If A and B are invertible matrices, then which of the following is not correct? (A) operatornameadj A = |A| ·

NCERT Exemplar · Q34

A

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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 while the inverse of a product reverses the order, the inverse of a sum does not distribute like that — so option (D) is the false statement.

We need to test each option against known properties of invertible matrices. Let’s go through them one by one.

Option (A): $\operatorname{adj} A = |A| \cdot A^{-1}$

This is a standard formula. For any invertible matrix $A$ , the adjugate satisfies $A \cdot \operatorname{adj} A = |A| I$ . Multiplying both sides by $A^{-1}$ gives $\operatorname{adj} A = |A| A^{-1}$ . So this is correct.
Option (B): $\det(A)^{-1} = [\det(A)]^{-1}$

This is just notation. The left side means the determinant of $A^{-1}$ , and the right side means the reciprocal of $\det(A)$ . Since $\det(A^{-1}) = 1/\det(A)$ for invertible $A$ , they are equal. So this is correct.
Option (C): $(AB)^{-1} = B^{-1} A^{-1}$

This is the fundamental reversal property for inverses of products. Multiply $AB$ on the right by $B^{-1} A^{-1}$ : $(AB)(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1} = A I A^{-1} = I$ . So indeed $B^{-1} A^{-1}$ is the inverse of $AB$ . Correct.
Option (D): $(A + B)^{-1} = B^{-1} + A^{-1}$ …

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Q.If $A$ and $B$ are invertible matrices, then which of the following is not correct?
(A) $\operatorname{adj} A = |A| \cdot A^{-1}$
(B) $\det(A)^{-1} = [\det(A)]^{-1}$
(C) $(AB)^{-1} = B^{-1} A^{-1}$
(D) $(A + B)^{-1} = B^{-1} + A^{-1}$

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

Method: Testing Matrix-Inverse Identity Statements

Steps

Common Mistakes

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