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If A=[10−17]A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, then the value of k so that A2=8A+kIA^2 = 8A + kI 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.

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