If A be a square matrix of order 3 such that \(|A| = 2\), then \(|\mathrm{adj}\,(2A)|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 256
For a square matrix of order n: \(|kA| = k^n|A|\) and \(|\mathrm{adj}\,B| = |B|^{n-1}\). Here n = 3:
\(|2A| = 2^3|A| = 8 \times 2 = 16\)
\(|\mathrm{adj}\,(2A)| = |2A|^{3-1} = 16^2 = 256\)
Why the other options are incorrect
Option (1) 16 — Incorrect. This is \(|2A|\) itself, not the determinant of its adjoint.
Option (2) 32 — Incorrect. This is \(|2\,\mathrm{adj}\,A| = 2^3 \times |A|^2 = 8 \times 4\). But \(\mathrm{adj}\,(kA) = k^{n-1}\,\mathrm{adj}\,A\), so \(\mathrm{adj}\,(2A) = 4\,\mathrm{adj}\,A\), not \(2\,\mathrm{adj}\,A\).
Option (3) 64 — Incorrect. \(|4\,\mathrm{adj}\,A| = 4^3 \times |\mathrm{adj}\,A| = 64 \times 4 = 256\); 64 leaves out the factor \(|\mathrm{adj}\,A| = |A|^2 = 4\).