Determinants JEE Main previous year questions with solutions

5 solved JEE Main questions on Determinants, free to read — no sign-in needed. The full chapter has 242 questions; sign in to attempt the remaining 237 in the exam simulator.

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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.$529$
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   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 (03 Apr, Shift 1)Differentiation and Integration of Determinants
   If $\quad y(x)=\left|\begin{array}{ccc}\sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1\end{array}\right|, x \in \mathbb{R}$, then $\frac{d^2 y}{d x^2}+y$ is equal to
   1. A.$-1$
   2. B.$28$
   3. C.$27$
   4. D.$1$
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   Answer: (A)

   $\begin{aligned} & C_3 \rightarrow C_3-C_1 \\ & y(x)=\left|\begin{array}{ccc}\sin x & \cos x & 1+\cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0\end{array}\right| \\ & y(x)=-(1+\cos x) \\ & \frac{d y}{d x}=\sin x \\ & \frac{d^2 y}{d x^2}=\cos x \\ & \frac{d^2 y}{d x^2}+y=-1\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 2)Expansion of Determinants
   If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$, then $\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$ is equal to:
   1. A.3
   2. B.0
   3. C.1
   4. D.2
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   Answer: (B)

   $\begin{aligned} & \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\ & \left|\begin{array}{ccc}\alpha-\mathrm{a} & \mathrm{b}-\beta & 0 \\ 0 & \beta-\mathrm{b} & \mathrm{c}-\gamma \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0 \\ & (\alpha-\mathrm{a})(\gamma(\beta-\mathrm{b})-\mathrm{b}(\mathrm{c}-\gamma))-(\mathrm{b}-\beta)(-\mathrm{a}(\mathrm{c}-\gamma))=0 \\ & \gamma(\alpha-\mathrm{a})(\beta-\mathrm{b})-\mathrm{b}(\alpha-\mathrm{a})(\mathrm{c}-\gamma)+\mathrm{a}(\mathrm{b}-\beta)(\mathrm{c}-\gamma) \\ & \frac{\gamma}{\gamma-\mathrm{c}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\mathrm{a}}{\alpha-\mathrm{a}}=0\end{aligned}$
4. Q4JEE Main 2023 (29 Jan, Shift 2)Inverse of a Matrix
   The set of all values of $t\in ℝ$, for which the matrix $\left[\begin{matrix}{e}^{t} & {e}^{-t}\left(\sin t-2\cos t\right) & {e}^{-t}\left(-2\sin t-\cos t\right) \\ {e}^{t} & {e}^{-t}\left(2\sin t+\cos t\right) & {e}^{-t}\left(\sin t-2\cos t\right) \\ {e}^{t} & {e}^{-t}\cos t & {e}^{-t}\sin t\end{matrix}\right]$ is invertible, is
   1. A.$\left\{\left(2k+1\right)\dfrac{\pi }{2},k\in ℤ\right\}$
   2. B.$\left\{k\pi +\dfrac{\pi }{4},k\in ℤ\right\}$
   3. C.$\left\{k\pi ,k\in ℤ\right\}$
   4. D.$ℝ$
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   Answer: (D)

   Let $A=\left[\begin{matrix}{e}^{t} & {e}^{-t}\left(\sin t-2\cos t\right) & {e}^{-t}\left(-2\sin t-\cos t\right) \\ {e}^{t} & {e}^{-t}\left(2\sin t+\cos t\right) & {e}^{-t}\left(\sin t-2\cos t\right) \\ {e}^{t} & {e}^{-t}\cos t & {e}^{-t}\sin t\end{matrix}\right]$ Given matrix $A$ is invertible if $\left|A\right|\neq 0$ $\Rightarrow \left|\begin{matrix}\begin{matrix}{e}^{t} & {e}^{-t}\left(\sin t-2\cos t\right) & {e}^{-t}\left(-2\sin t-\cos t\right) \\ {e}^{t} & {e}^{-t}\left(2\sin t+\cos t\right) & {e}^{-t}\left(\sin t-2\cos t\right) \\ {e}^{t} & {e}^{-t}\cos t & {e}^{-t}\sin t\end{matrix}\end{matrix}\right|\neq 0$ $\Rightarrow {e}^{t}\cdot {e}^{-t}\cdot {e}^{-t}\left|\begin{matrix}1 & \sin t-2\cos t & -2\sin t-\cos t \\ 1 & 2\sin t+\cos t & \sin t-2\cos t \\ 1 & \cos t & \sin t\end{matrix}\right|\neq 0$ Applying ${R}_{1}\rightarrow {R}_{1}-{R}_{2}$ then ${R}_{2}\rightarrow {R}_{2}-{R}_{3}$, ee get ${e}^{-t}\left|\begin{matrix}0 & -\sin t-3\cos t & -3\sin t+\cos t \\ 0 & 2\sin t & -2\cos t \\ 1 & \cos t & \sin t\end{matrix}\right|\neq 0$ By expanding we have, ${e}^{-t}\times \left(2\sin tcost+6{\cos }^{2}t+6{\sin }^{2}t-2\sin t\cos t\right)\neq 0$ $\Rightarrow {e}^{-t}\times 6\neq 0$ for $\forall t\in ℝ$
5. Q5JEE Main 2022 (29 Jul, Shift 2)System of Linear Equations
   If the system of equations $x+y+z=6$ $2x+5y+\alpha z=\beta$ $x+2y+3z=14$ has infinitely many solutions, then $\alpha +\beta$ is equal to
   1. A.$8$
   2. B.$36$
   3. C.$44$
   4. D.$48$
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   Answer: (C)

   $x+y+z=6....\left(1\right)$ $2x+5y+\alpha z=\beta ....\left(2\right)$ $x+2y+3z=14...\left(3\right)$ For infinite solution $\left|\begin{matrix}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3\end{matrix}\right|=0$ $\Rightarrow \left(15-2\alpha \right)-\left(6-\alpha \right)+\left(4-5\right)=0$ $\Rightarrow 8-\alpha =0\Rightarrow \alpha =8$ Also $\left|\begin{matrix}1 & 1 & 6 \\ 2 & 5 & \beta \\ 1 & 2 & 14\end{matrix}\right|=0$ $\Rightarrow \left(70-2\beta \right)-\left(28-\beta \right)+6\left(4-5\right)=0$ $\Rightarrow 36-\beta =0$ i.e. $\alpha =8,\beta =36$ $\Rightarrow \alpha +\beta =44$