Board Exam-Pattern Question Bank

Determinants is a guaranteed scorer in the CBSE paper — typically one 2-mark evaluation or singularity check, a 3-mark adjoint/cofactor problem, and the flagship 5-mark matrix-method system that has appeared in some form nearly every year. The bank below mirrors that structure: attempt each question fully before reading its solution, and present your work the way the marking scheme rewards — state $\vert A \vert$ before using $A^{-1}$, keep the $\pm$ in given-area problems, and finish system solutions with a verification line.

The questions below follow the pattern of recent board papers; they are original items written for practice.

2-mark questions

Q1. If $A = \begin{pmatrix} 1 & 2 \\ 4 & 2 \end{pmatrix}$, show that $\vert 2A \vert = 4\vert A \vert$.

Solution. $2A = \begin{pmatrix} 2 & 4 \\ 8 & 4 \end{pmatrix}$, so $\vert 2A \vert = 8 - 32 = -24$. Also $\vert A \vert = 2 - 8 = -6$, and $4\vert A \vert = -24$. Hence $\vert 2A \vert = 4\vert A \vert$. ∎

Q2. Find the values of $x$ for which $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$.

Solution. $x^2 - 36 = 36 - 36 = 0$, so $x^2 = 36$ and $x = \pm 6$. (Both values must be stated — the marking scheme allots half the marks to the negative root.)

Q3. Find the minor and cofactor of the element $a_{32}$ in $\Delta = \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$.

Solution. Deleting $R_3$ and $C_2$: $M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = -1 - 12 = -13$. Since $3 + 2 = 5$ is odd, $A_{32} = -M_{32} = 13$.

Q4. Find the values of $k$ if the area of the triangle with vertices $(k, 0)$, $(4, 0)$ and $(0, 2)$ is $4$ square units.

Solution. $\frac{1}{2}\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$. Expanding along $C_2$: the half-determinant equals $4 - k$, so $4 - k = \pm 4$, giving $k = 0$ or $k = 8$.

3-mark questions

Q5. Using cofactors of the elements of the second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.

Solution. $A_{21} = -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = 7$, $A_{22} = \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 7$, $A_{23} = -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -7$. Then $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) + 0(7) + 1(-7) = 7$.

Q6. For $A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{pmatrix}$, find $\operatorname{adj} A$ and verify that $A(\operatorname{adj} A) = \vert A \vert I$.

Solution. Computing all nine cofactors and transposing: $\operatorname{adj} A = \begin{pmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{pmatrix}$. Expanding along $R_1$: $\vert A \vert = 1(3) + 1(12) + 2(6) = 27$. Multiplying, every diagonal entry of $A(\operatorname{adj} A)$ works out to $27$ and every off-diagonal entry to $0$ — e.g. the $(1,1)$ entry is $1(3) + (-1)(-12) + 2(6) = 27$ — so $A(\operatorname{adj} A) = 27I = \vert A \vert I$. ∎

Q7. Find the inverse of $A = \begin{pmatrix} -1 & 5 \\ -3 & 2 \end{pmatrix}$, and verify that $AA^{-1} = I$.

Solution. $\vert A \vert = -2 + 15 = 13 \neq 0$, so $A^{-1}$ exists. By the shortcut, $\operatorname{adj} A = \begin{pmatrix} 2 & -5 \\ 3 & -1 \end{pmatrix}$ and $A^{-1} = \frac{1}{13}\begin{pmatrix} 2 & -5 \\ 3 & -1 \end{pmatrix}$. Verification: $AA^{-1} = \frac{1}{13}\begin{pmatrix} -2 + 15 & 5 - 5 \\ -6 + 6 & 15 - 2 \end{pmatrix} = \frac{1}{13}\begin{pmatrix} 13 & 0 \\ 0 & 13 \end{pmatrix} = I$. ∎

Q8. For $A = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$, and hence find $A^{-1}$.

Solution. $A^2 = \begin{pmatrix} 11 & 8 \\ 4 & 3 \end{pmatrix}$. The $(2,1)$ entry gives $4 + a = 0$, so $a = -4$; the $(1,1)$ entry then gives $11 - 12 + b = 0$, so $b = 1$. (The remaining entries confirm.) From $A^2 - 4A + I = O$: $I = 4A - A^2 = A(4I - A)$, so $A^{-1} = 4I - A = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}$.

5-mark questions

Q9. For $A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}$, find $A^{-1}$ and use it to solve $$2x - 3y + 5z = 11, \qquad 3x + 2y - 4z = -5, \qquad x + y - 2z = -3$$

Solution. $\vert A \vert = 2(0) + 3(-2) + 5(1) = -1 \neq 0$. Computing cofactors and transposing: $\operatorname{adj} A = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix}$, so $A^{-1} = -\operatorname{adj} A = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}$. Then $$X = A^{-1}B = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}\begin{pmatrix} 11 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$ Hence $x = 1$, $y = 2$, $z = 3$. Check: $2 - 6 + 15 = 11$. ✓

Q10. Solve by the matrix method: $$x - y + 2z = 7, \qquad 3x + 4y - 5z = -5, \qquad 2x - y + 3z = 12$$

Solution. $\vert A \vert = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8) = 7 + 19 - 22 = 4 \neq 0$ — unique solution. Computing $\operatorname{adj} A$ and dividing by $4$, then multiplying into $B = (7, -5, 12)^{\prime}$: $$X = A^{-1}B = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$ Hence $x = 2$, $y = 1$, $z = 3$. Check: $2 - 1 + 6 = 7$ ✓ and $4 - 1 + 9 = 12$ ✓.

Q11. For $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{pmatrix}$, show that $A^3 - 6A^2 + 5A + 11I = O$, and hence find $A^{-1}$.

Solution. $A^2 = \begin{pmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{pmatrix}$; computing $A^3 = A^2 A$ and substituting, every entry of $A^3 - 6A^2 + 5A + 11I$ vanishes. ∎ Rearranging: $A(A^2 - 6A + 5I) = -11I$, so $$A^{-1} = -\frac{1}{11}(A^2 - 6A + 5I) = \frac{1}{11}\begin{pmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{pmatrix}$$

Q12. Solve by the matrix method: $$2x + y + z = 1, \qquad x - 2y - z = \frac{3}{2}, \qquad 3y - 5z = 9$$

Solution. $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{pmatrix}$; $\vert A \vert = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) = 26 + 5 + 3 = 34 \neq 0$. The matrix method gives $$X = A^{-1}\begin{pmatrix} 1 \\ \frac{3}{2} \\ 9 \end{pmatrix} = \begin{pmatrix} 1 \\ \frac{1}{2} \\ -\frac{3}{2} \end{pmatrix}$$ Hence $x = 1$, $y = \frac{1}{2}$, $z = -\frac{3}{2}$. Check the third equation: $3\left(\frac{1}{2}\right) - 5\left(-\frac{3}{2}\right) = \frac{3}{2} + \frac{15}{2} = 9$. ✓ (Fractional answers are common in this pattern — present them as exact fractions, never decimals.)