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If A=[500050005]A=\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix} such that A5=λ2AA^5=\lambda^2 A, then the value of λ\lambda is

Solution

✅ Correct Option: 3

Here A=5IA=5I, so A5=55I=3125 IA^5=5^5I=3125\,I and λ2A=5λ2I\lambda^2A=5\lambda^2 I.

Comparing, 5λ2=31255\lambda^2=3125, so λ2=625\lambda^2=625 and λ=±25\lambda=\pm25.

Since the equation only fixes λ2\lambda^2, both signs are admissible.

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