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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q33
(a) If A = [0, 2, 1; -2, -1, -2; 1, -1, 0], find A^-1 and use it to solve the following system of equations: -2 y + z =7,2 x- y - z =8, x-2 y =10
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0\end{array}\right]$, find $\displaystyle \mathrm{A}^{-1}$ and use it to solve the following system of equations: $\displaystyle -2 \mathrm{y}+\mathrm{z}=7,2 x-\mathrm{y}-\mathrm{z}=8, x-2 \mathrm{y}=10$
If $\displaystyle \left[\begin{array}{ccc}3 & -1 & \sin 3 x \\ -7 & 4 & \cos 2 x \\ -11 & 7 & 2\end{array}\right]$ is a singular matrix, then find all values of $\displaystyle x$ where $\displaystyle x \in\left[0, \frac{\pi}{2}\right]$.
Official answer
(a)
A⁻¹ = [[-$\displaystyle 2$, -$\displaystyle 1$, -$\displaystyle 3$], [$\displaystyle 2$, $\displaystyle 1$, $\displaystyle 2$], [$\displaystyle 4$, $\displaystyle 2$, $\displaystyle 4$]]; x = $\displaystyle 2$, y = -$\displaystyle 1$, z = $\displaystyle 5$ OR (b) x = π/$\displaystyle 6$, $\displaystyle 5$π/$\displaystyle 6$
Marking-scheme solution
$\displaystyle |A|=-2(2)+1(3)=-1 \neq 0$
$\displaystyle \operatorname{adj} A=\begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix}$
$\displaystyle A^{-1}=\dfrac{1}{|A|} \operatorname{adj} A=\begin{bmatrix} 2 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix}$
Given system of equations is equivalent to $\displaystyle A^{T} X=B$, where $\displaystyle X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=\begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix}$
$\displaystyle X=\left(A^{T}\right)^{-1} B=\left(A^{-1}\right)^{T} B$
$\displaystyle \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix}\begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix}=\begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix}$
$\displaystyle \therefore x=0, y=-5, z=-3$
Given matrix is singular, hence, its determinant $\displaystyle =0$,
i.e., $\displaystyle \begin{vmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{vmatrix}=0$
$\displaystyle \Rightarrow 2 \cos 2 x+\sin 3 x=2$
$\displaystyle \Rightarrow 4 \sin^{3} x+4 \sin^{2} x-3 \sin x=0$
$\displaystyle \Rightarrow \sin x\left(4 \sin^{2} x+4 \sin x-3\right)=0$
$\displaystyle \Rightarrow \sin x(2 \sin x-1)(2 \sin x+3)=0$
Solving to get $\displaystyle x=0, \dfrac{\pi}{6}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.