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If AA and BB are square matrices of the same order, then which of the following statements are correct?

(A) ∣A−1∣=∣A∣−1|A^{-1}| = |A|^{-1}

(B) adj(A)=∣A∣A−1adj(A) = |A|A^{-1}

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

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

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

Starting from the identity every invertible matrix satisfies:

A⋅A−1=IA \cdot A^{-1} = I

Taking determinant on both sides using ∣XY∣=∣X∣⋅∣Y∣|XY| = |X| \cdot |Y|:

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

∣A−1∣=1∣A∣=∣A∣−1|A^{-1}| = \dfrac{1}{|A|} = |A|^{-1}

Statement (A) is correct. ✓


The inverse of a matrix is given by:

A−1=adj(A)∣A∣A^{-1} = \dfrac{\text{adj}(A)}{|A|}

∣A∣⋅A−1=adj(A)|A| \cdot A^{-1} = \text{adj}(A)

Statement (B) is correct. ✓


Inverse does not distribute over addition.

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

(I+I)−1=(2I)−1=12I(I + I)^{-1} = (2I)^{-1} = \dfrac{1}{2}I

I−1+I−1=I+I=2II^{-1} + I^{-1} = I + I = 2I

12I≠2I\dfrac{1}{2}I \neq 2I

Statement (C) is incorrect. ✗


To verify (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}, the product (AB)(B−1A−1)(AB)(B^{-1}A^{-1}) should equal II:

(AB)(B−1A−1)(AB)(B^{-1}A^{-1})

=A(B⋅B−1)A−1= A(B \cdot B^{-1})A^{-1}

=A⋅I⋅A−1= A \cdot I \cdot A^{-1}

=AA−1= AA^{-1}

=I= I

Statement (D) is correct. ✓


Statements (A), (B), and (D) are correct → Option 1

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