If \( A=\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix} \) such that \( A^5=\lambda^2A \), then the value of \( \lambda \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \pm 25 \)
A is a scalar matrix: \( A=5\mathrm{I} \). Since \( \mathrm{I}^n=\mathrm{I} \),
\( \displaystyle A^5=(5\mathrm{I})^5=5^5\,\mathrm{I} \)
\( \displaystyle =5^4\cdot(5\mathrm{I})=625A \)
\( \displaystyle \lambda^2A=625A\ \Rightarrow\ \lambda^2=625 \)
\( \displaystyle \lambda=\pm 25 \)
Nothing in the question says \( \lambda\gt 0 \), so both signs are kept.
Why the other options are incorrect
Option (1) 25 — Incorrect. \( \lambda=25 \) satisfies \( \lambda^2=625 \), but so does \( \lambda=-25 \); the equation has both roots.
Option (2) -25 — Incorrect. \( \lambda=-25 \) is only one of the two roots of \( \lambda^2=625 \).
Option (4) 125 — Incorrect. 125 gives \( \lambda^2=15625 \), not 625.