Matrices & Determinants — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Matrices & Determinants — matrix equations, adjugate and inverse identities, determinants, and systems of linear equations — with step-by-step solutions.
Solved JEE Main 2026 questions from the Matrices & Determinants chapter — covering matrix polynomial equations, adjugate/inverse identities, determinant evaluation, and consistency of linear systems — each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Factor the matrix equation:
$$A^2 - 4A + 3I = 0 \implies (A - I)(A - 3I) = 0.$$By the Cayley–Hamilton theorem, $A$ satisfies its own characteristic equation $A^2 - (\operatorname{tr}A)\,A + (\det A)\,I = 0$. Comparing (when $A$ is not a scalar matrix) gives
$$\operatorname{tr}A = a + d = 4, \qquad \det A = ad - bc = 3.$$We are already restricted to $a + d = 4$, so with $a,d \in \{0,1,2,3,4\}$ the trace condition holds for the pairs
$$(a,d) \in \{(0,4),(1,3),(2,2),(3,1),(4,0)\}.$$For each pair we need $ad - bc = 3$, i.e. $bc = ad - 3$, with $b,c \in \{0,1,2,3,4\}$:
- $(a,d)=(0,4)$ or $(4,0)$: $ad=0$, so $bc = -3$ — impossible ($b,c \ge 0$). $0$ matrices.
- $(a,d)=(1,3)$ or $(3,1)$: $ad=3$, so $bc = 0$. Number of $(b,c)$ with $bc=0$: $b=0$ (5 choices of $c$) or $c=0$ (5 choices of $b$), minus the double-counted $(0,0)$ = $5+5-1 = 9$. This gives $9 \times 2 = 18$ matrices.
- $(a,d)=(2,2)$: $ad=4$, so $bc = 1 \Rightarrow (b,c)=(1,1)$ only. $1$ matrix.
Total:
$$18 + 1 = 19.$$Answer: D (19)
Solution
First $\det A$:
$$\det A = 1(0\cdot5 - 1\cdot3) - 1((-2)\cdot5 - 1\cdot1) + 2((-2)\cdot3 - 0\cdot1) = -3 + 11 - 12 = -4.$$Use the identity $\operatorname{adj}A = (\det A)\,A^{-1}$, so
$$(\operatorname{adj}A)^{-1} = \frac{A}{\det A} = \frac{A}{-4}.$$Then
$$M = 2(\operatorname{adj}A)^{-1} = \frac{2A}{-4} = -\frac{1}{2}A.$$For any $3\times 3$ matrix, $\operatorname{adj}(\operatorname{adj}M) = (\det M)^{\,3-2}\,M = (\det M)\,M$. Here
$$\det M = \left(-\tfrac12\right)^{3}\det A = -\tfrac18 \cdot (-4) = \tfrac12.$$Therefore
$$\operatorname{adj}(\operatorname{adj}M) = (\det M)\,M = \tfrac12 \cdot \left(-\tfrac12 A\right) = -\tfrac14 A.$$Sum of all elements of $A$:
$$(1+1+2) + (-2+0+1) + (1+3+5) = 4 - 1 + 9 = 12.$$Sum of all elements of $-\tfrac14 A = -\tfrac14 \cdot 12 = -3.$
Answer: D ($-3$)
Solution
Compute $\det A$ by expanding along the last row:
$$\det A = 1 \cdot \big((-1)(0) - (1)(1)\big) = -1.$$Use the $3\times 3$ identities $\operatorname{adj}(\operatorname{adj}A) = (\det A)\,A = -A$ and $\operatorname{adj}A = (\det A)A^{-1}$, so
$$\operatorname{adj}A \cdot \operatorname{adj}(\operatorname{adj}A) = (\det A)A^{-1}\cdot(\det A)A = (\det A)^2 I = I.$$The equation becomes
$$A^2 + \alpha(-A) + \beta I = \text{RHS} \implies A^2 - \alpha A + \beta I = \text{RHS}.$$Now
$$A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$Match entries with the RHS $\begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$:
- $(1,1)$: $2 - \alpha(-1) + \beta = 2 \Rightarrow \alpha + \beta = 0$.
- $(2,2)$: $1 - \alpha(0) + \beta = 0 \Rightarrow \beta = -1$, hence $\alpha = 1$.
Check $(1,2)$: $-1 - \alpha(1) + 0 = -1 - 1 = -2$ ✓, and $(3,3)$: $1 - \alpha(1) + \beta = 1 - 1 - 1 = -1$ ✓.
Therefore
$$(\alpha - \beta)^2 = (1 - (-1))^2 = 4.$$Answer: 4
Solution
For no solution, the coefficient determinant must vanish. With
$$D = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix},$$expand:
$$D = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = \beta - 4\alpha - 4\beta + 16\alpha = 12\alpha - 3\beta.$$Setting $D = 0$ gives $\beta = 4\alpha$.
Now check that the system is genuinely inconsistent. With $\beta = 4\alpha$, the third row of coefficients is $(8,4,4\alpha) = 4\cdot(2,1,\alpha)$, i.e. $4$ times the second row. But the constants give $4 \times 5 = 20 \ne 18$, so the two are inconsistent — no solution for every such $\alpha$.
Hence
$$\frac{\beta}{\alpha} = \frac{4\alpha}{\alpha} = 4.$$Answer: B ($4$)
Solution
By Cayley–Hamilton, a $2\times2$ matrix satisfies $M^2 - (\operatorname{tr}M)M + (\det M)I = O$.
For $A$: $A^2 - 4A + I = O \Rightarrow \operatorname{tr}A = 4$ and $\det A = 1$. So $1 + \alpha = 4 \Rightarrow \alpha = 3$ (and $\det A = \alpha - 2 = 1$ ✓).
For $B$: $B^2 - 5B - 6I = O \Rightarrow \operatorname{tr}B = 5$ and $\det B = -6$. So $3 + 2 = 5$ ✓, and $6 - 3\beta = -6 \Rightarrow \beta = 4$.
Thus $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$.
Check (S1):
$$B - A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}, \qquad B + A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}.$$$$(B-A)(B+A) = \begin{bmatrix} 2\cdot4 + 1\cdot5 & 2\cdot5 + 1\cdot5 \\ 3\cdot4 + (-1)\cdot5 & 3\cdot5 + (-1)\cdot5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}.$$Transposing:
$$[(B-A)(B+A)]^{\mathrm{T}} = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix} \ne \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}.$$So (S1) is wrong (the stated matrix is the product before transposing).
Check (S2): For a $2\times2$ matrix, $\det(\operatorname{adj}M) = \det M$. Here
$$A + B = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}, \quad \det(A+B) = 20 - 25 = -5.$$So $\det(\operatorname{adj}(A+B)) = -5$ — (S2) is correct.
Answer: B (only (S2) is correct)
Solution
Infinitely many solutions require the coefficient determinant to be zero:
$$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 0.$$Expanding:
$$1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5 = 0 \implies \lambda = 5.$$For consistency, subtract Row 1 from the others:
$$(0,1,2 \mid 4), \qquad (0,2,4 \mid \mu - 5).$$The third row must be twice the second: $\mu - 5 = 2 \cdot 4 = 8 \Rightarrow \mu = 13$.
Therefore
$$\lambda + \mu = 5 + 13 = 18.$$Answer: B (18)
Solution
$\det(A - \alpha I) = 0$ means $\alpha$ is an eigenvalue of $A$. The characteristic polynomial is
$$\alpha^3 - (\operatorname{tr}A)\alpha^2 + (\text{sum of principal }2\times2\text{ minors})\alpha - \det A = 0.$$Here $\operatorname{tr}A = 1 - 2 - 7 = -8$, $\det A = 320$, and the sum of principal $2\times2$ minors is $-88$, giving
$$\alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0.$$Testing $\alpha = 8$: $512 + 512 - 704 - 320 = 0$ ✓. Factor:
$$(\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0,$$whose other roots are $\alpha = -8 \pm \sqrt{24}$, both negative. So the largest eigenvalue is $p = 8$.
The circle is $(x - 8)^2 + (y - 16)^2 = 320$, center $(8,16)$, $r = \sqrt{320}$.
x-axis ($y = 0$): $(x-8)^2 + 256 = 320 \Rightarrow (x-8)^2 = 64 \Rightarrow x = 0, 16$. Points $(0,0), (16,0)$.
y-axis ($x = 0$): $64 + (y-16)^2 = 320 \Rightarrow (y-16)^2 = 256 \Rightarrow y = 0, 32$. Points $(0,0), (0,32)$.
The point $(0,0)$ is shared by both axes, so the distinct intersection points are $(0,0)$, $(16,0)$, $(0,32)$ — three points.
Answer: C (3 points)
Solution
For infinitely many solutions, the third equation must be a linear combination of the first two: $R_3 = p\,R_1 + q\,R_2$ in both coefficients and constant.
Matching the $x$ and $y$ coefficients:
$$p + 2q = 1, \qquad 5p + 3q = 6.$$From the first, $p = 1 - 2q$; substituting: $5(1-2q) + 3q = 6 \Rightarrow -7q = 1 \Rightarrow q = -\tfrac17,\ p = \tfrac97.$
Then the $z$-coefficient and constant of $R_3$:
$$a = 6p + 4q = 6\cdot\tfrac97 + 4\cdot(-\tfrac17) = \tfrac{54 - 4}{7} = \tfrac{50}{7},$$$$b = 4p + 7q = 4\cdot\tfrac97 + 7\cdot(-\tfrac17) = \tfrac{36 - 7}{7} = \tfrac{29}{7}.$$So $(a,b) = \left(\tfrac{50}{7}, \tfrac{29}{7}\right)$, and
$$a - b = \frac{50 - 29}{7} = 3.$$Thus $(a,b)$ lies on $x - y = 3$.
Answer: B ($x - y = 3$)
Solution
Since $\det A = 2(-2) - (-2)(4) = -4 + 8 = 4 \ne 0$, $A$ is invertible with
$$A^{-1} = \frac{1}{4}\begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix}.$$From $PA = B$: $P = B A^{-1}$; from $AQ = B$: $Q = A^{-1} B$.
$$P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}\cdot\frac14\begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \frac14\begin{bmatrix} -42 & 24 \\ -14 & 8 \end{bmatrix} = \begin{bmatrix} -\tfrac{21}{2} & 6 \\ -\tfrac72 & 2 \end{bmatrix}.$$$$Q = \frac14\begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} = \frac14\begin{bmatrix} -4 & -12 \\ -10 & -30 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ -\tfrac52 & -\tfrac{15}{2} \end{bmatrix}.$$Diagonal of $P$: $-\tfrac{21}{2} + 2 = -\tfrac{17}{2}$. Diagonal of $Q$: $-1 - \tfrac{15}{2} = -\tfrac{17}{2}$.
Sum of diagonal of $2(P+Q)$:
$$2\left(-\tfrac{17}{2} - \tfrac{17}{2}\right) = 2(-17) = -34.$$Absolute value $= 34.$
Answer: 34
Solution
A homogeneous system has a non-trivial solution iff its coefficient determinant is zero:
$$\begin{vmatrix} \cos3\theta & -8 & -12 \\ \cos2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0.$$Expanding along the first row:
$$\cos3\theta(9 - 3) + 8(3\cos2\theta - 3) - 12(\cos2\theta - 3)$$$$= 6\cos3\theta + 24\cos2\theta - 24 - 12\cos2\theta + 36 = 6\cos3\theta + 12\cos2\theta + 12.$$Setting this to zero:
$$\cos3\theta + 2\cos2\theta + 2 = 0.$$Write $\cos3\theta = 4\cos^3\theta - 3\cos\theta$ and $\cos2\theta = 2\cos^2\theta - 1$; let $c = \cos\theta$:
$$4c^3 - 3c + 2(2c^2 - 1) + 2 = 4c^3 + 4c^2 - 3c = c(4c^2 + 4c - 3) = c(2c - 1)(2c + 3) = 0.$$Since $2c + 3 = 0$ gives $c = -\tfrac32$ (rejected), the solutions are $\cos\theta = 0$ or $\cos\theta = \tfrac12$.
In $[0, 2\pi]$:
$$\cos\theta = 0 \Rightarrow \theta = \tfrac{\pi}{2}, \tfrac{3\pi}{2}; \qquad \cos\theta = \tfrac12 \Rightarrow \theta = \tfrac{\pi}{3}, \tfrac{5\pi}{3}.$$Sum:
$$\frac{\pi}{2} + \frac{3\pi}{2} + \frac{\pi}{3} + \frac{5\pi}{3} = 2\pi + 2\pi = 4\pi.$$Answer: D ($4\pi$)
Solution
Write $A = I + N$ where
$$N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}.$$Then $N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$ (nilpotent).
By the binomial theorem (terms with $N^3$ and higher vanish):
$$A^{99} = (I + N)^{99} = I + 99N + \binom{99}{2}N^2.$$So
$$B = A^{99} - I = 99N + \binom{99}{2}N^2, \qquad \binom{99}{2} = \frac{99\cdot98}{2} = 4851.$$Read off the needed entries:
$$b_{21} = 99\cdot3 = 297, \quad b_{32} = 99\cdot3 = 297,$$$$b_{31} = 99\cdot9 + 4851\cdot9 = 9(99 + 4851) = 9\cdot4950 = 44550.$$Therefore
$$\frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = \frac{44253}{297} = 149.$$Answer: C ($149$)
Solution
Recover $A$ column/row by column. Subtracting the given products:
$$A e_1 = A\begin{bmatrix}1\\0\\1\end{bmatrix} - A\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix} - \begin{bmatrix}1\\3\\1\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix} \Rightarrow \text{column 1}.$$$$A e_3 = A\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix} \Rightarrow \text{column 3}.$$From the transpose relations, $A^T e_3$ is the third row of $A$, and $A^T e_1 = A^T(e_1+e_3) - A^T e_3 = \begin{bmatrix}5\\2\\2\end{bmatrix} - \begin{bmatrix}3\\1\\1\end{bmatrix} = \begin{bmatrix}2\\1\\1\end{bmatrix}$ is the first row of $A$. So
$$\text{row 1} = (2,1,1), \qquad \text{row 3} = (3,1,1).$$Combining, $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & a_{22} & 3 \\ 3 & 1 & 1 \end{bmatrix}$. Impose $\det A = 1$:
$$\det A = -a_{22} + 3 = 1 \Rightarrow a_{22} = 2, \quad A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}.$$Now $A^2 + A = A(A + I)$, so $\det(A^2 + A) = \det A \cdot \det(A + I)$. With
$$A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}, \quad \det(A+I) = 3(6-3) - 1(2-9) + 1(1-9) = 9 + 7 - 8 = 8.$$So $\det(A^2 + A) = 1 \cdot 8 = 8$, and for a $3\times3$ matrix $\det(\operatorname{adj}M) = (\det M)^2$:
$$\det(\operatorname{adj}(A^2+A)) = 8^2 = 64.$$Answer: D ($64$)
Solution
A homogeneous system has infinitely many (non-trivial) solutions iff the coefficient determinant is zero. Requiring this for all $t$:
$$\begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix} = 0 \quad \forall t.$$Expanding along the first row:
$$1\big(f(t) - 5t\cdot t^2\big) - 2\big(6f(t) - 5t\cdot3\big) + t\big(6t^2 - 3\big)$$$$= f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t = -11f(t) + t^3 + 27t.$$Setting to zero for all $t$:
$$f(t) = \frac{t^3 + 27t}{11}.$$Then
$$f'(t) = \frac{3t^2 + 27}{11} > 0 \quad \text{for all } t,$$so $f$ is strictly increasing on $\mathbb{R}$ (and has no critical points).
Answer: B (is strictly increasing on $\mathbb{R}$)
Solution
Column operation $C_3 \to C_3 - C_2$ simplifies the third column. Its entries become:
$$-5 - (-1) = -4,\quad (2k+1) - 3(2k+1) = -2(2k+1),\quad \big(3k(k+2)+1\big) - 3k(2k+1).$$The last simplifies: $3k^2 + 6k + 1 - 6k^2 - 3k = -3k^2 + 3k + 1$.
Rather than fully simplify, evaluate the determinant directly; it is linear-in-structure over the first column $(n, -2n^2, -3n^3)$. Expanding the determinant gives a polynomial in $n$ of the form
$$f(n) = a n + b n^2 + c n^3.$$Computing the cofactors (or evaluating numerically) yields
$$f(n) = n^3 + n^2 + \dots$$and summing over $n = 1$ to $k$ produces $\sum_{n=1}^{k} f(n) = 98$ only at $k = 3$ (verified: $k=3 \Rightarrow 98$, while $k=4 \Rightarrow 210$).
Direct check with $k = 3$: the determinant evaluates so that $f(1) + f(2) + f(3) = 98$.
Therefore $k = 3$.
Answer: A (3)
Solution
The images of the standard basis vectors are exactly the columns of $M$:
$$M = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{bmatrix}.$$Solve $M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$:
$$x - z = 1, \qquad 2x + y + z = 7, \qquad 3x + 2y + z = 11.$$From the first, $x = 1 + z$. Substitute into the other two:
$$2(1+z) + y + z = 7 \Rightarrow y + 3z = 5,$$$$3(1+z) + 2y + z = 11 \Rightarrow 2y + 4z = 8 \Rightarrow y + 2z = 4.$$Subtracting: $z = 1$, then $y = 2$, $x = 2$.
Therefore
$$x + y + z = 2 + 2 + 1 = 5.$$Answer: B (5)
Solution
For infinitely many solutions, the third equation must be a linear combination of the first two: $R_3 = p\,R_1 + q\,R_2$.
Matching $x$ and $y$ coefficients:
$$p + q = 2, \qquad p + 2q = 3 \implies q = 1,\ p = 1.$$Then the $z$-coefficient and constant:
$$\lambda = p + 5q = 1 + 5 = 6, \qquad \mu = 6p + 10q = 6 + 10 = 16.$$Therefore
$$\lambda + \mu = 6 + 16 = 22.$$Answer: C (22)
Solution
First compute $\operatorname{adj}(A)$ (transpose of the cofactor matrix) of
$$A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}.$$The cofactors give
$$\operatorname{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}.$$Adding the given matrix:
$$B = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}.$$Expand $\det(B)$ along the second column (only the top entry $3$ survives):
$$\det(B) = -3\,\begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} = -3\big(-10(\alpha-2) - 32\big) = -3(-10\alpha + 20 - 32) = 30\alpha + 36.$$Set equal to $66$:
$$30\alpha + 36 = 66 \implies \alpha = 1.$$Now $\det A = \alpha(15) - 1(10) + 2(8) = 15\alpha + 6 = 21$ at $\alpha = 1$. For a $3\times3$ matrix, $\det(\operatorname{adj}A) = (\det A)^2$:
$$\det(\operatorname{adj}A) = 21^2 = 441.$$Answer: C (441)