Let A be a non-singular square matrix of order 3. Consider the following statements:
(A) \(|A^{T}| = |A|\)
(B) \(|-A| = |A|\)
(C) \(|\mathrm{adj}\,A| = |A|^{3}\)
(D) \(\displaystyle |A^{-1}| = \frac{1}{|A|}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
For a square matrix A of order n = 3:
\(|A^{T}| = |A|\)
\(|kA| = k^{3}|A|\), so \(|-A| = (-1)^{3}|A| = -|A|\)
\(|\mathrm{adj}\,A| = |A|^{n-1} = |A|^{2}\)
From \(AA^{-1} = I\), the product rule \(|AB| = |A|\,|B|\) gives \(|A|\,|A^{-1}| = 1\).
So \(|A^{T}| = |A|\), \(|-A| = -|A|\), \(|\mathrm{adj}\,A| = |A|^{2}\) and \(|A^{-1}| = 1/|A|\).
Checking the statements
(A) \(|A^{T}| = |A|\) — Correct. Transposing does not change a determinant.
(B) \(|-A| = |A|\) — Incorrect. \(|-A| = (-1)^{3}|A| = -|A|\), and \(|A| \ne 0\).
(C) \(|\mathrm{adj}\,A| = |A|^{3}\) — Incorrect. The power is \(n - 1 = 2\), so \(|\mathrm{adj}\,A| = |A|^{2}\).
(D) \(|A^{-1}| = 1/|A|\) — Correct. \(|A|\,|A^{-1}| = |I| = 1\).