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If A2=8A+kI,A=[10−17],I=[1001]A^2 = 8A + kI, A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} then value of kk is:

Solution

✅ Correct Option: 3

A2=[10−849]A^2=\begin{bmatrix}1&0\\-8&49\end{bmatrix} and 8A=[80−856]8A=\begin{bmatrix}8&0\\-8&56\end{bmatrix}.

So A2−8A=[−700−7]=−7IA^2-8A=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}=-7I, giving k=−7k=-7.

This also matches Cayley-Hamilton: trace =8=8, det⁡=7\det=7, so A2−8A+7I=OA^2-8A+7I=O.

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