Board 2025 Class 12 Maths Q15 Solution: Which property of invertible matrices is not correct?

The question

If A and B are invertible matrices, then which of the following is not correct ?

(A) $(A + B)^{-1} = B^{-1} + A^{-1}$

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

(C) $\text{adj}\,(A) = |A|\,A^{-1}$

(D) $|A|^{-1} = |A^{-1}|$

Exam answer

(B):

\begin{aligned} &(AB)(B^{-1}A^{-1}) \\ &= A(BB^{-1})A^{-1} \\ &= AIA^{-1} = AA^{-1} = I \end{aligned}

So $(AB)^{-1} = B^{-1}A^{-1}$ ⁠: (B) is correct.

Why?

To show a matrix is the inverse of AB, multiply it by AB and get I. Matrix multiplication is associative, so we may regroup the brackets, and $BB^{-1} = I$ ⁠.

The order reverses: the last matrix applied is the first one undone.

(C): $A^{-1} = \frac{1}{|A|}\,\text{adj}\,A$ ⁠, so $\text{adj}\,A = |A|\,A^{-1}$ ⁠. (C) is correct.

Why?

The formula for the inverse comes from $A\,(\text{adj}\,A) = |A|\,I$ ⁠, which holds for every square matrix. Multiplying both sides of the formula by $|A|$ gives (C).

(D):

\begin{aligned} AA^{-1} = I &\Rightarrow |A|\,|A^{-1}| = |I| = 1 \\ &\Rightarrow |A^{-1}| = \frac{1}{|A|} = |A|^{-1} \end{aligned}

So (D) is correct.

Why?

The determinant of a product is the product of the determinants, $|PQ| = |P|\,|Q|$ ⁠, and $|I| = 1$ ⁠. Since A is invertible, $|A| \ne 0$ ⁠, so we may divide by it.

(A): take $A = B = I$ ⁠, which are invertible.

\begin{aligned} (A + B)^{-1} &= (2I)^{-1} = \frac{1}{2}I \\ B^{-1} + A^{-1} &= I + I = 2I \end{aligned}

$\frac{1}{2}I \ne 2I$ ⁠, so (A) is not correct.

Why?

To show a general rule is false, one counterexample is enough. $(2I)(\frac12 I) = I$ ⁠, so $(2I)^{-1} = \frac12 I$ ⁠.

In fact $A + B$ need not even be invertible: for $B = -A$ ⁠, $A + B = O$ ⁠.

Answer: (A) $(A + B)^{-1} = B^{-1} + A^{-1}$

Where this comes from

Chapter 4, Determinants: Properties of the inverse and adjoint.

Question type: One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

How to spot it: click to see

Type 10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Here: No matrix is given, so test each option against the cards: $(AB)^{-1} = B^{-1}A^{-1}$ ⁠, $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ and $|A^{-1}| = \frac{1}{|A|}$ all hold. (A) is no card, and $A = B = I$ breaks it.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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Q15: Which property of invertible matrices is not correct?