If A is a non singular matrix of order 3, then
(A) \( |\,\mathrm{adj}\,\mathrm{A}|=|\,\mathrm{A}\,|^{2} \)
(B) If AB = AC, then B \( \ne \) C, where B and C are square matrices of order 3.
(C) \( (\mathrm{AB})^{-1}=\mathrm{B}^{-1}\mathrm{A}^{-1} \), where B is a non singular matrix of order 3.
(D) \( \left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}\mathrm{I}_{3} \) (Where \( \mathrm{I}_{3} \) is an identity matrix of order 3)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
For a matrix of order n, \( |\mathrm{adj}\,\mathrm{A}|=|\mathrm{A}|^{n-1} \); here n = 3.
Since A is non singular, \( \mathrm{A}^{-1} \) exists. Multiplying AB = AC by \( \mathrm{A}^{-1} \) on the left,
\( \displaystyle \mathrm{B}=\mathrm{C} \)
For non singular A and B, the reversal law gives \( (\mathrm{AB})^{-1}=\mathrm{B}^{-1}\mathrm{A}^{-1} \), and \( (\mathrm{A}^{-1})^{-1}=\mathrm{A} \).
So \( |\mathrm{adj}\,\mathrm{A}|=|\mathrm{A}|^{2} \), AB = AC forces B = C, \( (\mathrm{AB})^{-1}=\mathrm{B}^{-1}\mathrm{A}^{-1} \) and \( (\mathrm{A}^{-1})^{-1}=\mathrm{A} \).
Checking the statements
(A) \( |\,\mathrm{adj}\,\mathrm{A}|=|\,\mathrm{A}\,|^{2} \) — Correct. \( |\mathrm{A}|^{n-1}=|\mathrm{A}|^{2} \) for n = 3.
(B) If AB = AC, then B \( \ne \) C — Incorrect. Multiplying by \( \mathrm{A}^{-1} \) gives B = C.
(C) \( (\mathrm{AB})^{-1}=\mathrm{B}^{-1}\mathrm{A}^{-1} \) — Correct. This is the reversal law, and A and B are both non singular.
(D) \( \left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}\mathrm{I}_{3} \) — Correct. \( (\mathrm{A}^{-1})^{-1}=\mathrm{A} \), and \( \mathrm{A}\mathrm{I}_{3}=\mathrm{A} \).