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 176 questions; sign in to attempt the remaining 172 in the exam simulator.

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1. Q1JEE Main 2026 (05 Apr, Shift 1)Adjoint and its Properties
   Let $A$ be a $3 \times 3$ matrix such that $A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix}$, $A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix}$, $A \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix}$ and $A \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix}$. If $\det(A) = 1$, then $\det(\operatorname{adj}(A^2 + A))$ is equal to:
   1. A.$16$
   2. B.$25$
   3. C.$49$
   4. D.$64$
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   Answer: (D)

   Let $C_1, C_2, C_3$ be the columns of $A$ and $R_1, R_2, R_3$ be the rows of $A$. From the given equations, we have: $A \begin{bmatrix}0\\0\\1\end{bmatrix} = C_3 = \begin{bmatrix}1\\3\\1\end{bmatrix}$ $A \begin{bmatrix}1\\0\\1\end{bmatrix} = C_1 + C_3 = \begin{bmatrix}3\\4\\4\end{bmatrix} \Rightarrow C_1 = \begin{bmatrix}2\\1\\3\end{bmatrix}$ Similarly, for the transpose $A^T$, the columns correspond to the rows of $A$: $A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = R_3^T = \begin{bmatrix}3\\1\\1\end{bmatrix} \Rightarrow R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix}$ $A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = R_1^T + R_3^T = \begin{bmatrix}5\\2\\2\end{bmatrix} \Rightarrow R_1^T = \begin{bmatrix}2\\1\\1\end{bmatrix} \Rightarrow R_1 = \begin{bmatrix}2 & 1 & 1\end{bmatrix}$ Using the elements of $C_1, C_3, R_1,$ and $R_3$, we can construct the matrix $A$ with an unknown central element $b$: $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 \end{bmatrix}$ We are given that $\det(A) = 1$. Expanding along the first row: $\det(A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1$ $2b - 6 + 8 + 1 - 3b = 1$ $3 - b = 1 \Rightarrow b = 2$ Thus, the matrix $A$ is: $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ We need to find $\det(\operatorname{adj}(A^2 + A))$. Using the properties of determinants and adjoints: $\det(\operatorname{adj}(A^2 + A)) = (\det(A^2 + A))^2 = (\det(A) \cdot \det(A + I))^2$ First, find $A + I$: $A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}$ Now, calculate $\det(A + I)$: $\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9)$ $\det(A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8$ Since $\det(A) = 1$, we have: $\det(A^2 + A) = 1 \times 8 = 8$ Finally, for a $3 \times 3$ matrix: $\det(\operatorname{adj}(A^2 + A)) = 8^{3-1} = 8^2 = 64$ Answer: $64$
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.$10$
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   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 $4{\alpha }^{2}+{\beta }^{2}+{\lambda }^{2}=14$