Matrices JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Matrices, free to read — no sign-in needed. The full chapter has 41 questions; sign in to attempt the remaining 36 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $\mathbb{R}$ denote the set of all real numbers and let $i = \sqrt{-1}$. Consider the matrices $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Let $a, b, c, d$ be real numbers such that $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Let $H = \{x + iy : x, y \in \mathbb{R} \text{ and } y \gt 0\}$. Then which of the following statements is (are) TRUE ?
   1. A.$\dfrac{b + ia}{d + ic} = i$
   2. B.If $\omega = \dfrac{-1 + i\sqrt{3}}{2}$, then $\dfrac{a\omega + b}{c\omega + d} = \omega$
   3. C.If $m$ is an integer greater than $2$ such that $(ST)^2 = (ST)^m$, then $m$ is an integer multiple of $8$
   4. D.If $z \in H$, then $\dfrac{az + b}{cz + d} \in H$
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   Answer: B,D

   We are given the matrices $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. First, we compute the product $ST$: $ST = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$ Comparing this with $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we get $a = 0$, $b = -1$, $c = 1$, and $d = 1$. Evaluating Option (A): $\dfrac{b + ia}{d + ic} = \dfrac{-1 + i(0)}{1 + i(1)} = \dfrac{-1}{1 + i} = \dfrac{-1(1 - i)}{(1 + i)(1 - i)} = \dfrac{-1 + i}{2}$ This is not equal to $i$, so statement (A) is FALSE. Evaluating Option (B): Given $\omega = \dfrac{-1 + i\sqrt{3}}{2}$, which is a complex cube root of unity, satisfying $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$. $\dfrac{a\omega + b}{c\omega + d} = \dfrac{0 \cdot \omega - 1}{1 \cdot \omega + 1} = \dfrac{-1}{\omega + 1}$ Since $\omega + 1 = -\omega^2$, we have: $\dfrac{-1}{-\omega^2} = \dfrac{1}{\omega^2} = \dfrac{\omega^3}{\omega^2} = \omega$ Thus, statement (B) is TRUE. Evaluating Option (C): The characteristic equation of $ST$ is $\lambda^2 - \text{Tr}(ST)\lambda + \det(ST) = 0$, which gives $\lambda^2 - \lambda + 1 = 0$. By the Cayley-Hamilton theorem, $(ST)^2 - ST + I = 0$. Multiplying by $(ST + I)$, we get $(ST)^3 + I = 0 \Rightarrow (ST)^3 = -I$. Squaring both sides gives $(ST)^6 = I$. We are given $(ST)^2 = (ST)^m$, which implies $(ST)^{m-2} = I$. This means $m - 2$ must be a multiple of $6$, so $m = 6k + 2$ for some integer $k$. For $k = 1$, $m = 8$ (which is a multiple of $8$). For $k = 2$, $m = 14$ (which is NOT a multiple of $8$). Thus, statement (C) is FALSE. Evaluating Option (D): For $z \in H$, let $z = x + iy$ where $y \gt 0$. $\dfrac{az + b}{cz + d} = \dfrac{-1}{z + 1} = \dfrac{-1}{(x + 1) + iy}$ Multiplying the numerator and denominator by the conjugate $(x + 1) - iy$: $\dfrac{-1}{(x + 1) + iy} \times \dfrac{(x + 1) - iy}{(x + 1) - iy} = \dfrac{-(x + 1) + iy}{(x + 1)^2 + y^2}$ The imaginary part of this new complex number is $\dfrac{y}{(x + 1)^2 + y^2}$. Since $y \gt 0$ and $(x + 1)^2 + y^2 \gt 0$, the imaginary part is strictly positive. Therefore, $\dfrac{az + b}{cz + d} \in H$, making statement (D) TRUE. Answer: If $\omega = \dfrac{-1 + i\sqrt{3}}{2}$, then $\dfrac{a\omega + b}{c\omega + d} = \omega$; If $z \in H$, then $\dfrac{az + b}{cz + d} \in H$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Which one of the following matrices can be obtained by performing elementary row transformations on the $3 \times 3$ identity matrix?
   1. A.$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
   2. B.$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}$
   3. C.$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{bmatrix}$
   4. D.$\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{bmatrix}$
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   Answer: (B)

   Any matrix obtained by performing elementary row transformations on the identity matrix is equivalent to the identity matrix, which means it must be non-singular (invertible). Therefore, its determinant must be non-zero. Evaluating the determinant of the matrix in option (A): $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0$ Evaluating the determinant of the matrix in option (B): $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 1(3 - 8) - 1(2 - 4) + 1(4 - 3) = -5 + 2 + 1 = -2 \neq 0$ Evaluating the determinant of the matrix in option (C): $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{vmatrix} = 1(24 - 20) - 1(16 - 8) + 1(10 - 6) = 4 - 8 + 4 = 0$ Evaluating the determinant of the matrix in option (D): $\begin{vmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{vmatrix} = 1(3 - 4) - 1(-3 - 0) + 1(-2 - 0) = -1 + 3 - 2 = 0$ Since only the matrix in option (B) has a non-zero determinant, it is the only one that can be obtained from the identity matrix by elementary row transformations. Answer: $\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}$
3. Q3JEE Advanced Adv 2025 (Paper 2)
   Let $I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ and $P=\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right)$. Let $Q=\left(\begin{array}{ll}\mathrm{x} & \mathrm{y} \\ \mathrm{z} & 4\end{array}\right)$ for some non-zero real numbers $x, y$, and $z$, for which there is $2 \times 2$ matrix $R$ with all entries being non-zero real numbers, such that $Q R=R P$. Then which of the following statements is (are) TRUE?
   1. A.The determinant of $Q-2 I$ is zero
   2. B.The determinant of $\mathrm{Q}-6 I$ is 12
   3. C.The determinant of $Q-3 I$ is 15
   4. D.$y z=2$
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   Answer: A,B

   $\begin{aligned} & \mathrm{QR}=\mathrm{RP} \\ & P=\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right) \quad Q=\left(\begin{array}{ll} x & y \\ z & 4 \end{array}\right) \\ & \left(\begin{array}{ll} x & y \\ z & 4 \end{array}\right)\left(\begin{array}{ll} r_1 & r_2 \\ r_3 & r_4 \end{array}\right)=\left(\begin{array}{ll} r_1 & r_2 \\ r_3 & r_4 \end{array}\right)\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right) \\ & \left.\begin{array}{l} \mathrm{xr}_1+\mathrm{yr}_3=2 \mathrm{r}_1 \\ \mathrm{zr}_1+4 \mathrm{r}_3=2 \mathrm{r}_3 \end{array}\right\} \rightarrow \frac{\mathrm{x}-2}{-\mathrm{y}}=\frac{\mathrm{r}_3}{\mathrm{r}_1}=\frac{\mathrm{z}}{-2} \end{aligned}$ $2 x-4=y z$ $\left.\begin{array}{l}\mathrm{xr}_2+\mathrm{yr}_4=3 \mathrm{r}_2 \\ \mathrm{zr}_2+4 \mathrm{r}_4=3 \mathrm{r}_4\end{array}\right\} \rightarrow \frac{\mathrm{x}-3}{-\mathrm{y}}=\frac{\mathrm{r}_4}{\mathrm{r}_2}=-\mathrm{z} \quad$ $\mathrm{x}-3=\mathrm{yz}$ $\Rightarrow 2 x-4=x-3$ $\Rightarrow x=1 ~\&~ y z=-2$ $\begin{aligned} & \Rightarrow \mathrm{Q}-\lambda \mathrm{I}=\left(\begin{array}{cc} \mathrm{x}-\lambda & \mathrm{y} \\ \mathrm{z} & 4-\lambda \end{array}\right) \\ & |\mathrm{Q}-\lambda \mathrm{I}|=(\lambda-\mathrm{x})(\lambda-4)-\mathrm{yz} \\ & =\lambda^2-(\mathrm{x}+4) \lambda+4 \mathrm{x}-\mathrm{yz} \\ & |\mathrm{Q}-\lambda \mathrm{I}|=\lambda^2-5 \lambda+6 \end{aligned}$ Now verify the option
4. Q4JEE Advanced Adv 2025 (Paper 1)
   Consider the matrix, $P=\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right)$ Let the transpose of a matrix $X$ be denoted by $\mathrm{X}^T$. Then the number of $3 \times 3$ invertible matrices Q with integer entries, such that $Q^{-1}=Q^T$ and $P Q=Q P$ is
   1. A.32
   2. B.8
   3. C.16
   4. D.24
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   Answer: (C)

   $\begin{aligned} & \mathrm{PQ}=\mathrm{QP} \Rightarrow\left[\begin{array}{ccc} 2 \mathrm{a}_1 & 2 \mathrm{~b}_1 & 2 \mathrm{c}_1 \\ 2 \mathrm{a}_2 & 2 \mathrm{~b}_2 & 2 \mathrm{c}_2 \\ 3 \mathrm{a}_3 & 3 \mathrm{~b}_3 & 3 \mathrm{c}_3 \end{array}\right]=\left[\begin{array}{ccc} 2 \mathrm{a}_1 & 2 \mathrm{~b}_1 & 3 \mathrm{c}_1 \\ 2 \mathrm{a}_2 & 2 \mathrm{~b}_2 & 3 \mathrm{c}_2 \\ 2 \mathrm{a}_3 & 2 \mathrm{~b}_3 & 3 \mathrm{c}_3 \end{array}\right] \\ & \mathrm{c}_1=0, \mathrm{c}_2=0, \mathrm{a}_3=0, \mathrm{~b}_3=0 \\ & \mathrm{Q}=\left[\begin{array}{ccc} \mathrm{a}_1 & \mathrm{~b}_1 & 0 \\ \mathrm{a}_2 & \mathrm{~b}_2 & 0 \\ 0 & 0 & \mathrm{c}_3 \end{array}\right] \\ & \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0 \\ & \mathrm{a}_1^2+\mathrm{b}_1^2=1, \mathrm{a}_2^2+\mathrm{b}_2^2=1, \mathrm{c}_3^2=1 \end{aligned}$ $\begin{array}{lllll} a_1 & b_1 & a_2 & b_2 & c_3 \\ 1 & 0 & 0 & 1,-1 & +1,-1 \\ -1 & 0 & 0 & 1,-1 & 1,-1 \\ 0 & 1 & 1,-1 & 0 & 1,-1 \\ 0 & -1 & 1,-1 & 0 & 1,-1 \end{array}$ Total 16 matrices
5. Q5JEE Advanced Adv 2023 (Paper 2)
   Let $R=\left\{\left(\begin{matrix}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{matrix}\right):a,b,c,d\in \left\{0,3,5,7,11,13,17,19\right\}\right\}$. Then the number of invertible matrices in $R$ is
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   Answer: 3780

   Given, $R=\left[\begin{matrix}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{matrix}\right]$ $\Rightarrow |R|=-5\left|\begin{matrix}a & b \\ c & d\end{matrix}\right|$ We know that, for invertible matrices $\left|R\right|\neq 0$ Now $|R|$ can be zero in following cases: (i) Two of $a,b,c,d$ are zeroes which can be ($a$ and $b$),($b$ and $d$),($d$ and $c$) or ($c$ and $a$) $\rightarrow 4\times {7}^{2}\text{ways}=196$ (ii) Any three of $a,b,c,d$ are zeroes $\rightarrow C34\times 7=28$ (iii) All four of $a,b,c,d$ are zeroes $\rightarrow 1$ (iv) All four of $a,b,c,d$ are non-zero but same number $\rightarrow 7$ (v) When two are alike and $2$ other are alike (non-zero) $\rightarrow 7{C}_{2}\times 2\times 2=84$ Number of invertible matrices $={8}^{4}-196-28-1-7-84=3780$