For a square matrix \(A_{n \times n}\)
(A) \(|\mathrm{adj}\,A| = |A|^{n-1}\)
(B) \(|A| = |\mathrm{adj}\,A|^{n-1}\)
(C) \(A(\mathrm{adj}\,A) = |A|\)
(D) \(\left|A^{-1}\right| = \displaystyle\frac{1}{|A|}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
For a square matrix A of order n, the basic property of the adjoint is
\(A(\mathrm{adj}\,A) = (\mathrm{adj}\,A)A = |A|\,I\)
The right side is a matrix (|A| times the identity matrix), not a number. Taking determinants, with \(\left|\,|A|\,I\,\right| = |A|^n\):
\(|A|\,|\mathrm{adj}\,A| = |A|^n\)
\(|\mathrm{adj}\,A| = |A|^{n-1}\)
When \(A^{-1}\) exists, \(AA^{-1} = I\) gives
\(|A|\,\left|A^{-1}\right| = |I| = 1\)
\( \displaystyle \left|A^{-1}\right| = \frac{1}{|A|} \)
So \(|\mathrm{adj}\,A| = |A|^{n-1}\), \(A(\mathrm{adj}\,A) = |A|\,I\) (a matrix) and \(\left|A^{-1}\right| = \frac{1}{|A|}\).
Checking the statements
(A) \(|\mathrm{adj}\,A| = |A|^{n-1}\) — Correct. Taking determinants of \(A(\mathrm{adj}\,A) = |A|\,I\) gives \(|A|\,|\mathrm{adj}\,A| = |A|^n\).
(B) \(|A| = |\mathrm{adj}\,A|^{n-1}\) — Incorrect. \(|\mathrm{adj}\,A|^{n-1} = |A|^{(n-1)^2}\); for n = 3 and |A| = 2 this is 16, not 2.
(C) \(A(\mathrm{adj}\,A) = |A|\) — Incorrect. The left side is a matrix and |A| is a number; the correct result is \(A(\mathrm{adj}\,A) = |A|\,I\).
(D) \(\left|A^{-1}\right| = \frac{1}{|A|}\) — Correct. \(|A|\,\left|A^{-1}\right| = |I| = 1\).