Let \(A\) be a non singular matrix of order \(n \times n\), then |adj (3A)| is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3^{n(n-1)}|A|^{n-1}\)
Two standard results for a square matrix B of order n and a scalar k:
\(|\mathrm{adj}\,B| = |B|^{n-1}\)
\(|kB| = k^n|B|\)
Taking \(B = 3A\):
\(|\mathrm{adj}(3A)| = |3A|^{n-1}\)
\(= \left(3^n|A|\right)^{n-1}\)
\(= 3^{n(n-1)}|A|^{n-1}\)
Why the other options are incorrect
Option (1) \(3|A|^{n-1}\) — Incorrect. This is \(3|\mathrm{adj}\,A|\). The factor 3 multiplies every one of the n rows of 3A, so it does not come out as a single 3.
Option (3) \(3^n|A|^{n-1}\) — Incorrect. This is \(|3\,\mathrm{adj}\,A|\). But \(\mathrm{adj}(3A) = 3^{n-1}\mathrm{adj}\,A\), not \(3\,\mathrm{adj}\,A\).
Option (4) \(3^n|A|^n\) — Incorrect. Since \(|\mathrm{adj}\,B| = |B|^{n-1}\), the power of \(|A|\) in \(|\mathrm{adj}(3A)|\) is n − 1, not n.