(a) If $A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$, find $(AB)^{-1}$. OR (b) Solve the following system of equations by matrix method: $x + 2y + 3z = 6$, $2x - y + z = 2$, $3x + 2y - 2z = 3$.

(a) If A = 1 2 -2 \-1 3 0 \0 -2 1 and B⁻¹ = 3 -1 1 \-15 6 -5 \5 -2 2 , find (AB)⁻¹. OR (b) Solve the following

Question

A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}

(OR)

(b) Solve the following system of equations by matrix method:

x + 2y + 3z = 6

2x - y + z = 2

3x + 2y - 2z = 3

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Part (a)Concept understanding — Inverse of a Product

Inverse of a Product: The "Socks and Shoes" Principle

You put on your socks first, then your shoes. To take them off, you can't remove the socks while the shoes are still on — you must reverse the order: shoes off first, then socks.

That's exactly the inverse of a product of matrices. If you apply transformation $A$ first, then $B$ , the combined effect is $BA$ (read right-to-left: $A$ acts first, then $B$ ). To undo it, undo $B$ first, then $A$ :

(AB)^{-1} = B^{-1}A^{-1}

The order flips — forced by the logic of undoing.

Why the order must reverse

Check that $B^{-1}A^{-1}$ is the inverse of $AB$ . We need $(AB)(B^{-1}A^{-1}) = I$ and $(B^{-1}A^{-1})(AB) = I$ :

(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A I A^{-1} = AA^{-1} = I

$B$ and $B^{-1}$ cancel first, leaving $A$ and $A^{-1}$ to cancel. The other check works the same way:

(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1}B = I

If you tried $(AB)^{-1} = A^{-1}B^{-1}$ instead:

(AB)(A^{-1}B^{-1}) = A(BA^{-1})B^{-1}

and $BA^{-1}$ is not $I$ — the matrices are in the wrong order. So the reversal is essential.

Watch out

A common mistake is writing $(AB)^{-1} = A^{-1}B^{-1}$ . This is false unless $A$ and $B$ commute (which they almost never do). Always flip the order.

A concrete example with numbers

Let $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ , with inverses:

A^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}, \quad B^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}

Then:

AB = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}, \qquad (AB)^{-1} = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}

Now compute $B^{-1}A^{-1}$ :

B^{-1}A^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}

They match. Try $A^{-1}B^{-1}$ and you'll get a different matrix — the wrong answer.

--- …

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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Part (a): $(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}10&7&21\\-49&-34&-103\\17&12&36\end{bmatrix}$ . Part (b): solving $AX=B$ gives $x=1,\ y=1,\ z=1$ .

Part (a)

We use the reversal law $(AB)^{-1}=B^{-1}A^{-1}$ . We are given $B^{-1}$ , so we only need $A^{-1}$ .

1. $\det A$ . For $A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}$ ,

\det A=1(3\cdot1-0)-2((-1)\cdot1-0)+(-2)((-1)(-2)-0)=3+2-4=1.

2. Adjoint / inverse. The cofactor matrix is $\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}$ ; its transpose is the adjoint, and since $\det A=1$ ,

A^{-1}=\operatorname{adj}A=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}.

3. Multiply $B^{-1}A^{-1}$ . …

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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 65/2/11 markMCQ
$A$
›Reveal solutionSolution
The inverse of a scalar multiple of a matrix, $(kA)^{-1}$ , is equal to $\frac{1}{k}A^{-1}$ . For $(3A)^{-1}$ , this means the result is $\boxed{\frac{1}{3}A^{-1}}$ .

Concept and Intuition

The inverse of a square matrix $M$ , denoted $M^{-1}$ , is defined such that when $M$ is multiplied by $M^{-1}$ , the result is the identity matrix $I$ . That is, $MM^{-1} = M^{-1}M = I$ . The identity matrix acts like the number $1$ in scalar multiplication: $MI = IM = M$ .

When we consider a scalar multiple of a matrix, say $kA$ , we are essentially scaling every element of the matrix $A$ by the scalar $k$ . If we want to find the inverse of this new matrix $(kA)$ , we need to find a matrix that, when multiplied by $kA$ , yields the identity matrix $I$ .

Intuitively, if $A^{-1}$ "undoes" the operation of $A$ , and $k$ "scales" $A$ , then to "undo" $kA$ , we would need to "un-scale" by $\frac{1}{k}$ and then "un-matrix" by $A^{-1}$ . This suggests that the inverse of $kA$ should involve $\frac{1}{k}$ and $A^{-1}$ .

Let's verify this intuition formally.

Step-by-Step Derivation

Recall the definition of an inverse matrix:

For any invertible square matrix $M$ , its inverse $M^{-1}$ satisfies the property $MM^{-1} = I$ , where $I$ is the identity matrix of the same dimension.

Apply the definition to $(3A)$ :

We are looking for $(3A)^{-1}$ . Let's denote this unknown inverse as $X$ . By definition, $X$ must satisfy:

$(3A)X = I$

Propose a form for $X$ based on intuition:

As discussed in the concept section, we expect $X$ to be of the form $c A^{-1}$ for some scalar $c$ . Let's substitute this into the equation:

$(3A)(cA^{-1}) = I$

Use properties of scalar and matrix multiplication:

For any scalars $k_1, k_2$ and matrices $M_1, M_2$ , we know that $(k_1 M_1)(k_2 M_2) = (k_1 k_2)(M_1 M_2)$ . Applying this property:

$(3c)(AA^{-1}) = I$

Substitute $AA^{-1} = I$ : …
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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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CBSE 2022Set TERM11 markMCQ
$Q. If A and B are invertible matrices then$
(a) (AB)⁻¹ = B⁻¹A⁻¹
(b) (AB)⁻¹ = A⁻¹B⁻¹
(c) (AB)⁻¹ = (BA)⁻¹
(d) None of these
›Reveal solutionSolution

The inverse of a product reverses the order of the factors.

…

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Part (a)

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