Which of the following statements are correct ?
(A) If the product of 2 non-zero square matrices is a zero matrix, then either A is singular or B is singular
(B) If A is a non singular matrix, then \( |A^{-1}|=\displaystyle\frac{1}{|A|} \)
(C) If A is an invertible square matrix then \( \mathrm{adj}\,A^T=\displaystyle\frac{1}{\mathrm{adj}\,A} \)
(D) If A and B are invertible matrices then \( (AB)^{-1}=B^{-1}A^{-1} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Let AB = O with A ≠ O and B ≠ O. If A were non-singular, then
\( \displaystyle B=A^{-1}(AB)=A^{-1}O=O \)
This contradicts B ≠ O, so A is singular; in the same way B is singular.
For a non-singular A, taking determinants in \( AA^{-1}=I \):
\( \displaystyle |A|\,|A^{-1}|=|I|=1 \)
\( \displaystyle |A^{-1}|=\frac{1}{|A|} \)
For invertible A and B of the same order (reversal law for inverses):
\( \displaystyle (AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1} \)
\( \displaystyle =AA^{-1}=I \)
Also, division by a matrix is not defined; the correct result for the adjoint of a transpose is \( \mathrm{adj}(A^T)=(\mathrm{adj}\,A)^T \).
So if AB = O with A, B non-zero, both A and B are singular; \( |A^{-1}|=\frac{1}{|A|} \); \( (AB)^{-1}=B^{-1}A^{-1} \); and \( \frac{1}{\mathrm{adj}\,A} \) has no meaning.
Checking the statements
(A) If the product of 2 non-zero square matrices is a zero matrix, then either A is singular or B is singular — Correct. In fact both A and B are singular.
(B) If A is a non singular matrix, then \( |A^{-1}|=\frac{1}{|A|} \) — Correct. \( |A|\,|A^{-1}|=|I|=1 \).
(C) If A is an invertible square matrix then \( \mathrm{adj}\,A^T=\frac{1}{\mathrm{adj}\,A} \) — Incorrect. Division by a matrix is not defined; \( \mathrm{adj}(A^T)=(\mathrm{adj}\,A)^T \).
(D) If A and B are invertible matrices then \( (AB)^{-1}=B^{-1}A^{-1} \) — Correct. \( (AB)(B^{-1}A^{-1})=I \).