(a) If $A = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}$, then find $AB$ and use it to solve the following system of equations: $x - 2y = 3$ $2x - y - z = 2$ $-2y + z = 3$ OR (b) If $f(\alpha) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$, prove that $f(\alpha) \cdot f(-\beta) = f(\alpha - \beta)$.

(a) If A = -3 -2 -4 \2 1 2 \2 1 3 and B = 1 2 0 \-2 -1 -2 \0 -1 1 , then find AB and use it to solve the follo

Question

A = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}

(OR)

(b) If

f(\alpha) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}

, prove that

f(\alpha) \cdot f(-\beta) = f(\alpha - \beta)

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

Part (b)Concept understanding — Rotation Matrix Composition

Rotation Matrix Composition

Hold a book, rotate it so the cover faces you, then tilt it to look down at the spine. The final orientation depends on which turn you do first. That is the whole idea: composing rotations means applying one after another, and in 3D the order matters.

Each rotation is a matrix, and doing two in sequence corresponds to multiplying their matrices. The subtlety is that matrix multiplication is generally not commutative: $AB \neq BA$ .

The Rule

R_1

Read right to left: a rotation acts on a column vector as $R\mathbf{v}$ , so applying $R_1$ then $R_2$ gives $R_2(R_1\mathbf{v}) = (R_2 R_1)\mathbf{v}$ . The rotation done first sits on the right.

A 2D Example

In the plane, a counter-clockwise rotation by angle $\theta$ is

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.

Here $R(60^\circ)\,R(30^\circ) = R(90^\circ)$ : the angles simply add. In 2D rotations do commute, because they all share the same axis (the axis pointing out of the plane).

Why 3D Is Different

In 3D, rotations are about different axes, and swapping the order changes the result. Use the standard right-handed matrices

R_x(90^\circ) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \qquad R_y(90^\circ) = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix},

and track the vector $\mathbf{v} = (0,0,1)$ .

Order 1 — $R_x$ first, then $R_y$ (product $R_y R_x$ ):

$R_x(90^\circ)\mathbf{v} = (0,-1,0)$ — the point moves onto the $y$ -axis.
$R_y$ leaves the $y$ -axis fixed, so the result stays $(0,-1,0)$ .

Order 2 — $R_y$ first, then $R_x$ (product $R_x R_y$ ):

$R_y(90^\circ)\mathbf{v} = (1,0,0)$ — the point moves onto the $x$ -axis.
$R_x$ leaves the $x$ -axis fixed, so the result stays $(1,0,0)$ .

Watch out

Same two rotations, opposite order, different final point: $(0,-1,0)$ versus $(1,0,0)$ . So $R_x R_y \neq R_y R_x$ in general. (Notice each rotation leaves its own axis unmoved — a rotation about the $y$ -axis can never shift a vector that already lies on the $y$ -axis.) …

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Part (a): $AB=I$ , so $B=A^{-1}$ ; the system's coefficient matrix is $B^{T}$ , giving $X=A^{T}C$ and $x=1,\ y=-1,\ z=1$ . Part (b): multiplying $f(\alpha)$ by $f(-\beta)$ and applying the cosine/sine difference identities yields $f(\alpha-\beta)$ .

Part (a)

1. Compute $AB$ .

A=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix},\quad B=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}.

Row-by-column multiplication gives every diagonal entry $1$ and every off-diagonal entry $0$ :

AB=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I,

so $B=A^{-1}$ (equivalently $A=B^{-1}$ ).

2. Matrix form of the system.

\begin{cases}x-2y+0z=3\\2x-y-z=2\\0x-2y+z=3\end{cases}\Rightarrow MX=C,\quad M=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix},\ C=\begin{bmatrix}3\\2\\3\end{bmatrix}.

The rows of $M$ are the columns of $B$ , i.e. $M=B^{T}$ .

3. Solve using $AB=I$ . Since $M=B^{T}$ and $B^{-1}=A$ ,

X=M^{-1}C=(B^{T})^{-1}C=(B^{-1})^{T}C=A^{T}C.

With $A^{T}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}$ ,

$X=\begin{bmatrix}-9+4+6\\-6+2+3\\-12+4+9\end{bmatrix}=\begin{bmatrix}1\\-1\\1\end{bmatrix}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$
(a) 9
(b) 0
(c) 4
(d) 5
›Reveal solutionSolution

Two matrices are equal only if all corresponding entries are equal; adding all three entry-equations gives $x+y+z$ directly.

From $\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ , equating corresponding entries:

…

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CBSE 2026Set ANNUAL1 markMCQ
$A = \begin{bmatrix} x & 0 \\ 2 & 3 \end{bmatrix}$
(a) $x = 4$
(b) $x = \pm3$
(c) $x = -3$
(d) $x = -4$
›Reveal solutionSolution

Computing $A^2$ and matching it to $9I$ forces both $x^2=9$ and $2x+6=0$ ; only $x=-3$ satisfies both.

$A=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}$ , so

$A^2=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}=\begin{bmatrix}x^2 & 0\\ 2x+6 & 9\end{bmatrix}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If [[x-2y, 0], [5, x]] = [[-3, 0], [5, 3]], then y is equal to:$
(a) 1
(b) 3
(c) 2
(d) 4
›Reveal solutionSolution

Two matrices are equal only if all corresponding entries are equal; comparing the $(2,2)$ entries gives $x=3$ , then the $(1,1)$ entries give $y$ .

Given:

$\begin{bmatrix}x-2y & 0\\ 5 & x\end{bmatrix} = \begin{bmatrix}-3 & 0\\ 5 & 3\end{bmatrix}$

Comparing the $(2,2)$ entries: $x = 3$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(i) $-2$
(ii) $-1$
(iii) $2$
(iv) $1$
›Reveal solutionSolution

Multiply out the matrices and compare entries.

$\begin{bmatrix}1&3\\2&1\end{bmatrix}\begin{bmatrix}1\\x\end{bmatrix}=\begin{bmatrix}1(1)+3(x)\\2(1)+1(x)\end{bmatrix}=\begin{bmatrix}1+3x\\2+x\end{bmatrix}$

Setting this equal to $\begin{bmatrix}7\\4\end{bmatrix}$ :

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $x = -\dfrac{1}{3}, y = 7$
(b) Not possible to find
(c) $x = -\dfrac{2}{3}, y = 7$
(d) $x = -\dfrac{1}{3}, y = -\dfrac{2}{3}$
›Reveal solutionSolution

Equating corresponding entries gives two different equations for x that contradict each other, so no consistent solution exists.

For $\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$ , equating each entry:

$3x+7 = 0 \Rightarrow x = -\dfrac{7}{3}$

$5 = y-2 \Rightarrow y = 7$

$y+1 = 8 \Rightarrow y = 7$ (consistent with above)

$2-3x = 4 \Rightarrow x = -\dfrac{2}{3}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If [[x-2y, 0], [5, x]] = [[-5, 0], [5, 3]], then y is equal to:$
(a) 1
(b) 3
(c) 2
(d) 4
›Reveal solutionSolution

Equal matrices have equal corresponding entries — match the (2,2) entries first to get $x$ , then use the (1,1) entry to get $y$ .

Given $\begin{pmatrix} x-2y & 0 \\ 5 & x\end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 5 & 3\end{pmatrix}$ .

Comparing the (2,2) entries: $x = 3$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. If [[a+4, 3b], [8, -14]] = [[2a+2, b+4], [8, a-8b]], then find the value of a + b.$
›Reveal solutionSolution

Equate corresponding entries of the two equal matrices to get $a=2$ , $b=2$ , so $a+b=4$ .

Two matrices are equal only if every corresponding entry is equal. Comparing entries of

$\begin{bmatrix} a+4 & 3b \\ 8 & -14 \end{bmatrix} = \begin{bmatrix} 2a+2 & b+4 \\ 8 & a-8b \end{bmatrix}$ :

From the (1,1) entries: $a+4 = 2a+2 \Rightarrow 2 = a \Rightarrow a = 2$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If A = [[2x, 0], [x, x]] and A⁻¹ = [[1, 0], [-1, 2]], then x equals -$
(i) 1
(ii) 2
(iii) 1/2
(iv) −2
›Reveal solutionSolution

Compute $A^{-1}$ from $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$ using the $2\times2$ inverse formula and match it to the given $A^{-1}$ .

For $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$ , $\det A = (2x)(x)-(0)(x)=2x^2$ .

Using $A^{-1}=\dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ :

$A^{-1}=\dfrac{1}{2x^2}\begin{pmatrix}x&0\\-x&2x\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2x}&0\\-\tfrac{1}{2x}&\tfrac{1}{x}\end{pmatrix}$ .

…

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CBSE 20241 markMCQ
$\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X$
›Reveal solutionSolution
We solve the matrix equation $A = BX$ by left-multiplying both sides by $B^{-1}$ , giving $X = B^{-1}A$ . Computing the inverse of $B = \begin{bmatrix}1&2\\3&1\end{bmatrix}$ and multiplying yields $X = \begin{bmatrix}2&0\\3&7\end{bmatrix}$ , which matches option (C).

The core idea here is that a matrix equation like $A = BX$ is solved exactly like the scalar equation $a = bx$ — you isolate $X$ by multiplying both sides by the inverse of $B$ . But because matrix multiplication is not commutative, you must multiply on the left by $B^{-1}$ , not on the right. That single detail is the entire key.

Let’s walk through it.

Set up the equation clearly. We are given

$\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X.$
Call the left matrix $A$ and the coefficient matrix $B$ , so $A = BX$ . Our job is to find $X$ .

Why left-multiplication by $B^{-1}$ works. If $B$ is invertible, then $B^{-1}B = I$ , the identity matrix. Multiplying both sides of $A = BX$ on the left by $B^{-1}$ gives

$B^{-1}A = B^{-1}(BX) = (B^{-1}B)X = IX = X.$
So $X = B^{-1}A$ . Notice: if we had multiplied on the right instead, we’d get $AB^{-1}$ , which is a completely different (and wrong) matrix.

Watch out

A common mistake is to write $X = A B^{-1}$ by analogy with scalars. But matrix multiplication is not commutative — $B^{-1}A \neq A B^{-1}$ in general. Always multiply on the side where the inverse cancels the original matrix.

Find $B^{-1}$ . For a $2 \times 2$ matrix $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the inverse is

$B^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix},$
provided the determinant $ad - bc \neq 0$ .

Here $a=1$ , $b=2$ , $c=3$ , $d=1$ . The determinant is
$\det(B) = (1)(1) - (2)(3) = 1 - 6 = -5.$
So
$B^{-1} = \frac{1}{-5} \begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix}.$

Multiply $B^{-1}A$ . Now $A = \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}$ . Compute $X = B^{-1}A$ :

$X = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix} \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}.$
Multiply entry by entry: …
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CBSE 2024Set D1 markMCQ
$2A + B + X = 0$
(a) $\begin{bmatrix} 1 & 2 \\ -7 & -13 \end{bmatrix}$
(b) $\begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix}$
(c) $\begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}$
(d) $\begin{bmatrix} -1 & -2 \\ 7 & 13 \end{bmatrix}$
›Reveal solutionSolution

From $2A + B + X = 0$ , solve $X = -2A - B$ .

$2A = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix}$ , so …

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CBSE 2024Set D1 markMCQ
$[\,x\ \ y\,] = [\,2x - 1\ \ 9\,] \Rightarrow$
(a) $x = 3,\ y = 9$
(b) $x = 1,\ y = 9$
(c) $x = 0,\ y = 9$
(d) $x = 3,\ y = 4$
›Reveal solutionSolution

Equal matrices have equal corresponding entries.

…

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CBSE 2024Set ANNUAL1 markMCQ
$\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
(a) $5$
(b) $7$
(c) $9$
(d) none of these
›Reveal solutionSolution

Matching the first row of the given matrix equation reads off x+y+z directly, no further algebra needed.

The matrix equation $\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}$ means corresponding entries are equal:

Row 1: $x+y+z=9$ …

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Part (a)

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