If $A$ is a matrix of order $3 \times 3$, then $(A^2)^{-1} = $ ________ .

> [!TLDR] > The key idea is that the inverse of a product reverses the order, and the inverse of a square is the square of the inverse. So $(A^2)^{-1} = (A^{-1})^2$. ## Why this approach works The problem asks for the inverse of $A^2$. When you square a matrix and then take its inverse, you're essentially undoing two applications of $A$ at once. The natural question: does the inverse of $A^2$ equal $(A^{-1})^2$? Yes — and the reason comes from the fundamental property of inverses: the inverse of a product is the product of inverses in reverse order. > [!FORMULA] > For any invertible matrices $P$ and $Q$, $(PQ)^{-1} = Q^{-1} P^{-1}$. Apply this to $A^2 = A \cdot A$. Since both factors are the same matrix, the reverse order doesn't change anything — you get $A^{-1} A^{-1}$, which is just $(A^{-1})^2$. ## Step-by-step reasoning 1. **Write $A^2$ as a product.** $A^2 = A \cdot A$. This is just the matrix multiplied by itself. 2. **Apply the inverse-of-product rule.** Using $(PQ)^{-1} = Q^{-1} P^{-1}$ with $P = A$ and $Q = A$, we get: $$(A^2)^{-1} = (A \cdot A)^{-1} = A^{-1} \cdot A^{-1}$$ 3. **Simplify the result.** $A^{-1} \cdot A^{-1}$ is exactly $(A^{-1})^2$, the square of the inverse matrix. …

If A is a matrix of order 3 × 3, then (A²)⁻¹ = ____ .

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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.

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The key idea is that the inverse of a product reverses the order, and the inverse of a square is the square of the inverse. So $(A^2)^{-1} = (A^{-1})^2$ .

Why this approach works

The problem asks for the inverse of $A^2$ . When you square a matrix and then take its inverse, you're essentially undoing two applications of $A$ at once. The natural question: does the inverse of $A^2$ equal $(A^{-1})^2$ ? Yes — and the reason comes from the fundamental property of inverses: the inverse of a product is the product of inverses in reverse order.

P

Apply this to $A^2 = A \cdot A$ . Since both factors are the same matrix, the reverse order doesn't change anything — you get $A^{-1} A^{-1}$ , which is just $(A^{-1})^2$ .

Step-by-step reasoning

Write $A^2$ as a product.

$A^2 = A \cdot A$ . This is just the matrix multiplied by itself.
Apply the inverse-of-product rule.

Using $(PQ)^{-1} = Q^{-1} P^{-1}$ with $P = A$ and $Q = A$ , we get:

(A^2)^{-1} = (A \cdot A)^{-1} = A^{-1} \cdot A^{-1}

Simplify the result. $A^{-1} \cdot A^{-1}$ is exactly $(A^{-1})^2$ , the square of the inverse matrix. …

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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 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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Q.If $A$ is a matrix of order $3 \times 3$ , then $(A^2)^{-1} =$ ________ .

Concept understanding — Inverse of a Product

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

Why the order must reverse

A concrete example with numbers

Why this approach works

Step-by-step reasoning

Method: Applying the Inverse-Reversal Rule to a Power of a Matrix

Steps

Concept and Intuition

Step-by-Step Derivation

Q.If AAA is a matrix of order 3×33 \times 33×3, then (A2)−1=(A^2)^{-1} = (A2)−1= ________ .

Why this approach works

Step-by-step reasoning

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Q.If $A$ is a matrix of order $3 \times 3$ , then $(A^2)^{-1} =$ ________ .

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