(a) Given $A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}$, find $AB$. Hence, solve the system of equations: $x - y + z = 4$, $\ x - 2y - 2z = 9$, $\ 2x + y + 3z = 1$. OR (b) If $A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}$, find $A^{-1}$. Hence, solve the system of equations: $x - 2y = 10$, $\ 2x - y - z = 8$, $\ -2y + z = 7$.

(a) Given A = -4 4 4 \-7 1 3 \5 -3 -1 and B = 1 -1 1 \1 -2 -2 \2 1 3 , find AB. Hence, solve the system of equ

Question

A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}

(OR)

(b) If

A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}

, find

A^{-1}

. Hence, solve the system of equations:

x - 2y = 10

\ 2x - y - z = 8

\ -2y + z = 7

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Part (a)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

Part (b)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

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$AB=8I$ , so $B^{-1}=\tfrac18A$ and the system $BX=D$ gives $x=3,\ y=-2,\ z=-1$ .
$\det A=1$ and $A^{-1}=\begin{bmatrix}-3&2&4\\-2&1&2\\-2&1&3\end{bmatrix}$ ; the given system solves to $x=0,\ y=-5,\ z=-3$ .

Part (a)

The inverse-matrix method solves $CX=D$ as $X=C^{-1}D$ . The hint "find $AB$ , hence solve" signals that $AB$ collapses to a scalar multiple of $I$ , handing us the inverse cheaply.

Compute $AB$ . Row-by-column multiplication gives, for example, entry $(1,1)=(-4)(1)+(4)(1)+(4)(2)=8$ ; carrying this out for all entries,

AB=\begin{bmatrix}8&0&0\\0&8&0\\0&0&8\end{bmatrix}=8I.

Use it. The system

x-y+z=4,\quad x-2y-2z=9,\quad 2x+y+3z=1

is $BX=D$ where $B=\begin{bmatrix}1&-1&1\\1&-2&-2\\2&1&3\end{bmatrix}$ and $D=\begin{bmatrix}4\\9\\1\end{bmatrix}$ . From $AB=8I$ , multiply on the right by $B^{-1}$ : $A=8B^{-1}$ , hence $B^{-1}=\tfrac18A$ .

Solve. …

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

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) $\det(A)$
(b) $\frac{1}{\det(A)}$
(c) $1$
(d) $0$
›Reveal solutionSolution

$\det(A^{-1})=\dfrac{1}{\det A}$ .

From $AA^{-1}=I$ , $\det(A)\det(A^{-1})=\det(I)=1$ , so $\det(A^{-1})=\dfrac{1}{\det(A)}$ . (This holds for a …

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CBSE 2026Set ANNUAL1 mark
$AX = B$
›Reveal solutionSolution

The unique solution of $AX=B$ (when $|A| e0$ ) is $X=A^{-1}B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
(a) $19$
(b) $\dfrac{1}{19}$
(c) $-19$
(d) $-\dfrac{1}{19}$
›Reveal solutionSolution

Compute |A| and adj(A), form A⁻¹, and compare it entry-by-entry with kA.

$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ , $|A| = 2(-2) - 3(5) = -4-15 = -19$

$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

…

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CBSE 2024Set D1 markMCQ
$A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$
(a) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(b) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(c) $\begin{bmatrix} 4 & 8 \\ 6 & 12 \end{bmatrix}$
(d) $\begin{bmatrix} 4 & 8 \\ -6 & 12 \end{bmatrix}$
›Reveal solutionSolution

For a $2\times2$ matrix $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Here $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ , so $\det A = (2)(6) - (-3)(4) = 12 + 12 = 24.$

The adjoint (swap diagonal, negate off-diagonal) is $\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}$ . Hence …

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Part (a)

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