Let A be any square matrix of order n, then which of the following are true?
(A) \(|\mathrm{adj}\, A| = |A|^{n-1}\)
(B) \(\displaystyle|A^{-1}|=\frac{1}{|A|}\)
(C) \(|\mathrm{adj}\, A| = |A|^n\)
(D) \((A^T)^{-1}= (A^{-1})^T\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
Statements (B) and (D) involve \(A^{-1}\), so they are about an invertible matrix A, that is, \(|A| \ne 0\).
Checking the statements
(A) \(|\mathrm{adj}\, A| = |A|^{n-1}\) — Correct. Taking determinants in \(A(\mathrm{adj}\,A) = |A|\,I\), where \(|A|\,I\) is a scalar matrix of order n:
\( \displaystyle |A|\,|\mathrm{adj}\,A| = |A|^n \)
\( \displaystyle |\mathrm{adj}\,A| = |A|^{n-1} \)
(B) \(\displaystyle|A^{-1}|=\frac{1}{|A|}\) — Correct. Taking determinants in \(AA^{-1} = I\):
\(|A|\,|A^{-1}| = |I| = 1\)
(C) \(|\mathrm{adj}\, A| = |A|^n\) — Incorrect. The power is n − 1, as in (A). For example, \(|\mathrm{adj}\,(2I_2)| = |2I_2| = 4\), while \(|2I_2|^2 = 16\).
(D) \((A^T)^{-1}= (A^{-1})^T\) — Correct. By the reversal law for transposes:
\(A^T(A^{-1})^T = (A^{-1}A)^T = I^T = I\)
So \((A^{-1})^T\) is the inverse of \(A^T\).