gamma u + \alpha v + \beta w = 0γu+αv+βw=0 has non-trivial solution, then the value of a2b\dfrac{a^{2}}{b}ba2​ is
  1. (A)555
  2. (B)333
  3. (C)111
  4. (D)000

Correct answer: (B)

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Q250·MathematicsNumericalJEE Main 2021
If 111, log⁡10(4x−2)\log_{10}(4^{x}-2)log10​(4x−2) and log⁡10(4x+185)\log_{10}\left(4^{x}+\frac{18}{5}\right)log10​(4x+518​) are in arithmetic progression for a real number xxx, then the value of the determinant ∣2(x−12)x−1x210xx10∣\begin{vmatrix} 2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{vmatrix}​2(x−21​)1x​x−101​x2x0​​ is equal to :

Correct answer: 2

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Q251·MathematicsNumericalJEE Main 2021
Let A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​] and B=[αβ]≠[00]B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}B=[αβ​]=[00​] such that AB=BAB = BAB=B and a+d=2021a + d = 2021a+d=2021, then the value of ad−bcad - bcad−bc is equal to ________.

Correct answer: 2020

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Q252·MathematicsSingle correctJEE Main 2021
If A=(0sin⁡αsin⁡α0)A = \begin{pmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{pmatrix}A=(0sinα​sinα0​) and det⁡(A2−12I)=0\det\left(A^2 - \frac{1}{2}I\right) = 0det(A2−21​I)=0, then a possible value of α\alphaα is
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π3\frac{\pi}{3}3π​
  3. (C)π4\frac{\pi}{4}4π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (C)

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Q253·MathematicsSingle correctJEE Main 2021
If x,y,zx, y, zx,y,z are in arithmetic progression with common difference ddd, x≠3dx \neq 3dx=3d, and the determinant of the matrix [342x452y5kz]\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}​345​42​52​k​xyz​​ is zero, then the value of k2k^{2}k2 is
  1. (A)727272
  2. (B)121212
  3. (C)363636
  4. (D)666

Correct answer: (A)

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Q254·MathematicsNumericalJEE Main 2021
If A=[230−1]A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}A=[20​3−1​], then the value of det⁡(A4)+det⁡(A10−(Adj(2A))10)\det(A^4) + \det\left(A^{10} - (\text{Adj}(2A))^{10}\right)det(A4)+det(A10−(Adj(2A))10) is _________ .

Correct answer: 16

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Q255·MathematicsSingle correctJEE Main 2021
The system of equations kx+y+z=1kx + y + z = 1kx+y+z=1, x+ky+z=kx + ky + z = kx+ky+z=k and x+y+zk=k2x + y + zk = k^2x+y+zk=k2 has no solution if k is equal to :
  1. (A)0
  2. (B)1
  3. (C)−1-1−1
  4. (D)−2-2−2

Correct answer: (D)

Step-by-step solution →
Q256·MathematicsSingle correctJEE Main 2021
The maximum value of f(x)=∣sin⁡2x1+cos⁡2xcos⁡2x1+sin⁡2xcos⁡2xcos⁡2xsin⁡2xcos⁡2xsin⁡2x∣f(x)=\begin{vmatrix} \sin^{2}x & 1+\cos^{2}x & \cos 2x \\ 1+\sin^{2}x & \cos^{2}x & \cos 2x \\ \sin^{2}x & \cos^{2}x & \sin 2x \end{vmatrix}f(x)=​sin2x1+sin2xsin2x​1+cos2xcos2xcos2x​cos2xcos2xsin2x​​, x∈Rx \in Rx∈R is:
  1. (A)7\sqrt{7}7​
  2. (B)34\frac{3}{4}43​
  3. (C)5\sqrt{5}5​
  4. (D)555

Correct answer: (C)

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Q257·MathematicsNumericalJEE Main 2021
The total number of 3 × 3 matrices A having enteries from the set (0, 1, 2, 3) such that the sum of all the diagonal entries of AATAA^{T}AAT is 9, is equal to _____.

Correct answer: 766

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Q258·MathematicsNumericalJEE Main 2021
Let A=[a1a2]A=\begin{bmatrix} a_{1} \\ a_{2} \end{bmatrix}A=[a1​a2​​] and B=[b1b2]B=\begin{bmatrix} b_{1} \\ b_{2} \end{bmatrix}B=[b1​b2​​] be two 2×12 \times 12×1 matrices with real entries such that A=XBA = XBA=XB, where X=13[1−11k]X=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix}X=3​1​[11​−1k​], and k∈Rk \in Rk∈R. If a12+a22=23(b12+b22)a_{1}^{2}+a_{2}^{2}=\frac{2}{3}\left(b_{1}^{2}+b_{2}^{2}\right)a12​+a22​=32​(b12​+b22​) and (k2+1)b22≠−2b1b2(k^{2}+1)b_{2}^{2} \neq -2b_{1}b_{2}(k2+1)b22​=−2b1​b2​, then the value of k is ________.

Correct answer: 1

Step-by-step solution →
Q259·MathematicsSingle correctJEE Main 2021
Let A=[i−i−ii]A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}A=[i−i​−ii​], i=−1i = \sqrt{-1}i=−1​. Then, the system of linear equations A8[xy]=[864]A^{8}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}A8[xy​]=[864​] has :
  1. (A)A unique solution
  2. (B)Infinitely many solutions
  3. (C)No solution
  4. (D)Exactly two solutions

Correct answer: (C)

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Q260·MathematicsNumericalJEE Main 2021
Let P=[−302056901401121206014]P = \begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}P=​−3090120​2014060​5611214​​ and A=[27ω2−1−ω10−ω−ω+1]A = \begin{bmatrix} 2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1 \end{bmatrix}A=​2−10​7−ω−ω​ω21−ω+1​​ where ω=−1+i32\omega = \frac{-1+i\sqrt{3}}{2}ω=2−1+i3​​, and I₃ be the identity matrix of order 3. If the determinant of the matrix (P−1AP−I3)2(P^{-1}AP - I_{3})^{2}(P−1AP−I3​)2 is αω2\alpha\omega^{2}αω2, then the value of α is equal to ______ .

Correct answer: 36

Step-by-step solution →
Q261·MathematicsNumericalJEE Main 2021
If the matrix A=[10002030−1]A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}A=​103​020​00−1​​ satisfies the equation A20+αA19+βA=[100040001]A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}A20+αA19+βA=​100​040​001​​ for some real numbers α\alphaα and β\betaβ, then β−α\beta - \alphaβ−α is equal to________________.

Correct answer: 4

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Q262·MathematicsSingle correctJEE Main 2021
Consider the following system of equations : x + 2y − 3z = a 2x + 6y − 11 z = b x − 2y + 7z = c, Where a, b and c are real constants. Then the system of equations :
  1. (A)has a unique solution when 5a = 2b + c
  2. (B)has infinite number of solutions when 5a = 2b +c
  3. (C)has no solution for all a, b and c
  4. (D)has a unique solution for all a, b and c

Correct answer: (B)

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Q263·MathematicsSingle correctJEE Main 2021
Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2A^{2}A2 is 1, then the possible number of such matrices is:
  1. (A)6
  2. (B)1
  3. (C)4
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q264·MathematicsSingle correctJEE Main 2021
The value of ∣(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41∣\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}​(a+1)(a+2)(a+2)(a+3)(a+3)(a+4)​a+2a+3a+4​111​​ is
  1. (A)−2
  2. (B)(a+1) (a+2) (a+3)
  3. (C)0
  4. (D)(a+2) (a+3) (a+4)

Correct answer: (A)

Step-by-step solution →
Q265·MathematicsNumericalJEE Main 2021
If A=[0−tan⁡(θ2)tan⁡(θ2)0]A = \begin{bmatrix} 0 & -\tan\left(\frac{\theta}{2}\right) \\ \tan\left(\frac{\theta}{2}\right) & 0 \end{bmatrix}A=[0tan(2θ​)​−tan(2θ​)0​] and (I2+A)(I2−A)−1=[a−bba](I_2 + A)(I_2 - A)^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}(I2​+A)(I2​−A)−1=[ab​−ba​], then 13(a2+b2)13(a^2 + b^2)13(a2+b2) is equal to _______.

Correct answer: 13

Step-by-step solution →
Q266·MathematicsNumericalJEE Main 2021
Let A=[xyzyzxzxy]A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}A=​xyz​yzx​zxy​​, where xxx, yyy and zzz are real numbers such that x+y+z>0x + y + z > 0x+y+z>0 and xyz=2xyz = 2xyz=2. If A2=I3A^2 = I_3A2=I3​, then the value of x3+y3+z3x^3 + y^3 + z^3x3+y3+z3 is ________.

Correct answer: 7

Step-by-step solution →
Q267·MathematicsSingle correctJEE Main 2021
The following system of linear equations 3x + 3y + 2z = 9 3x + 2y + 2z = 9 x – y + 4z = 8
  1. (A)does not have any solution
  2. (B)has a unique solution
  3. (C)has a solution (α, β, γ) satisfying α + β2^{2}2 + γ3^{3}3 = 12
  4. (D)has infinitely many solutions

Correct answer: (B)

Step-by-step solution →
Q268·MathematicsSingle correctJEE Main 2021
If for the matrix, A = [1α–βα], AAT^{T}T = I2, then the value of α4^{4}4 + β4^{4}4 is:
  1. (A)1
  2. (B)3
  3. (C)2
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q269·MathematicsNumericalJEE Main 2021
If the system of equations kx+y+2z=1kx + y + 2z = 1kx+y+2z=1 3x−y−2z=23x - y - 2z = 23x−y−2z=2 −2x−2y−4z=3-2x - 2y - 4z = 3−2x−2y−4z=3 has infinitely many solutions, then kkk is equal to ___________.

Correct answer: 21

Step-by-step solution →
Q270·MathematicsSingle correctJEE Main 2021
Let A be a 3 × 3 matrix with det(A) = 4. Let Ri denote the ith^{th}th row of A. If a matrix B is obtained by performing the operation R2→ 2R2 + 5R3 on 2A, then det(B) is equal to:
  1. (A)64
  2. (B)16
  3. (C)80
  4. (D)128

Correct answer: (A)

Step-by-step solution →
Q271·MathematicsSingle correctJEE Main 2021
For the system of linear equations: x−2y=1, x−y+kz=−2, ky+4z=6, k∈Rx-2y=1,\ x-y+kz=-2,\ ky+4z=6,\ k\in \mathbf{R}x−2y=1, x−y+kz=−2, ky+4z=6, k∈R consider the following statements: (A) The system has unique solution if k≠2,k≠−2k\neq 2, k\neq -2k=2,k=−2. (B) The system has unique solution if k=−2k=-2k=−2. (C) The system has unique solution if k=2k=2k=2. (D) The system has no-solution if k=2k=2k=2. (E) The system has infinite number of solutions if k≠−2k\neq -2k=−2. Which of the following statements are correct?
  1. (A)(B) and (E) only
  2. (B)(C) and (D) only
  3. (C)(A) and (D) only
  4. (D)(A) and (E) only

Correct answer: (C)

Step-by-step solution →
Q272·MathematicsSingle correctJEE Main 2021
The system of linear equations 3x−2y−kz=103x - 2y - kz = 103x−2y−kz=10 2x−4y−2z=62x - 4y - 2z = 62x−4y−2z=6 x+2y−z=5mx + 2y - z = 5mx+2y−z=5m is inconsistent if :
  1. (A)k=3,m=45k = 3, m = \frac{4}{5}k=3,m=54​
  2. (B)k≠3,m∈Rk \neq 3, m \in Rk=3,m∈R
  3. (C)k≠3,m≠45k \neq 3, m \neq \frac{4}{5}k=3,m=54​
  4. (D)k=3,m≠45k = 3, m \neq \frac{4}{5}k=3,m=54​

Correct answer: (D)

Step-by-step solution →
Q273·MathematicsNumericalJEE Main 2021
Let P=[3−1−220α3−50]P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}P=​323​−10−5​−2α0​​, where α∈R\alpha \in Rα∈R. Suppose Q=[qij]Q = [q_{ij}]Q=[qij​] is a matrix satisfying PQ=kI3PQ = kI_3PQ=kI3​ for some non-zero k∈Rk \in Rk∈R. If q23=−k8q_{23} = -\frac{k}{8}q23​=−8k​ and ∣Q∣=k22|Q| = \frac{k^2}{2}∣Q∣=2k2​, then α2+k2\alpha^2 + k^2α2+k2 is equal to _______

Correct answer: 17

Step-by-step solution →
Q274·MathematicsSingle correctJEE Main 2021
Let A and B be 3×33\times 33×3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2−B2A2)X=O\left(\mathrm{A}^{2}\mathrm{B}^{2}-\mathrm{B}^{2}\mathrm{A}^{2}\right)\mathrm{X}=\mathrm{O}(A2B2−B2A2)X=O, where X is a 3×13\times 13×1 column matrix of unknown variables and O is a 3×13\times 13×1 null matrix, has :
  1. (A)a unique solution
  2. (B)exactly two solutions
  3. (C)infinitely many solutions
  4. (D)no solution

Correct answer: (C)

Step-by-step solution →
Q275·MathematicsMultiple correctJEE Advanced 2020
Let M be a 3×33 \times 33×3 invertible matrix with real entries and let I denote the 3×33 \times 33×3 identity matrix. If M−1=adj⁡(adj⁡M)M^{-1} = \operatorname{adj}(\operatorname{adj} M)M−1=adj(adjM), then which of the following statement is/are ALWAYS TRUE ?
  1. (A)M=IM = IM=I
  2. (B)det⁡M=1\det M = 1detM=1
  3. (C)M2=IM^{2} = IM2=I
  4. (D)(adj⁡M)2=I(\operatorname{adj} M)^{2} = I(adjM)2=I

Correct answer: (B), (C), (D)

Step-by-step solution →
Q276·MathematicsIntegerJEE Advanced 2020
The trace of a square matrix is defined to be the sum of its diagonal entries. If AAA is a 2×22 \times 22×2 matrix such that the trace of AAA is 3 and the trace of A3A^{3}A3 is −18-18−18, then the value of the determinant of AAA is ________

Correct answer: 5

Step-by-step solution →
Q277·MathematicsSingle correctJEE Main 2020
The value of λ\lambdaλ and μ\muμ for which the system of linear equations x+y+z=2x+y+z=2x+y+z=2, x+2y+3z=5x+2y+3z=5x+2y+3z=5, x+3y+λz=μx+3y+\lambda z=\mux+3y+λz=μ has infinitely many solutions are, respectively:
  1. (A)6 and 8
  2. (B)5 and 7
  3. (C)5 and 8
  4. (D)4 and 9

Correct answer: (C)

Step-by-step solution →
Q278·MathematicsSingle correctJEE Main 2020
Let θ=π5\theta = \frac{\pi}{5}θ=5π​ and A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}A=[cosθ−sinθ​sinθcosθ​]. If B=A+A4B = A + A^{4}B=A+A4, then det (B):
  1. (A)is one.
  2. (B)lies in (2, 3).
  3. (C)is zero.
  4. (D)lies in (1, 2).

Correct answer: (D)

Step-by-step solution →
Q279·MathematicsSingle correctJEE Main 2020
Let m and M be respectively the minimum and maximum values of ∣cos⁡2x1+sin⁡2xsin⁡2x1+cos⁡2xsin⁡2xsin⁡2xcos⁡2xsin⁡2x1+sin⁡2x∣\begin{vmatrix}\cos^{2}x & 1+\sin^{2}x & \sin 2x\\ 1+\cos^{2}x & \sin^{2}x & \sin 2x\\ \cos^{2}x & \sin^{2}x & 1+\sin 2x\end{vmatrix}​cos2x1+cos2xcos2x​1+sin2xsin2xsin2x​sin2xsin2x1+sin2x​​. Then the ordered pair (m, M) is equal to:
  1. (A)(−3,3)(-3,3)(−3,3)
  2. (B)(−3,−1)(-3,-1)(−3,−1)
  3. (C)(−4,−1)(-4,-1)(−4,−1)
  4. (D)(1,3)(1,3)(1,3)

Correct answer: (B)

Step-by-step solution →
Q280·MathematicsNumericalJEE Main 2020
The sum of distinct values of λ\lambdaλ for which the system of equations (λ−1)x+(3λ+1)y+2λz=0(\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0(λ−1)x+(3λ+1)y+2λz=0 (λ−1)x+(4λ−2)y+(λ+3)z=0(\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0(λ−1)x+(4λ−2)y+(λ+3)z=0 2x+(3λ+1)y+3(λ−1)z=0,2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0,2x+(3λ+1)y+3(λ−1)z=0, has non-zero solutions, is __________.

Correct answer: 3

Step-by-step solution →
Q281·MathematicsSingle correctJEE Main 2020
If the system of linear equations x+y+3z=0x+y+3z=0x+y+3z=0 x+3y+k2z=0x+3y+k^2z=0x+3y+k2z=0 3x+y+3z=03x+y+3z=03x+y+3z=0 has a non-zero solution (x,y,z)(x,y,z)(x,y,z) for some k∈Rk\in Rk∈R, then x+(yz)x+\left(\dfrac{y}{z}\right)x+(zy​) is equal to:
  1. (A)333
  2. (B)999
  3. (C)−3-3−3
  4. (D)−9-9−9

Correct answer: (C)

Step-by-step solution →
Q282·MathematicsSingle correctJEE Main 2020
Let λ∈R\lambda \in Rλ∈R. The system of linear equations. 2x1−4x2+λx3=12x_1 - 4x_2 + \lambda x_3 = 12x1​−4x2​+λx3​=1 x1−6x2+x3=2x_1 - 6x_2 + x_3 = 2x1​−6x2​+x3​=2 λx1−10x2+4x3=3\lambda x_1 - 10x_2 + 4x_3 = 3λx1​−10x2​+4x3​=3 Is inconsistent for:
  1. (A)exactly one positive value of λ\lambdaλ
  2. (B)exactly one negative value of λ\lambdaλ
  3. (C)every value of λ\lambdaλ
  4. (D)exactly two values of λ\lambdaλ

Correct answer: (B)

Step-by-step solution →
Q283·MathematicsSingle correctJEE Main 2020
If a + x = b + y = c + z + 1, where a, b, c, x, y, x are non-zero distinct real numbers then ∣xa+yx+ayb+yy+bzc+yz+c∣\begin{vmatrix} x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c \end{vmatrix}​xyz​a+yb+yc+y​x+ay+bz+c​​ is equal to:
  1. (A)y(a - b)
  2. (B)y(b - a)
  3. (C)0
  4. (D)y(a - c)

Correct answer: (A)

Step-by-step solution →
Q284·MathematicsSingle correctJEE Main 2020
If the system of equations x + y + z = 2 2x + 4y − z = 6 3x+2y+λz=μ3x + 2y + \lambda z = \mu3x+2y+λz=μ has infinitely many solutions, then:
  1. (A)λ+2μ=14\lambda + 2\mu = 14λ+2μ=14
  2. (B)2λ+μ=142\lambda + \mu = 142λ+μ=14
  3. (C)2λ−μ=52\lambda - \mu = 52λ−μ=5
  4. (D)λ−2μ=−5\lambda - 2\mu = -5λ−2μ=−5

Correct answer: (B)

Step-by-step solution →
Q285·MathematicsSingle correctJEE Main 2020
If A=[cos⁡θisin⁡θisin⁡θcos⁡θ]A = \begin{bmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{bmatrix}A=[cosθisinθ​isinθcosθ​], (θ=π24)\left( \theta = \frac{\pi}{24} \right)(θ=24π​) and A5=[abcd]A^{5} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A5=[ac​bd​], where i=−1i = \sqrt{-1}i=−1​, then, which one of the following is not true?
  1. (A)a2−c2=1a^{2} - c^{2} = 1a2−c2=1
  2. (B)a2−b2=12a^{2} - b^{2} = \frac{1}{2}a2−b2=21​
  3. (C)0≤a2+b2≤10 \le a^{2} + b^{2} \le 10≤a2+b2≤1
  4. (D)a2−d2=0a^{2} - d^{2} = 0a2−d2=0

Correct answer: (B)

Step-by-step solution →
Q286·MathematicsNumericalJEE Main 2020
If the system of equations x−2y+3z=9x - 2y + 3z = 9x−2y+3z=9 2x+y+z=b2x + y + z = b2x+y+z=b x−7y+az=24x - 7y + az = 24x−7y+az=24, has infinitely many solutions, then a−ba - ba−b is equal to __________.

Correct answer: 5

Step-by-step solution →
Q287·MathematicsNumericalJEE Main 2020
Let A=[x110]A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}A=[x1​10​], x∈Rx \in Rx∈R and A4=[aij]A^{4} = [a_{ij}]A4=[aij​]. If a11=109a_{11} = 109a11​=109, then a22a_{22}a22​ is equal to __________.

Correct answer: 10

Step-by-step solution →
Q288·MathematicsSingle correctJEE Main 2020
If Δ=∣x−22x−33x−42x−33x−44x−53x−55x−810x−17∣=Ax3+Bx2+Cx+D\Delta=\begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ 2x-3 & 3x-4 & 4x-5 \\ 3x-5 & 5x-8 & 10x-17 \end{vmatrix}=Ax^{3}+Bx^{2}+Cx+DΔ=​x−22x−33x−5​2x−33x−45x−8​3x−44x−510x−17​​=Ax3+Bx2+Cx+D, then B + C is equal to:
  1. (A)−3
  2. (B)9
  3. (C)−1
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q289·MathematicsNumericalJEE Main 2020
Let S be the set of all integer solutions, (x, y, z), of the system of equation x−2y+5z=0x - 2y + 5z = 0x−2y+5z=0 −2x+4y+z=0-2x + 4y + z = 0−2x+4y+z=0 −7x+14y+9z=0-7x + 14y + 9z = 0−7x+14y+9z=0 such that 15≤x2+y2+z2≤15015 \le x^2 + y^2 + z^2 \le 15015≤x2+y2+z2≤150. Then, the number of elements in the set S is equal to _______.

Correct answer: 08.00

Step-by-step solution →
Q290·MathematicsSingle correctJEE Main 2020
Let A be a 3×33 \times 33×3 matrix such that adj A=[2−11−1021−2−1]\mathrm{adj}\,A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{bmatrix}adjA=​2−11​−10−2​12−1​​ and B=adj(adj A)B = \mathrm{adj}\left( \mathrm{adj}\,A \right)B=adj(adjA). If ∣A∣=λ\left| A \right| = \lambda∣A∣=λ and ∣(B−1)T∣=μ\left| \left( B^{-1} \right)^{T} \right| = \mu​(B−1)T​=μ, then the ordered pair, (∣λ∣.μ)\left( \left| \lambda \right| . \mu \right)(∣λ∣.μ) is equal to:
  1. (A)(9,19)\left( 9, \frac{1}{9} \right)(9,91​)
  2. (B)(3,81)(3, 81)(3,81)
  3. (C)(9,181)\left( 9, \frac{1}{81} \right)(9,811​)
  4. (D)(3,181)\left( 3, \frac{1}{81} \right)(3,811​)

Correct answer: (D)

Step-by-step solution →
Q291·MathematicsSingle correctJEE Main 2020
If the matrices A=[1121341−13]A=\begin{bmatrix}1 & 1 & 2\\1 & 3 & 4\\1 & -1 & 3\end{bmatrix}A=​111​13−1​243​​, B=adj AB=\text{adj}\,AB=adjA and C=3AC=3AC=3A, then ∣adj B∣∣C∣\dfrac{|\text{adj}\,B|}{|C|}∣C∣∣adjB∣​ is equal to:
  1. (A)72
  2. (B)8
  3. (C)16
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q292·MathematicsSingle correctJEE Main 2020
If for some α\alphaα and β\betaβ in R, the intersection of the following three planes x+4y−2z=1x+4y-2z=1x+4y−2z=1 x+7y−5z=βx+7y-5z=\betax+7y−5z=β x+5y+αz=5x+5y+\alpha z=5x+5y+αz=5 is a line in R3R^{3}R3, then α+β\alpha+\betaα+β is equal to:
  1. (A)000
  2. (B)222
  3. (C)101010
  4. (D)−10-10−10

Correct answer: (C)

Step-by-step solution →
Q293·MathematicsSingle correctJEE Main 2020
Let a−2b+c=1a - 2b + c = 1a−2b+c=1. If f(x)=∣x+ax+2x+1x+bx+3x+2x+cx+4x+3∣f(x) = \begin{vmatrix} x + a & x + 2 & x + 1 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix}f(x)=​x+ax+bx+c​x+2x+3x+4​x+1x+2x+3​​, then:
  1. (A)f(−50)=501f(-50) = 501f(−50)=501
  2. (B)f(50)=1f(50) = 1f(50)=1
  3. (C)f(50)=−501f(50) = -501f(50)=−501
  4. (D)f(−50)=−1f(-50) = -1f(−50)=−1

Correct answer: (B)

Step-by-step solution →
Q294·MathematicsNumericalJEE Main 2020
The number of all 3×33\times33×3 matrices, A, with enteries from the set {−1,0,1}\{-1,0,1\}{−1,0,1} such that the sum of the diagonal elements of AATAA^{T}AAT is 3, is

Correct answer: 672

Step-by-step solution →
Q295·MathematicsSingle correctJEE Main 2020
The system of linear equations λx+2y+2z=5\lambda x+2y+2z=5λx+2y+2z=5, 2λx+3y+5z=82\lambda x+3y+5z=82λx+3y+5z=8, 4x+λy+6z=104x+\lambda y+6z=104x+λy+6z=10 has:
  1. (A)no solution when λ=8\lambda=8λ=8
  2. (B)infinitely many solutions when λ=2\lambda=2λ=2
  3. (C)no solution when λ=2\lambda=2λ=2
  4. (D)a unique solution when λ=−8\lambda=-8λ=−8

Correct answer: (C)

Step-by-step solution →
Q296·MathematicsSingle correctJEE Main 2020
If A=(2294)A=\begin{pmatrix}2 & 2\\9 & 4\end{pmatrix}A=(29​24​) and I=(1001)I=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}I=(10​01​), then 10A−110A^{-1}10A−1 is equal to:
  1. (A)4I−A4I-A4I−A
  2. (B)A−6IA-6IA−6I
  3. (C)A−4IA-4IA−4I
  4. (D)6I−A6I-A6I−A

Correct answer: (B)

Step-by-step solution →
Q297·MathematicsSingle correctJEE Main 2020
For which of the following ordered pairs (μ,δ)(\mu,\delta)(μ,δ), the system of linear equations x+2y+3z=1x+2y+3z=1x+2y+3z=1 3x+4y+5z=μ3x+4y+5z=\mu3x+4y+5z=μ is inconsistent? 4x+4y+4z=δ4x+4y+4z=\delta4x+4y+4z=δ
  1. (A)(3, 4)
  2. (B)(1, 0)
  3. (C)(4, 3)
  4. (D)(4, 6)

Correct answer: (C)

Step-by-step solution →
Q298·MathematicsSingle correctJEE Main 2020
Let α\alphaα be a root of the equation x2+x+1=0x^{2}+x+1=0x2+x+1=0 and the matrix A=13[1111αα21α2α4]A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4} \end{bmatrix}A=3​1​​111​1αα2​1α2α4​​, then the matrix A31A^{31}A31 is equal to:
  1. (A)A3A^{3}A3
  2. (B)A
  3. (C)I3I_{3}I3​
  4. (D)A2A^{2}A2

Correct answer: (A)

Step-by-step solution →
Q299·MathematicsNumericalJEE Main 2020
If the system of linear equations, x+y+z=6x+y+z=6x+y+z=6, x+2y+3z=10x+2y+3z=10x+2y+3z=10, 3x+2y+λz=μ3x+2y+\lambda z=\mu3x+2y+λz=μ has more than two solutions, then μ−λ2\mu-\lambda^{2}μ−λ2 is equal to __________.

Correct answer: 13

Step-by-step solution →
Q300·MathematicsSingle correctJEE Main 2020
Let A=[aij]A=[a_{ij}]A=[aij​] and B=[bij]B=[b_{ij}]B=[bij​] be two 3×33\times 33×3 real matrices such that bij=(3)(i+j−2)ajib_{ij}=(3)^{(i+j-2)}a_{ji}bij​=(3)(i+j−2)aji​, where i,j=1,2,3i,j=1,2,3i,j=1,2,3. If the determinant of B is 81, then the determinant of A is:
  1. (A)13\dfrac{1}{3}31​
  2. (B)19\dfrac{1}{9}91​
  3. (C)181\dfrac{1}{81}811​
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q301·MathematicsSingle correctJEE Main 2020
If the system of linear equations 2x+2ay+az=02x+2ay+az=02x+2ay+az=0, 2x+3by+bz=0,2x+4cy+cz=02x+3by+bz=0,2x+4cy+cz=02x+3by+bz=0,2x+4cy+cz=0 where a,b,c ∈R\in R∈R are non-zero and distinct; has a non-zero solution, then:
  1. (A)a+b+c=0a+b+c=0a+b+c=0
  2. (B)a, b, c are in A.P.
  3. (C)1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c}a1​,b1​,c1​ are in A.P.
  4. (D)a, b, c are in G.P.

Correct answer: (C)

Step-by-step solution →
Q302·MathematicsMultiple correctJEE Advanced 2019
Let M=[01a1233b1]M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}M=​013​12b​a31​​ and adj M=[−11−18−62−53−1]M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}M=​−18−5​1−63​−12−1​​ where a and b area real numbers. Which of the following options is/are correct?
  1. (A)(adj M)−1+adj M−1=−M(\mathrm{adj}\ M)^{-1} + \mathrm{adj}\ M^{-1} = -M(adj M)−1+adj M−1=−M
  2. (B)If M[αβγ]=[123]M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}M​αβγ​​=​123​​, then α−β+γ=3\alpha - \beta + \gamma = 3α−β+γ=3
  3. (C)det⁡(adj M2)=81\det(\mathrm{adj}\ M^2) = 81det(adj M2)=81
  4. (D)a+b=3a + b = 3a+b=3

Correct answer: (A), (B), (D)

Step-by-step solution →
Q303·MathematicsSingle correctJEE Advanced 2019
Let M=[sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ]=αI+βM−1M = \begin{bmatrix} \sin^4\theta & -1-\sin^2\theta \\ 1+\cos^2\theta & \cos^4\theta \end{bmatrix} = \alpha I + \beta M^{-1}M=[sin4θ1+cos2θ​−1−sin2θcos4θ​]=αI+βM−1, where α=α(θ)\alpha = \alpha(\theta)α=α(θ) and β=β(θ)\beta = \beta(\theta)β=β(θ) are real numbers, and I is the 2×22 \times 22×2 identity matrix. If α∗\alpha^*α∗ is the minimum of the set {α(θ):θ∈[0,2π)}\{\alpha(\theta) : \theta \in [0, 2\pi)\}{α(θ):θ∈[0,2π)} and β∗\beta^*β∗ is the minimum of the set {β(θ):θ∈[0,2π)}\{\beta(\theta) : \theta \in [0, 2\pi)\}{β(θ):θ∈[0,2π)}, then the value of α∗+β∗\alpha^* + \beta^*α∗+β∗ is
  1. (A)−3716-\frac{37}{16}−1637​
  2. (B)−3116-\frac{31}{16}−1631​
  3. (C)−1716-\frac{17}{16}−1617​
  4. (D)−2916-\frac{29}{16}−1629​

Correct answer: (D)

Step-by-step solution →
Q304·MathematicsMultiple correctJEE Advanced 2019
Let x∈Rx \in \mathbb{R}x∈R and let P=[111022003]P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}P=​100​120​123​​, Q=[2xx040xx6]Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix}Q=​20x​x4x​x06​​ and R=PQP−1R = PQP^{-1}R=PQP−1 Then which of the following options is/are correct?
  1. (A)For x=0x = 0x=0, if R[1ab]=6[1ab]R\begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6\begin{bmatrix} 1 \\ a \\ b \end{bmatrix}R​1ab​​=6​1ab​​, then a+b=5a + b = 5a+b=5
  2. (B)For x=1x = 1x=1, there exists a unit vector αi^+βj^+γk^\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}αi^+βj^​+γk^ for which R[αβγ]=[000]R\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}R​αβγ​​=​000​​
  3. (C)det⁡R=det⁡[2xx040xx5]+8\det R = \det\begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8detR=det​20x​x4x​x05​​+8, for all x∈Rx \in \mathbb{R}x∈R
  4. (D)There exists a real number xxx such that PQ=QPPQ = QPPQ=QP

Correct answer: (A), (C)

Step-by-step solution →
Q305·MathematicsMultiple correctJEE Advanced 2019
Let P1=I=[100010001]P_1 = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}P1​=I=​100​010​001​​, P2=[100001010]P_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}P2​=​100​001​010​​ P3=[010100001]P_3 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}P3​=​010​100​001​​ P4=[010001100]P_4 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}P4​=​001​100​010​​, P5=[001100010]P_5 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}P5​=​010​001​100​​, P6=[001010100]P_6 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}P6​=​001​010​100​​ and X=∑k=16Pk[213102321]PkTX = \displaystyle\sum_{k=1}^{6} P_k \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} P_k^{T}X=k=1∑6​Pk​​213​102​321​​PkT​ where PkTP_k^{T}PkT​ denotes the transpose of the matrix PkP_kPk​. Then which of the following options is/are correct?
  1. (A)X−30IX - 30IX−30I is an invertible matrix
  2. (B)X is a symmetric matrix
  3. (C)The sum of diagonal entries of X is 18
  4. (D)If X[111]=α[111]X\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \alpha\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}X​111​​=α​111​​, then α=30\alpha = 30α=30

Correct answer: (B), (C), (D)

Step-by-step solution →
Q306·MathematicsSingle correctJEE Main 2019
A value of θ∈(0,π/3)\theta \in (0, \pi/3)θ∈(0,π/3), for which ∣1+cos⁡2θsin⁡2θ4cos⁡6θcos⁡2θ1+sin⁡2θ4cos⁡6θcos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\begin{vmatrix} 1+\cos^{2}\theta & \sin^{2}\theta & 4\cos 6\theta \\ \cos^{2}\theta & 1+\sin^{2}\theta & 4\cos 6\theta \\ \cos^{2}\theta & \sin^{2}\theta & 1+4\cos 6\theta \end{vmatrix} = 0​1+cos2θcos2θcos2θ​sin2θ1+sin2θsin2θ​4cos6θ4cos6θ1+4cos6θ​​=0, is:
  1. (A)π18\dfrac{\pi}{18}18π​
  2. (B)π9\dfrac{\pi}{9}9π​
  3. (C)7π36\dfrac{7\pi}{36}367π​
  4. (D)7π24\dfrac{7\pi}{24}247π​

Correct answer: (B)

Step-by-step solution →
Q307·MathematicsSingle correctJEE Main 2019
If A is a symmetric matrix and B is a skew-symmetrix matrix such that A + B = [235−1]\begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}[25​3−1​], then AB is equal to :
  1. (A)[4−21−4]\begin{bmatrix} 4 & -2 \\ 1 & -4 \end{bmatrix}[41​−2−4​]
  2. (B)[4−2−1−4]\begin{bmatrix} 4 & -2 \\ -1 & -4 \end{bmatrix}[4−1​−2−4​]
  3. (C)[−4214]\begin{bmatrix} -4 & 2 \\ 1 & 4 \end{bmatrix}[−41​24​]
  4. (D)[−4−2−14]\begin{bmatrix} -4 & -2 \\ -1 & 4 \end{bmatrix}[−4−1​−24​]

Correct answer: (B)

Step-by-step solution →
Q308·MathematicsSingle correctJEE Main 2019
If B = [52α1021α3−1]\begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}​50α​2α23​11−1​​ is the inverse of a 3×33 \times 33×3 matrix A, then the sum of all values of α\alphaα for which det(A) + 1 = 0, is :
  1. (A)0
  2. (B)−1-1−1
  3. (C)1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q309·MathematicsSingle correctJEE Main 2019
If [x] denotes the greatest integer ≤\leq≤ x, then the system of linear equations [sin θ\thetaθ] x + [−cos⁡θ-\cos\theta−cosθ] y = 0 [cot θ\thetaθ] x + y = 0
  1. (A)have infinitely many solutions if θ∈(π2,2π3)\theta \in \left(\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right)θ∈(2π​,32π​) and has a unique solution if θ∈(π,7π6)\theta \in \left(\pi, \dfrac{7\pi}{6}\right)θ∈(π,67π​)
  2. (B)have infinitely many solutions if θ∈(π2,2π3)∪(π,7π6)\theta \in \left(\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right) \cup \left(\pi, \dfrac{7\pi}{6}\right)θ∈(2π​,32π​)∪(π,67π​)
  3. (C)has a unique solution if θ∈(π2,2π3)\theta \in \left(\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right)θ∈(2π​,32π​) and have infinitely many solutions if θ∈(π,7π6)\theta \in \left(\pi, \dfrac{7\pi}{6}\right)θ∈(π,67π​)
  4. (D)has a unique solution if θ∈(π2,2π3)∪(π,7π6)\theta \in \left(\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right) \cup \left(\pi, \dfrac{7\pi}{6}\right)θ∈(2π​,32π​)∪(π,67π​)

Correct answer: (A)

Step-by-step solution →
Q310·MathematicsSingle correctJEE Main 2019
Let λ\lambdaλ be a real number for which the system of linear equations x+y+z=6x + y + z = 6x+y+z=6 4x+λy−λz=λ−24x + \lambda y - \lambda z = \lambda - 24x+λy−λz=λ−2 3x+2y−4z=−53x + 2y - 4z = -53x+2y−4z=−5 Has indefinitely many solutions. Then λ\lambdaλ is a root of the quadratic equation
  1. (A)λ2−λ−6=0\lambda^{2} - \lambda - 6 = 0λ2−λ−6=0
  2. (B)λ2−3λ−4=0\lambda^{2} - 3\lambda - 4 = 0λ2−3λ−4=0
  3. (C)λ2+3λ−4=0\lambda^{2} + 3\lambda - 4 = 0λ2+3λ−4=0
  4. (D)λ2+λ−6=0\lambda^{2} + \lambda - 6 = 0λ2+λ−6=0

Correct answer: (A)

Step-by-step solution →
Q311·MathematicsSingle correctJEE Main 2019
The sum of the real roots of the equation ∣x−6−12−3xx−3−32xx=2∣=0\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x - 3 \\ -3 & 2x & x = 2 \end{vmatrix} = 0​x2−3​−6−3x2x​−1x−3x=2​​=0 is equal to
  1. (A)-4
  2. (B)0
  3. (C)6
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q312·MathematicsSingle correctJEE Main 2019
If the system of equations 2x+3y−z=02x + 3y - z = 02x+3y−z=0, x+ky−2z=0x + ky - 2z = 0x+ky−2z=0 and 2x−y+z=02x - y + z = 02x−y+z=0 has a non-trivial solution (x,y,z)(x, y, z)(x,y,z), then xy+yz+zx+k\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} + kyx​+zy​+xz​+k is equal to
  1. (A)34\dfrac{3}{4}43​
  2. (B)−4-4−4
  3. (C)12\dfrac{1}{2}21​
  4. (D)−14-\dfrac{1}{4}−41​

Correct answer: (C)

Step-by-step solution →
Q313·MathematicsSingle correctJEE Main 2019
Let α\alphaα and β\betaβ be the roots of the equation x2+x+1=0x^{2}+x+1=0x2+x+1=0. Then for y≠0y\neq 0y=0 in R, ∣y+1αβαy+β1β1y+α∣\begin{vmatrix}y+1 & \alpha & \beta\\ \alpha & y+\beta & 1\\ \beta & 1 & y+\alpha\end{vmatrix}​y+1αβ​αy+β1​β1y+α​​ is equal to:
  1. (A)y(y2−3)y(y^{2}-3)y(y2−3)
  2. (B)y3−1y^{3}-1y3−1
  3. (C)y3y^{3}y3
  4. (D)y(y2−1)y(y^{2}-1)y(y2−1)

Correct answer: (C)

Step-by-step solution →
Q314·MathematicsSingle correctJEE Main 2019
If [1101][1201][1301]…………[1n−101]=[17801]\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1&3\\0&1\end{bmatrix}\ldots\ldots\ldots\ldots\begin{bmatrix}1&n-1\\0&1\end{bmatrix}=\begin{bmatrix}1&78\\0&1\end{bmatrix}[10​11​][10​21​][10​31​]…………[10​n−11​]=[10​781​], then the inverse of [1n01]\begin{bmatrix}1&n\\0&1\end{bmatrix}[10​n1​] is:
  1. (A)[1−1201]\begin{bmatrix}1&-12\\0&1\end{bmatrix}[10​−121​]
  2. (B)[10131]\begin{bmatrix}1&0\\13&1\end{bmatrix}[113​01​]
  3. (C)[10121]\begin{bmatrix}1&0\\12&1\end{bmatrix}[112​01​]
  4. (D)[1−1301]\begin{bmatrix}1&-13\\0&1\end{bmatrix}[10​−131​]

Correct answer: (D)

Step-by-step solution →
Q315·MathematicsSingle correctJEE Main 2019
Let A=(cos⁡α−sin⁡αsin⁡αcos⁡α)A=\begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}A=(cosαsinα​−sinαcosα​), (α∈R)(\alpha \in R)(α∈R) such that A32=(0−110)A^{32}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}A32=(01​−10​). Then a value of α\alphaα is:
  1. (A)000
  2. (B)π16\frac{\pi}{16}16π​
  3. (C)π32\frac{\pi}{32}32π​
  4. (D)π64\frac{\pi}{64}64π​

Correct answer: (D)

Step-by-step solution →
Q316·MathematicsSingle correctJEE Main 2019
If the system of linear equations x−2y+kz=1x-2y+kz=1x−2y+kz=1, 2x+y+z=22x+y+z=22x+y+z=2, 3x−y−kz=33x-y-kz=33x−y−kz=3 has a solution (x,y,z)≠0(x,y,z)\ne 0(x,y,z)=0, then (x,y)(x,y)(x,y) lies on the straight line whose equation is:
  1. (A)3x−4y−1=03x-4y-1=03x−4y−1=0
  2. (B)4x−3y−4=04x-3y-4=04x−3y−4=0
  3. (C)4x−3y−1=04x-3y-1=04x−3y−1=0
  4. (D)3x−4y−4=03x-4y-4=03x−4y−4=0

Correct answer: (B)

Step-by-step solution →
Q317·MathematicsSingle correctJEE Main 2019
The greatest value of c∈Rc\in Rc∈R for which the system of linear equations x−cy−cz=0x-cy-cz=0x−cy−cz=0 cx−y+cz=0cx-y+cz=0cx−y+cz=0 cx+cy−z=0cx+cy-z=0cx+cy−z=0 has a non-trivial solution, is:
  1. (A)-1
  2. (B)2
  3. (C)12\dfrac{1}{2}21​
  4. (D)0

Correct answer: (C)

Step-by-step solution →
Q318·MathematicsSingle correctJEE Main 2019
Let P=[100310931]P = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}P=​139​013​001​​ and Q=[qij]Q = [q_{ij}]Q=[qij​] be two 3×33 \times 33×3 matrices such that Q−P5=I3Q - P^{5} = I_{3}Q−P5=I3​. Then q21+q31q32\dfrac{q_{21} + q_{31}}{q_{32}}q32​q21​+q31​​ is equal to:
  1. (A)10
  2. (B)135
  3. (C)15
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q319·MathematicsSingle correctJEE Main 2019
If A=[1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1]A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}A=​1−sinθ−1​sinθ1−sinθ​1sinθ1​​; then for all θ∈(3π4,5π4)\theta \in \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right)θ∈(43π​,45π​), det (A) lies in the interval :
  1. (A)(1,52]\left(1, \frac{5}{2}\right](1,25​]
  2. (B)[52,4)\left[\frac{5}{2}, 4\right)[25​,4)
  3. (C)(0,32]\left(0, \frac{3}{2}\right](0,23​]
  4. (D)(32,3]\left(\frac{3}{2}, 3\right](23​,3]

Correct answer: (D)

Step-by-step solution →
Q320·MathematicsSingle correctJEE Main 2019
An ordered pair (α,β)(\alpha, \beta)(α,β) for which the system of linear equations (1+α)x+βy+z=2(1 + \alpha)x + \beta y + z = 2(1+α)x+βy+z=2 αx+(1+β)y+z=3\alpha x + (1 + \beta)y + z = 3αx+(1+β)y+z=3 αx+βy+2z=2\alpha x + \beta y + 2z = 2αx+βy+2z=2 has a unique solution, is:
  1. (A)(2,4)(2, 4)(2,4)
  2. (B)(−3,1)(-3, 1)(−3,1)
  3. (C)(−4,2)(-4, 2)(−4,2)
  4. (D)(1,−3)(1, -3)(1,−3)

Correct answer: (A)

Step-by-step solution →
Q321·MathematicsSingle correctJEE Main 2019
Let A and B be two invertible matrices of order 3×33\times 33×3. If det⁡(ABAT)=8\det(ABA^{T})=8det(ABAT)=8 and det⁡(AB−1)=8\det(AB^{-1})=8det(AB−1)=8, then det⁡(BA−1BT)\det(BA^{-1}B^{T})det(BA−1BT) is equal to:
  1. (A)14\frac{1}{4}41​
  2. (B)1
  3. (C)116\frac{1}{16}161​
  4. (D)16

Correct answer: (C)

Step-by-step solution →
Q322·MathematicsSingle correctJEE Main 2019
If the system of linear equations 2x+2y+3z=a2x + 2y + 3z = a2x+2y+3z=a 3x−y+5z=b3x - y + 5z = b3x−y+5z=b x−3y+2z=cx - 3y + 2z = cx−3y+2z=c Where a, b, c are non zero real numbers, has more than one solution, then:
  1. (A)b−c+a=0b-c+a=0b−c+a=0
  2. (B)b−c−a=0b-c-a=0b−c−a=0
  3. (C)a+b+c=0a+b+c=0a+b+c=0
  4. (D)b+c−a=0b+c-a=0b+c−a=0

Correct answer: (B)

Step-by-step solution →
Q323·MathematicsSingle correctJEE Main 2019
Let A=(02qrpq−rp−qr)A=\begin{pmatrix}0 & 2q & r\\ p & q & -r\\ p & -q & r\end{pmatrix}A=​0pp​2qq−q​r−rr​​. If AAT=I3AA^{T}=I_{3}AAT=I3​, ∣P∣|P|∣P∣ then ∣p∣|p|∣p∣ is:
  1. (A)15\dfrac{1}{\sqrt{5}}5​1​
  2. (B)13\dfrac{1}{\sqrt{3}}3​1​
  3. (C)12\dfrac{1}{\sqrt{2}}2​1​
  4. (D)16\dfrac{1}{\sqrt{6}}6​1​

Correct answer: (C)

Step-by-step solution →
Q324·MathematicsSingle correctJEE Main 2019
If the system of equations x+y+z=5x+y+z=5x+y+z=5, x+2y+3z=9x+2y+3z=9x+2y+3z=9, x+3y+αz=βx+3y+\alpha z=\betax+3y+αz=β has infinitely many solutions, then β−α\beta-\alphaβ−α equals:
  1. (A)21
  2. (B)8
  3. (C)18
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q325·MathematicsSingle correctJEE Main 2019
If the system of linear equation x−4y+7z=g,3y−5z=h,−2x+5y−9z=kx-4y+7z=g, 3y-5z=h, -2x+5y-9z=kx−4y+7z=g,3y−5z=h,−2x+5y−9z=k is consistent, then:
  1. (A)g+h+k=0g+h+k=0g+h+k=0
  2. (B)2g+h+k=02g+h+k=02g+h+k=0
  3. (C)g+h+2k=0g+h+2k=0g+h+2k=0
  4. (D)g+2h+k=0g+2h+k=0g+2h+k=0

Correct answer: (B)

Step-by-step solution →
Q326·MathematicsSingle correctJEE Main 2019
If A=[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcos⁡tet2e−tsin⁡t−2e−tcos⁡t]A = \begin{bmatrix} e^{t} & e^{-t}\cos t & e^{-t}\sin t \\ e^{t} & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^{t} & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}A=​etetet​e−tcost−e−tcost−e−tsint2e−tsint​e−tsint−e−tsint+e−tcost−2e−tcost​​ Then A is
  1. (A)Invertible only if t=π2t = \frac{\pi}{2}t=2π​
  2. (B)not invertible for any t∈Rt \in Rt∈R
  3. (C)invertible for all t∈Rt \in Rt∈R
  4. (D)invertible only if t=πt = \pit=π

Correct answer: (C)

Step-by-step solution →
Q327·MathematicsSingle correctJEE Main 2019
If A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}A=[cosθsinθ​−sinθcosθ​], then the matrix A−50A^{-50}A−50 when θ=π12\theta=\dfrac{\pi}{12}θ=12π​, is equal to
  1. (A)[12−323212]\begin{bmatrix}\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{bmatrix}​21​23​​​−23​​21​​​
  2. (B)[32−121232]\begin{bmatrix}\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\\[4pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{bmatrix}​23​​21​​−21​23​​​​
  3. (C)[3212−1232]\begin{bmatrix}\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\\[4pt] -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{bmatrix}​23​​−21​​21​23​​​​
  4. (D)[1232−3212]\begin{bmatrix}\dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\\[4pt] -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{bmatrix}​21​−23​​​23​​21​​​

Correct answer: (C)

Step-by-step solution →
Q328·MathematicsSingle correctJEE Main 2019
The system of linear equation x+y+z=2x + y + z = 2x+y+z=2, 2x+3y+2z=52x + 3y + 2z = 52x+3y+2z=5, 2x+3y+(a2−1)z=a+12x + 3y + (a^2 - 1)z = a + 12x+3y+(a2−1)z=a+1 then
  1. (A)is inconsistent when a=4a = 4a=4
  2. (B)has a unique solution for ∣a∣=3|a| = \sqrt{3}∣a∣=3​
  3. (C)has infinitely many solutions for a=4a = 4a=4
  4. (D)inconsistent when ∣a∣=3|a| = \sqrt{3}∣a∣=3​

Correct answer: (D)

Step-by-step solution →
Q329·MathematicsNumericalJEE Advanced 2018
Let P be a matrix of order 3×33 \times 33×3 such that all the entries in P are from the set {−1,0,1}\{-1, 0, 1\}{−1,0,1}. Then, the maximum possible value of the determinant of P is ______ .

Correct answer: 4

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Q330·MathematicsMultiple correctJEE Advanced 2018
Let S be the set of all column matrices [b1b2b3]\begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \end{bmatrix}​b1​b2​b3​​​ such that b1b_{1}b1​, b2b_{2}b2​, b3∈Rb_{3} \in Rb3​∈R and the system of equations (in real variables) −x+2y+5z=b1-x + 2y + 5z = b_{1}−x+2y+5z=b1​ 2x−4y+3z=b22x - 4y + 3z = b_{2}2x−4y+3z=b2​ x−2y+2z=b3x - 2y + 2z = b_{3}x−2y+2z=b3​ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each [b1b2b3]∈S\begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \end{bmatrix} \in S​b1​b2​b3​​​∈S ?
  1. (A)x+2y+3z=b1x + 2y + 3z = b_{1}x+2y+3z=b1​, 4y+5z=b24y + 5z = b_{2}4y+5z=b2​ and x+2y+6z=b3x + 2y + 6z = b_{3}x+2y+6z=b3​
  2. (B)x+y+3z=b1x + y + 3z = b_{1}x+y+3z=b1​, 5x+2y+6z=b25x + 2y + 6z = b_{2}5x+2y+6z=b2​ and −2x−y−3z=b3-2x - y - 3z = b_{3}−2x−y−3z=b3​
  3. (C)−x+2y−5z=b1-x + 2y - 5z = b_{1}−x+2y−5z=b1​, 2x−4y+10z=b22x - 4y + 10z = b_{2}2x−4y+10z=b2​ and x−2y+5z=b3x - 2y + 5z = b_{3}x−2y+5z=b3​
  4. (D)x+2y+5z=b1x + 2y + 5z = b_{1}x+2y+5z=b1​, 2x+3z=b22x + 3z = b_{2}2x+3z=b2​ and x+4y−5z=b3x + 4y - 5z = b_{3}x+4y−5z=b3​

Correct answer: (A), (D)

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Q331·MathematicsMultiple correctJEE Advanced 2017
Which of the following is(are) NOT the square of a 3×33 \times 33×3 matrix with real entries ?
  1. (A)[10001000−1]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}​100​010​00−1​​
  2. (B)[−1000−1000−1]\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}​−100​0−10​00−1​​
  3. (C)[100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}​100​010​001​​
  4. (D)[1000−1000−1]\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}​100​0−10​00−1​​

Correct answer: (A), (B)

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Q332·MathematicsMultiple correctJEE Advanced 2017
If f(x)=∣cos⁡(2x)cos⁡(2x)sin⁡(2x)−cos⁡xcos⁡x−sin⁡xsin⁡xsin⁡xcos⁡x∣f(x) = \begin{vmatrix} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix}f(x)=​cos(2x)−cosxsinx​cos(2x)cosxsinx​sin(2x)−sinxcosx​​, then
  1. (A)f′(x)=0f'(x) = 0f′(x)=0 at exactly three points in (−π,π)(-\pi, \pi)(−π,π)
  2. (B)f′(x)=0f'(x) = 0f′(x)=0 at more than three points in (−π,π)(-\pi, \pi)(−π,π)
  3. (C)f(x)f(x)f(x) attains its maximum at x=0x = 0x=0
  4. (D)f(x)f(x)f(x) attains its minimum at x=0x = 0x=0

Correct answer: (B), (C)

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Q333·MathematicsIntegerJEE Advanced 2017
For a real number α\alphaα, if the system [1αα2α1αα2α1][xyz]=[1−11]\begin{bmatrix} 1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}​1αα2​α1α​α2α1​​​xyz​​=​1−11​​ of linear equations, has infinitely many solutions, then 1+α+α2=1 + \alpha + \alpha^{2} =1+α+α2=

Correct answer: 1

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Q334·MathematicsSingle correctJEE Advanced 2016
Let P=[1004101641]P = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix}P=​1416​014​001​​ and I be the identity matrix of order 3. If Q=[qij]Q = \left[ q_{ij} \right]Q=[qij​] is a matrix such that P50−Q=IP^{50} - Q = IP50−Q=I, then q31+q32q21\frac{q_{31} + q_{32}}{q_{21}}q21​q31​+q32​​ equals
  1. (A)52
  2. (B)103
  3. (C)201
  4. (D)205

Correct answer: (B)

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Q335·MathematicsIntegerJEE Advanced 2016
The total number of distinct x∈Rx \in \mathbb{R}x∈R for which ∣xx21+x32x4x21+8x33x9x21+27x3∣=10\begin{vmatrix} x & x^{2} & 1 + x^{3} \\ 2x & 4x^{2} & 1 + 8x^{3} \\ 3x & 9x^{2} & 1 + 27x^{3} \end{vmatrix} = 10​x2x3x​x24x29x2​1+x31+8x31+27x3​​=10 is

Correct answer: 2

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Q336·MathematicsMultiple correctJEE Advanced 2016
Let P=[3−1−220α3−50]P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}P=​323​−10−5​−2α0​​, where α∈R\alpha \in \mathbb{R}α∈R. Suppose Q=[qij]Q = [q_{ij}]Q=[qij​] is a matrix such that PQ=kIPQ = kIPQ=kI, where k∈Rk \in \mathbb{R}k∈R, k≠0k \neq 0k=0 and III is the identity matrix of order 3. If q23=−k8q_{23} = -\frac{k}{8}q23​=−8k​ and det⁡(Q)=k22\det(Q) = \frac{k^{2}}{2}det(Q)=2k2​, then
  1. (A)α=0,k=8\alpha = 0, k = 8α=0,k=8
  2. (B)4α−k+8=04\alpha - k + 8 = 04α−k+8=0
  3. (C)det⁡(P adj(Q))=29\det(P\,\mathrm{adj}(Q)) = 2^{9}det(Padj(Q))=29
  4. (D)det⁡(Q adj(P))=213\det(Q\,\mathrm{adj}(P)) = 2^{13}det(Qadj(P))=213

Correct answer: (B), (C)

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Q337·MathematicsMultiple correctJEE Advanced 2016
Let a,λ,μ∈Ra, \lambda, \mu \in \mathbb{R}a,λ,μ∈R. Consider the system of linear equations ax+2y=λax + 2y = \lambdaax+2y=λ 3x−2y=μ3x - 2y = \mu3x−2y=μ Which of the following statement(s) is(are) correct?
  1. (A)If a=−3a = -3a=−3, then the system has infinitely many solutions for all values of λ\lambdaλ and μ\muμ
  2. (B)If a≠−3a \neq -3a=−3, then the system has a unique solution for all values of λ\lambdaλ and μ\muμ
  3. (C)If λ+μ=0\lambda + \mu = 0λ+μ=0, then the system has infinitely many solutions for a=−3a = -3a=−3
  4. (D)If λ+μ≠0\lambda + \mu \neq 0λ+μ=0, then the system has no solution for a=−3a = -3a=−3

Correct answer: (B), (C), (D)

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Q338·MathematicsMultiple correctJEE Advanced 2015
Which of the following values of α\alphaα satisfy the equation ∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣=−648α\begin{vmatrix} (1+\alpha)^{2} & (1+2\alpha)^{2} & (1+3\alpha)^{2} \\ (2+\alpha)^{2} & (2+2\alpha)^{2} & (2+3\alpha)^{2} \\ (3+\alpha)^{2} & (3+2\alpha)^{2} & (3+3\alpha)^{2} \end{vmatrix} = -648\alpha​(1+α)2(2+α)2(3+α)2​(1+2α)2(2+2α)2(3+2α)2​(1+3α)2(2+3α)2(3+3α)2​​=−648α ?
  1. (A)−4-4−4
  2. (B)999
  3. (C)−9-9−9
  4. (D)444

Correct answer: (B), (C)

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Q339·MathematicsMultiple correctJEE Advanced 2014
Let MMM be a 2×22 \times 22×2 symmetric matrix with integer entries. Then MMM is invertible if
  1. (A)the first column of MMM is the transpose of the second row of MMM
  2. (B)the second row of MMM is the transpose of the first column of MMM
  3. (C)MMM is a diagonal matrix with non-zero entries in the main diagonal
  4. (D)the product of entries in the main diagonal of MMM is not the square of an integer

Correct answer: (C), (D)

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Q340·MathematicsMultiple correctJEE Advanced 2014
Let MMM and NNN be two 3×33 \times 33×3 matrices such that MN=NMMN = NMMN=NM. Further, if M≠N2M \neq N^2M=N2 and M2=N4M^2 = N^4M2=N4, then
  1. (A)determinant of (M2+MN2)(M^2 + MN^2)(M2+MN2) is 000
  2. (B)there is a 3×33 \times 33×3 non-zero matrix UUU such that (M2+MN2)U(M^2 + MN^2)U(M2+MN2)U is the zero matrix
  3. (C)determinant of (M2+MN2)≥1(M^2 + MN^2) \geq 1(M2+MN2)≥1
  4. (D)for a 3×33 \times 33×3 matrix UUU, if (M2+MN2)U(M^2 + MN^2)U(M2+MN2)U equals the zero matrix then UUU is the zero matrix

Correct answer: (A), (B)

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Q341·MathematicsMultiple correctJEE Advanced 2013
Let ω\omegaω be a complex cube root of unity with ω≠1\omega \neq 1ω=1 and P=[pij]P = [p_{ij}]P=[pij​] be a n×nn \times nn×n matrix with pij=ωi+jp_{ij} = \omega^{i+j}pij​=ωi+j. Then P2≠0P^{2} \neq 0P2=0, when n=n =n=
  1. (A)575757
  2. (B)555555
  3. (C)585858
  4. (D)565656

Correct answer: (B), (C), (D)

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Q342·MathematicsMultiple correctJEE Advanced 2013
For 3×33\times 33×3 matrices M and N, which of the following statement(s) is (are) NOT correct ?
  1. (A)NTMNN^TMNNTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric
  2. (B)MN−NMMN-NMMN−NM is skew symmetric for all symmetric matrices M and N
  3. (C)MNMNMN is symmetric for all symmetric matrices M and N
  4. (D)(adj M)(adj N)=adj(MN)(\text{adj }M)(\text{adj }N)=\text{adj}(MN)(adj M)(adj N)=adj(MN) for all invertible matrices M and N

Correct answer: (C), (D)

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Matrices and Determinants — frequently asked

How many questions from Matrices and Determinants appear in JEE?

Matrices and Determinants has appeared in 180 of the last 186 JEE Main and JEE Advanced papers — about 97% of them — contributing 342 questions in total across those papers.

Is Matrices and Determinants an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 97% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Matrices and Determinants questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

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