If \( A=\begin{bmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{bmatrix} \), \( \lambda\ne 0 \), then \( |\mathrm{adj}\,A| \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \lambda^{6} \)
A is a diagonal (scalar) matrix, so its determinant is the product of the diagonal entries:
\( \displaystyle |A|=\lambda\cdot\lambda\cdot\lambda=\lambda^3 \)
For a square matrix of order n, \( |\mathrm{adj}\,A|=|A|^{n-1} \). Here n = 3:
\( \displaystyle |\mathrm{adj}\,A|=(\lambda^3)^{2}=\lambda^6 \)
Why the other options are incorrect
Option (1) \( \lambda^{27} \) — Incorrect. \( \lambda^{27}=(\lambda^3)^9 \) uses the power 9 instead of n − 1 = 2.
Option (2) \( \lambda^{3} \) — Incorrect. \( \lambda^3 \) is |A|, not |adj A|.
Option (3) \( \lambda^{9} \) — Incorrect. \( \lambda^9=(\lambda^3)^3 \) uses the power n = 3 instead of n − 1 = 2.