Matrices JEE Main previous year questions with solutions

4 solved JEE Main questions on Matrices, free to read — no sign-in needed. The full chapter has 221 questions; sign in to attempt the remaining 217 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Adjoint and its Properties
   Let $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B)=66$, then $\det(\text{adj}(A))$ equals:
   1. A.$289$
   2. B.$361$
   3. C.$441$
   4. D.$529$
   Show answer & solution

   Answer: (C)

   The cofactor matrix of $A$ is calculated as follows: $C_{11} = 15$, $C_{12} = -10$, $C_{13} = 8$ $C_{21} = 3$, $C_{22} = 5\alpha$, $C_{23} = -4\alpha$ $C_{31} = -6$, $C_{32} = 4$, $C_{33} = 3\alpha - 2$ The adjoint of $A$ is the transpose of the cofactor matrix: $\text{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$ The matrix $B$ is given by: $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}$ Expanding the determinant of $B$ along the second column: $\det(B) = -3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} = -3(-10(\alpha - 2) - 32) = -3(-10\alpha - 12) = 30\alpha + 36$ Given $\det(B) = 66$, we get: $30\alpha + 36 = 66 \Rightarrow 30\alpha = 30 \Rightarrow \alpha = 1$ The determinant of $A$ is: $\det(A) = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) = 15\alpha + 6$ Substituting $\alpha = 1$: $\det(A) = 15(1) + 6 = 21$ Using the property $\det(\text{adj}(A)) = (\det(A))^{n-1}$ for a $3 \times 3$ matrix: $\det(\text{adj}(A)) = (\det(A))^2 = (21)^2 = 441$ Answer: $441$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Basic Algebra of Matrices
   <p>Let $\alpha$ be a solution of $x^2+x+1=0$, and for some $a$ and $b$ in $\mathbb{R},\left[\begin{array}{lll}4 &amp; \mathrm{a} &amp; \mathrm{b}\end{array}\right]\left[\begin{array}{ccc}1 &amp; 16 &amp; 13 \\ -1 &amp; -1 &amp; 2 \\ -2 &amp; -14 &amp; -8\end{array}\right]=\left[\begin{array}{ccc}0 &amp; 0 &amp; 0\end{array}\right]$. If $\frac{4}{\alpha^4}$ $+\frac{\mathrm{m}}{\alpha^{\mathrm{a}}}+\frac{\mathrm{n}}{\alpha^{\mathrm{b}}}=3$, then $\mathrm{m}+\mathrm{n}$ is equal to</p>
   1. A.3
   2. B.11
   3. C.7
   4. D.<p>8</p>
   Show answer & solution

   Answer: (B)

   <p>$x^2+x+1=0$ $\alpha$ is root $\begin{aligned} &amp; \therefore \alpha^2+\alpha+1=0 \\ &amp; \Rightarrow \alpha=\omega \text { as } \omega^2 \text { [cube root of unity] } \end{aligned}$ also $\left.\begin{array}{l} {\left[\begin{array}{ll} 4-a-2 b &amp; 64-a-14 b \end{array} 52+2 a-8 b\right.} \end{array}\right]$ $\begin{aligned} &amp; =\left[\begin{array}{ll}0 &amp; 0&amp;0\end{array}\right] \\ \therefore &amp; a+2 b=4 \\ &amp; a+14 b=64 \\ \Rightarrow &amp; 12 b=60 \Rightarrow b=5 \\ \Rightarrow &amp; a=-6 \\ \therefore &amp; \frac{4}{\alpha^4}+\frac{m}{\alpha^{-6}}+\frac{n}{\alpha^5}=3 \\ \Rightarrow &amp; \frac{4}{\omega}+\frac{m}{1}+\frac{n}{\omega^2}=3 \\ \Rightarrow &amp; 4 \omega^2+m+n \omega=3\end{aligned}$ $\begin{aligned} &amp; \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=3 \\ &amp; \therefore-2+\mathrm{m}-\frac{\mathrm{n}}{2}=3....(1) \\ &amp; \&amp; \frac{-4 \sqrt{3}}{2}+\frac{\mathrm{n} \sqrt{3}}{2}=0 \\ &amp; \therefore \mathrm{n}=4 \\ &amp; \mathrm{~m}=7 \\ &amp; \therefore \mathrm{~m}+\mathrm{n}=11 \end{aligned}$</p>
3. Q3JEE Main 2024 (05 Apr, Shift 2)Formation and Basic of Matrix
   <p>Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{rrr}\beta &amp; \alpha &amp; 3 \\ \alpha &amp; \alpha &amp; \beta \\ -\beta &amp; \alpha &amp; 2 \alpha\end{array}\right]$. If $B=\left[\begin{array}{rrr}3 \alpha &amp; -9 &amp; 3 \alpha \\ -\alpha &amp; 7 &amp; -2 \alpha \\ -2 \alpha &amp; 5 &amp; -2 \beta\end{array}\right]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}(A B)$ is equal to :</p>
   1. A.64
   2. B.216
   3. C.343
   4. D.<p>125</p>
   Show answer & solution

   Answer: (B)

   <p>Equating co-factor fo $\mathrm{A}_{21}$ $\begin{aligned} &amp; \left(2 \alpha^2-3 \alpha\right)=\alpha \\ &amp; \alpha=0,2 \text { (accept) } \end{aligned}$ Now, $2 \alpha^2-\alpha \beta=3 \alpha$ $\begin{aligned} &amp; \alpha=2 \quad \beta=1 \\ &amp; |\mathrm{AB}|=|\mathrm{A} \operatorname{cof}(\mathrm{A})|=|\mathrm{A}|^3 \\ &amp; \mathrm{~A}=\left|\begin{array}{lll} 1 &amp; 2 &amp; 3 \\ 2 &amp; 2 &amp; 1 \\ -1 &amp; 2 &amp; 4 \end{array}\right|=6-2(9)+3(6)=6 \end{aligned}$</p>
4. Q4JEE Main 2023 (08 Apr, Shift 2)Inverse of a Matrix
   If $A=\left[\begin{matrix}1 & 5 \\ \lambda & 10\end{matrix}\right],{A}^{-1}=\alpha A+\beta I$ and $\alpha +\beta =-2$, then $4{\alpha }^{2}+{\beta }^{2}+{\lambda }^{2}$ is equal to :
   1. A.$12$
   2. B.$19$
   3. C.$14$
   4. D.$10$
   Show answer & solution

   Answer: (C)

   Given, $A=\left[\begin{matrix}1 & 5 \\ \lambda & 10\end{matrix}\right],{A}^{-1}=\alpha A+\beta I$ and $\alpha +\beta =-2$, Now solving by using characteristic equation, we get $|A-k|=0\Rightarrow \left|\begin{matrix}1-k & 5 \\ \lambda & 10-k\end{matrix}\right|=0$ $\Rightarrow {k}^{2}-11k+10-5\lambda =0$ $\Rightarrow {A}^{2}-11A+(10-5\lambda )I=0$ {by putting $k=A$} $\Rightarrow {A}^{-1}=\dfrac{1}{10-5\lambda }(-A+11I)$ Now on comparing with ${A}^{-1}=\alpha A+\beta I$ we get, $\Rightarrow \alpha =-\dfrac{1}{10-5\lambda },\beta =\dfrac{11}{10-5\lambda }$ And given, $\alpha +\beta =-2\Rightarrow \lambda =3,\alpha =\dfrac{1}{5},\beta =\dfrac{-11}{5}$ Hence, the value of $4{\alpha }^{2}+{\beta }^{2}+{\lambda }^{2}=14$