Let A be a square matrix of order 3. If \(\mathrm{adj}(3A) = k\,\mathrm{adj}\,A\), then the value of k is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
Each entry of the adjoint is a cofactor, and for a matrix of order 3 every cofactor is a 2 × 2 determinant.
When every entry of A is multiplied by 3, each 2 × 2 determinant gets two factors of 3:
\(\text{cofactor in } 3A = 3^{2} \times \text{cofactor in } A\)
\(\mathrm{adj}(3A) = 3^{2}\,\mathrm{adj}\,A = 9\,\mathrm{adj}\,A\)
In general \(\mathrm{adj}(kA) = k^{\,n - 1}\,\mathrm{adj}\,A\) for a matrix of order n; here n = 3, so k = 9.
Why the other options are incorrect
Option (1) 3 — Incorrect. Each cofactor is a 2 × 2 determinant, so the factor 3 appears twice, not once.
Option (2) 27 — Incorrect. \(3^{3} = 27\) is the factor for the determinant \(|3A|\), not for the adjoint.
Option (4) \(\displaystyle\frac{1}{3}\) — Incorrect. Multiplying the entries by 3 makes the cofactors larger, not smaller.