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If A and B are two invertible matrices, then which of the following statements are correct?

(A) ∣A−1∣=∣A∣−1|A^{-1}| = |A|^{-1}

(B) adjA=∣A∣A−1adjA = |A|A^{-1}

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

(D) (A+B)−1=A−1+B−1(A + B)^{-1} = A^{-1} + B^{-1}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

For ∣A−1∣=∣A∣−1|A^{-1}| = |A|^{-1}:

Since A⋅A−1=IA \cdot A^{-1} = I, taking determinants:

∣A⋅A−1∣=∣I∣|A \cdot A^{-1}| = |I|

∣A∣⋅∣A−1∣=1|A| \cdot |A^{-1}| = 1

∣A−1∣=1∣A∣|A^{-1}| = \frac{1}{|A|}

∣A−1∣=∣A∣−1|A^{-1}| = |A|^{-1}

Statement (A) is correct.


For adjA=∣A∣A−1adjA = |A|A^{-1}:

The fundamental relation is A⋅(adjA)=∣A∣⋅IA \cdot (adjA) = |A| \cdot I

Multiplying both sides by A−1A^{-1} from the left:

A−1⋅A⋅(adjA)=A−1⋅∣A∣⋅IA^{-1} \cdot A \cdot (adjA) = A^{-1} \cdot |A| \cdot I

I⋅(adjA)=∣A∣⋅A−1I \cdot (adjA) = |A| \cdot A^{-1}

adjA=∣A∣A−1adjA = |A|A^{-1}

Statement (B) is correct.


For (AB)−1=A−1B−1(AB)^{-1} = A^{-1}B^{-1}:

The correct formula is (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} (reverse order).

Checking: (AB)⋅(B−1A−1)=A⋅(B⋅B−1)⋅A−1(AB) \cdot (B^{-1}A^{-1}) = A \cdot (B \cdot B^{-1}) \cdot A^{-1}

=A⋅I⋅A−1= A \cdot I \cdot A^{-1}

=A⋅A−1= A \cdot A^{-1}

=I= I

However, (AB)⋅(A−1B−1)=A⋅(B⋅A−1)⋅B−1(AB) \cdot (A^{-1}B^{-1}) = A \cdot (B \cdot A^{-1}) \cdot B^{-1}, which does not simplify to II since BB and A−1A^{-1} do not commute.

Statement (C) is incorrect.


For (A+B)−1=A−1+B−1(A + B)^{-1} = A^{-1} + B^{-1}:

Taking A=IA = I and B=IB = I:

A+B=2IA + B = 2I

(A+B)−1=(2I)−1=12I(A + B)^{-1} = (2I)^{-1} = \frac{1}{2}I

But A−1+B−1=I+I=2IA^{-1} + B^{-1} = I + I = 2I

Since 12I≠2I\frac{1}{2}I \neq 2I, the formula is incorrect.

Statement (D) is incorrect.


Only statements (A) and (B) are correct.

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