Determinants JEE Main previous year questions with solutions

4 solved JEE Main questions on Determinants, free to read — no sign-in needed. The full chapter has 197 questions; sign in to attempt the remaining 193 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Expansion of Determinants
   If $f: \mathbf{N} \rightarrow \mathbf{Z}$ is defined by $f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$, $k \in \mathbf{N}$, and $\sum_{n=1}^{k} f(n) = 98$, then $k$ is equal to :
   1. A.$3$
   2. B.$4$
   3. C.$5$
   4. D.$6$
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   Answer: (A)

   Let $S = \displaystyle\sum_{n=1}^{k} f(n)$, where $f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$ Since only the first column depends on $n$, the summation can be taken inside that column: $S = \begin{vmatrix} \displaystyle\sum_{n=1}^{k} n & -1 & -5 \\ \displaystyle\sum_{n=1}^{k} (-2n^2) & 3(2k+1) & 2k+1 \\ \displaystyle\sum_{n=1}^{k} (-3n^3) & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$ Using standard summation formulas: $\displaystyle\sum_{n=1}^{k} n = \dfrac{k(k+1)}{2}$ $\displaystyle\sum_{n=1}^{k} (-2n^2) = -\dfrac{k(k+1)(2k+1)}{3}$ $\displaystyle\sum_{n=1}^{k} (-3n^3) = -\dfrac{3k^2(k+1)^2}{4}$ Substituting: $S = \begin{vmatrix} \dfrac{k(k+1)}{2} & -1 & -5 \\ -\dfrac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\ -\dfrac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$ Taking $\dfrac{k(k+1)}{12}$ common from $C_1$: $S = \dfrac{k(k+1)}{12} \begin{vmatrix} 6 & -1 & -5 \\ -4(2k+1) & 3(2k+1) & 2k+1 \\ -9k(k+1) & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$ Applying $C_1 \to C_1 + C_2 + C_3$: Row 1: $6 - 1 - 5 = 0$ Row 2: $-8k - 4 + 6k + 3 + 2k + 1 = 0$ Row 3: $-9k^2 - 9k + 6k^2 + 3k + 3k^2 + 6k + 1 = 1$ $S = \dfrac{k(k+1)}{12} \begin{vmatrix} 0 & -1 & -5 \\ 0 & 3(2k+1) & 2k+1 \\ 1 & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$ Expanding along $C_1$: $S = \dfrac{k(k+1)}{12} \cdot 1 \cdot \begin{vmatrix} -1 & -5 \\ 3(2k+1) & 2k+1 \end{vmatrix}$ $S = \dfrac{k(k+1)}{12}\left[(-1)(2k+1) - (-5)(3(2k+1))\right]$ $S = \dfrac{k(k+1)}{12}\left[-(2k+1) + 15(2k+1)\right]$ $S = \dfrac{k(k+1)}{12} \cdot 14(2k+1) = 7 \cdot \dfrac{k(k+1)(2k+1)}{6}$ Note that $\dfrac{k(k+1)(2k+1)}{6} = \displaystyle\sum_{n=1}^{k} n^2$. Given $S = 98$: $7 \cdot \dfrac{k(k+1)(2k+1)}{6} = 98$ $\dfrac{k(k+1)(2k+1)}{6} = 14$ Checking integer values of $k$: $k = 1$: sum $= 1$ $k = 2$: sum $= 5$ $k = 3$: sum $= 14$ Hence, $k = 3$, and the correct option is $(1)\ 3$.
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 (09 Apr, Shift 2)System of Linear Equations
   Consider the matrices : $A=\left[\begin{array}{ll}2 & -5 \\ 3 & m\end{array}\right], B=\left[\begin{array}{l}20 \\ m\end{array}\right]$ and $X=\left[\begin{array}{l}x \\ y\end{array}\right]$. Let the set of all $m$, for which the system of equations $A X=B$ has a negative solution (i.e., $x \lt 0$ and $y \lt 0$ ), be the interval $(a, b)$. Then $8 \int_a^b|A| d m$ is equal to_________
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   Answer: 450

   $\begin{aligned} & A=\left(\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right), B=\left(\begin{array}{c}20 \\ m\end{array}\right) \\ & X=\left(\begin{array}{l}x \\ y\end{array}\right) \\ & 2 x-5 y=20 \ldots(1)\\ & 3 x+m y=m \ldots(2)\\ & \Rightarrow y=\frac{2 m-60}{2 m+15} \\ & y \lt 0 \Rightarrow m \in\left(\frac{-15}{2}, 30\right)\end{aligned}$ $\begin{aligned} & \mathrm{x}=\frac{25 \mathrm{~m}}{2 \mathrm{~m}+15} \\ & \mathrm{x} \lt 0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \\ & \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \\ & |\mathrm{A}|=2 \mathrm{~m}+15 \end{aligned}$ Now, $\begin{aligned} & 8 \int_{\frac{-15}{2}}^0(2 \mathrm{~m}+15) \mathrm{dm}=8\left\{\mathrm{~m}^2+15 \mathrm{~m}\right\}_{\frac{-15}{2}}^0 \\ & \Rightarrow 8\left\{-\left(\frac{225}{4}-\frac{225}{2}\right)\right\} \\ & =8 \times \frac{225}{4}=450 \end{aligned}$
4. Q4JEE Main 2026 (23 Jan, Shift 1)Expansion of Determinants
   Among the statements : I: If $\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$, then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=\frac{3}{2}$, and II : If $\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q}$, then $\mathrm{p}^{2}=196 \mathrm{q}^{2}$,
   1. A.only II is true
   2. B.both are false
   3. C.both are true
   4. D.only I is true
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   Answer: (B)

   Statement I: Let $D_1 = \begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma$. $D_2 = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix} = 2\cos\alpha\cos\beta\cos\gamma$. $D_1 = D_2 \Rightarrow 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma = 0 \Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \neq \frac{3}{2}$. Statement I is false. Statement II: Evaluate by substituting $x = 0$: determinant $= -12$, so $q = -12$. At $x = 1$: determinant $= 12$, so $p + q = 12 \Rightarrow p = 24$. $p^2 = 576$ and $196q^2 = 196 \times 144 = 28224$. Since $p^2 \neq 196q^2$, Statement II is false. Both statements are false.