$(A^3)^{-1} = (A^{-1})^3$, where $A$ is a square matrix and $|A| \neq 0$.

> [!TLDR] > The property $(A^3)^{-1} = (A^{-1})^3$ holds for any invertible square matrix because matrix inversion reverses the order of multiplication, and raising a matrix to a power commutes with inversion when the exponent is a positive integer. The statement is **true**. The core idea here is simple once you see it: the inverse of a product is the product of inverses in reverse order. That rule, $(AB)^{-1} = B^{-1}A^{-1}$, is the engine behind this entire problem. When you cube a matrix, you're just multiplying it by itself three times — so the inverse of that cube must undo each multiplication, one at a time, in the opposite order. Let’s walk through it carefully. 1. **Start with the definition of $A^3$.** $A^3$ means $A \cdot A \cdot A$. Since matrix multiplication is associative, we can group it as $(A \cdot A) \cdot A$ or $A \cdot (A \cdot A)$ — both are the same. 2. **Apply the inverse-of-a-product rule.** For any two invertible matrices $X$ and $Y$, we know $(XY)^{-1} = Y^{-1}X^{-1}$. This is not commutative — order matters. So for $A^3 = A \cdot A \cdot A$, treat it as $(A \cdot A) \cdot A$ first: $$(A^3)^{-1} = ( (A \cdot A) \cdot A )^{-1} = A^{-1} \cdot (A \cdot A)^{-1}$$ 3. **Now apply the rule again to $(A \cdot A)^{-1}$.** $$(A \cdot A)^{-1} = A^{-1} \cdot A^{-1}$$ Substituting back: $$(A^3)^{-1} = A^{-1} \cdot (A^{-1} \cdot A^{-1}) = A^{-1} \cdot A^{-1} \cdot A^{-1}$$ 4. **Recognize the result.** The product $A^{-1} \cdot A^{-1} \cdot A^{-1}$ is exactly $(A^{-1})^3$, by definition of a matrix power. So we have: $$(A^3)^{-1} = (A^{-1})^3$$ …

(A³)⁻¹ = (A⁻¹)³, where A is a square matrix and |A| ≠ 0.

NCERT Exemplar · Q48

(A^3)^{-1} = (A^{-1})^3

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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 property $(A^3)^{-1} = (A^{-1})^3$ holds for any invertible square matrix because matrix inversion reverses the order of multiplication, and raising a matrix to a power commutes with inversion when the exponent is a positive integer. The statement is true.

The core idea here is simple once you see it: the inverse of a product is the product of inverses in reverse order. That rule, $(AB)^{-1} = B^{-1}A^{-1}$ , is the engine behind this entire problem. When you cube a matrix, you're just multiplying it by itself three times — so the inverse of that cube must undo each multiplication, one at a time, in the opposite order.

Let’s walk through it carefully.

Start with the definition of $A^3$ .

$A^3$ means $A \cdot A \cdot A$ . Since matrix multiplication is associative, we can group it as $(A \cdot A) \cdot A$ or $A \cdot (A \cdot A)$ — both are the same.
Apply the inverse-of-a-product rule.

For any two invertible matrices $X$ and $Y$ , we know $(XY)^{-1} = Y^{-1}X^{-1}$ . This is not commutative — order matters.

So for $A^3 = A \cdot A \cdot A$ , treat it as $(A \cdot A) \cdot A$ first:

(A^3)^{-1} = ( (A \cdot A) \cdot A )^{-1} = A^{-1} \cdot (A \cdot A)^{-1}

Now apply the rule again to $(A \cdot A)^{-1}$ .

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

Substituting back:

(A^3)^{-1} = A^{-1} \cdot (A^{-1} \cdot A^{-1}) = A^{-1} \cdot A^{-1} \cdot A^{-1}

Recognize the result. The product $A^{-1} \cdot A^{-1} \cdot A^{-1}$ is exactly $(A^{-1})^3$ , by definition of a matrix power. So we have: $(A^3)^{-1} = (A^{-1})^3$ …

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Q. $(A^3)^{-1} = (A^{-1})^3$ , where $A$ is a square matrix and $|A| \neq 0$ .

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

Method: Proving an Identity Involving Powers and Inverses of a Matrix

Steps

Common Mistakes

Q.(A3)−1=(A−1)3(A^3)^{-1} = (A^{-1})^3(A3)−1=(A−1)3, where AAA is a square matrix and ∣A∣≠0|A| \neq 0∣A∣=0.

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Q. $(A^3)^{-1} = (A^{-1})^3$ , where $A$ is a square matrix and $|A| \neq 0$ .

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