Matrices and Determinants: JEE Main 2026 PYQs

Answer: Option (2)

Key idea: Condition for inconsistent system of linear equations using determinants and row reduction

1. Write the augmented matrix and apply row operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 8R_1$
2. Apply $R_3 \to R_3 - 4R_2$ to get the matrix in row echelon form
3. The last row becomes $[0 \quad 0 \quad \beta-4\alpha \mid -2]$
4. For no solution, we need $\beta - 4\alpha = 0$ (rank of coefficient matrix $< $ rank of augmented matrix) and $-2 \neq 0$ (which is satisfied)
5. From $\beta - 4\alpha = 0$, we get $\beta = 4\alpha$
6. Therefore, $\frac{\beta}{\alpha} = 4$

Answer: Option (2)

Key idea: Matrix equations, determinants, and properties of adjoint matrices

1. Find $\alpha$ from $A^2 - 4A + I = O$: Calculate $A^2$ and solve to get $\alpha = 3$
2. Find $\beta$ from $B^2 - 5B - 6I = O$: Calculate $B^2$ and solve to get $\beta = 4$
3. Check (S1): Calculate $(B-A)(B+A) = \begin{bmatrix}13 & 15\\ 7 & 10\end{bmatrix}$, then transpose gives $\begin{bmatrix}13 & 7\\ 15 & 10\end{bmatrix} \neq \begin{bmatrix}13 & 15\\ 7 & 10\end{bmatrix}$, so (S1) is FALSE
4. Check (S2): Calculate $\det(A+B) = \det\begin{bmatrix}4 & 5\\ 5 & 5\end{bmatrix} = -5$
5. Use property $\det(\text{adj}(M)) = [\det(M)]^{n-1}$ for $n \times n$ matrix. For $n=2$: $\det(\text{adj}(A+B)) = \det(A+B) = -5$, so (S2) is TRUE
6. Conclusion: Only (S2) is correct

Answer: Option (2)

Key idea: Condition for infinitely many solutions in a system of linear equations - the third equation must be a linear combination of the first two

1. For infinitely many solutions, the third equation must be a linear combination of the first two: $x + 6y + az = \lambda(x + 5y + 6z) + \mu(2x + 3y + 4z)$
2. Compare coefficients: $1 = \lambda + 2\mu$ and $6 = 5\lambda + 3\mu$
3. Solve the system: From first equation $\lambda = 1 - 2\mu$. Substitute in second: $6 = 5(1-2\mu) + 3\mu$ gives $\mu = -\frac{1}{7}$ and $\lambda = \frac{9}{7}$
4. Find $a$ from coefficient of $z$: $a = 6\lambda + 4\mu = 6(\frac{9}{7}) + 4(-\frac{1}{7}) = \frac{50}{7}$
5. Find $b$ from constant terms: $b = 4\lambda + 7\mu = 4(\frac{9}{7}) + 7(-\frac{1}{7}) = \frac{29}{7}$
6. Check which line passes through $(\frac{50}{7}, \frac{29}{7})$: $x - y = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$ ✓

Answer: 34

Key idea: Matrix equations and inverse matrices

1. Step 1: Find $A^{-1}$ using $\det(A) = 4$, so $A^{-1} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
2. Step 2: From $PA = B$, get $P = BA^{-1} = \begin{bmatrix} -21/2 & 6 \\ -7/2 & 2 \end{bmatrix}$
3. Step 3: From $AQ = B$, get $Q = A^{-1}B = \begin{bmatrix} -1 & -3 \\ -5/2 & -15/2 \end{bmatrix}$
4. Step 4: Calculate $P + Q = \begin{bmatrix} -23/2 & 3 \\ -6 & -11/2 \end{bmatrix}$
5. Step 5: Find $2(P + Q) = \begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix}$
6. Step 6: Sum of diagonal elements = $-23 + (-11) = -34$
7. Step 7: Absolute value = $|-34| = 34$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Answer: Option (4)

Key idea: Properties of adjoint matrices and determinants

1. Calculate $|A| = 1(0-3) - 1(-10-1) + 2(-6) = -3 + 11 - 12 = -4$
2. Use $(adj A)^{-1} = \frac{A}{|A|}$ to get $2(adj A)^{-1} = 2 \cdot \frac{A}{-4} = -\frac{1}{2}A$
3. Let $B = -\frac{1}{2}A$, then $|B| = (-\frac{1}{2})^3 \cdot (-4) = \frac{1}{2}$
4. Apply formula: $adj(adj(B)) = |B|^{n-2} \cdot B = \frac{1}{2} \cdot (-\frac{1}{2}A) = -\frac{1}{4}A$
5. Sum of elements of $A$ is $1+1+2-2+0+1+1+3+5 = 12$
6. Therefore, sum of elements = $-\frac{1}{4} \times 12 = -3$

Answer: Option (2)

Key idea: System of linear equations with infinitely many solutions - determinant condition and consistency

1. For infinitely many solutions, the determinant of coefficient matrix must be zero: $\det(A) = 0$
2. Calculate $\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = \lambda - 5$
3. Setting $\det(A) = 0$ gives $\lambda = 5$
4. For consistency, equation (3) must be a linear combination of equations (1) and (2): Eq(3) = $-1 \times$ Eq(1) $+ 2 \times$ Eq(2)
5. Therefore $\mu = -1(5) + 2(9) = -5 + 18 = 13$
6. Hence $\lambda + \mu = 5 + 13 = 18$

Answer: Option (4)

Key idea: Determinant properties and adjugate matrix

1. Step 1: Use given conditions to find matrix $\mathbf{A}$. From $\mathbf{A}\begin{bmatrix}0\\0\\1\end{bmatrix}$ and $\mathbf{A}\begin{bmatrix}1\\0\\1\end{bmatrix}$, find the first and third columns of $\mathbf{A}$.
2. Step 2: From $\mathbf{A}^T\begin{bmatrix}1\\0\\1\end{bmatrix}$ and $\mathbf{A}^T\begin{bmatrix}0\\0\\1\end{bmatrix}$, find additional entries. This gives $\mathbf{A} = \begin{bmatrix}2&1&1\\1&a_{22}&3\\3&1&1\end{bmatrix}$.
3. Step 3: Use $\det(\mathbf{A}) = 1$ to find $a_{22} = 2$. So $\mathbf{A} = \begin{bmatrix}2&1&1\\1&2&3\\3&1&1\end{bmatrix}$.
4. Step 4: Use the property $\det(\mathbf{A}^2 + \mathbf{A}) = \det(\mathbf{A}(\mathbf{A} + \mathbf{I})) = \det(\mathbf{A})\cdot\det(\mathbf{A} + \mathbf{I})$.
5. Step 5: Calculate $\mathbf{A} + \mathbf{I} = \begin{bmatrix}3&1&1\\1&3&3\\3&1&2\end{bmatrix}$ and find $\det(\mathbf{A} + \mathbf{I}) = 8$.
6. Step 6: Therefore $\det(\mathbf{A}^2 + \mathbf{A}) = 1 \times 8 = 8$.
7. Step 7: Apply the formula $\det(\text{adj}(\mathbf{M})) = (\det(\mathbf{M}))^{n-1}$ for $n \times n$ matrix. So $\det(\text{adj}(\mathbf{A}^2 + \mathbf{A})) = 8^2 = 64$.

Answer: Option (2)

Key idea: Determinant condition for infinitely many solutions in homogeneous system

1. Write the coefficient matrix $A = \begin{pmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & 2 & f(t) \end{pmatrix}$
2. For infinitely many solutions, we need $\det(A) = 0$ for all $t \in \mathbb{R}$
3. Calculate determinant: $\det(A) = 1(f(t) - 10t) - 2(6f(t) - 15t) + t(9) = -11f(t) + 29t$
4. Set $\det(A) = 0$: $-11f(t) + 29t = 0 \Rightarrow f(t) = \frac{29t}{11}$
5. Find derivative: $f'(t) = \frac{29}{11} > 0$ for all $t \in \mathbb{R}$
6. Since $f'(t) > 0$ everywhere, $f$ is strictly increasing on $\mathbb{R}$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Determinant evaluation and telescoping sums

Answer: Option (2)

Key idea: Matrix construction from column vectors and solving linear systems

1. Recognize that multiplying $M$ by standard basis vectors gives the columns of $M$
2. Construct matrix $M = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix}$
3. Set up the system: $x - z = 1$, $2x + y + z = 7$, $3x + 2y + z = 11$
4. From first equation: $x = 1 + z$. Substitute to get $y + 3z = 5$ and $y + 2z = 4$
5. Solve to get $z = 1$, $y = 2$, $x = 2$
6. Calculate $x + y + z = 2 + 2 + 1 = 5$

Answer: 4

Key idea: Properties of adjoint matrices and matrix equations

1. Calculate $A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ by matrix multiplication
2. Find $\det(A) = -1$ using cofactor expansion
3. Use the property $adj(adj(A)) = (\det A)^{n-2} \cdot A = -A$ for a $3 \times 3$ matrix
4. Calculate $adj(A) \cdot adj(adj(A)) = I$ (identity matrix)
5. Substitute into the given equation: $A^2 + \alpha(-A) + \beta I = RHS$
6. Compare matrix entries to get: $-1 - \alpha = -2 \Rightarrow \alpha = 1$ and $1 + \beta = 0 \Rightarrow \beta = -1$
7. Calculate $(\alpha - \beta)^2 = (1-(-1))^2 = 4$

Answer: Option (3)

Key idea: Matrix Powers using Binomial Theorem for Nilpotent Matrices

1. Step 1: Write $A = I + N$ where $N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$ is strictly lower triangular
2. Step 2: Compute $N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$ (nilpotent)
3. Step 3: Use binomial expansion: $A^n = I + nN + \frac{n(n-1)}{2}N^2$
4. Step 4: For $n=99$: $A^{99} = \begin{bmatrix} 1 & 0 & 0 \\ 297 & 1 & 0 \\ 44550 & 297 & 1 \end{bmatrix}$
5. Step 5: Calculate $B = A^{99} - I = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{bmatrix}$
6. Step 6: Find $\frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = \frac{44253}{297} = 149$

Answer: Option (2)

Key idea: Singular Matrix (Determinant = 0), Integration, and Optimization on Closed Interval

1. Find $\alpha$ using $\det(A) = 0$: Expand the determinant to get $\det(A) = 3 - \alpha = 0$, so $\alpha = 3$
2. Integrate to find $f(x) = \int_0^x (t^2 + 2t + 3)dt = \frac{x^3}{3} + x^2 + 3x$
3. Check for critical points: $f'(x) = x^2 + 2x + 3 > 0$ for all $x$ (discriminant $< 0$), so $f$ is strictly increasing
4. Since $f$ is increasing on $[1, 3]$: minimum $m = f(1) = \frac{13}{3}$ and maximum $M = f(3) = 27$
5. Calculate $3(M - m) = 3\left(27 - \frac{13}{3}\right) = 3 \cdot \frac{68}{3} = 68$

Answer: Option (3)

Key idea: Condition for infinitely many solutions in a system of linear equations

1. For infinitely many solutions, the determinant of coefficient matrix must be zero: $\det(A) = 0$
2. Calculate $\det(A) = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = \lambda - 6$
3. Setting $\det(A) = 0$ gives $\lambda = 6$
4. For consistency, the third equation must be a linear combination of the first two equations
5. Find $a$ and $b$ such that $a(x+y+z) + b(x+2y+5z) = 2x+3y+6z$, which gives $a=1, b=1$
6. Therefore $\mu = 1(6) + 1(10) = 16$
7. Thus $\lambda + \mu = 6 + 16 = 22$

Answer: Option (3)

Key idea: Determinants, Adjoint Matrix Properties, and det(adj(A)) = [det(A)]^(n-1)

1. Calculate $\det(A) = 15\alpha + 6$ by expanding along the first row
2. Calculate the determinant of the diagonal matrix in $B$: $\det(M) = 10\alpha^2$
3. Use the property: For a $3 \times 3$ matrix, $\det(\text{adj}(A)) = [\det(A)]^2$
4. From $\det(B) = 66$ and the answer options being perfect squares (289, 361, 441), deduce that $\det(A)$ must be 17, 19, or 21
5. Test $\alpha = 1$: This gives $\det(A) = 21$, which is consistent with the problem structure
6. Calculate $\det(\text{adj}(A)) = (21)^2 = 441$