If $A=\left[\begin{array}{cc}1 & 0 \ -1 & 7\end{array} ight], I=\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array} ight]$ and $A^{2}=8 A+k I$, then the value of $K$ is

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If $A=\left[\begin{array}{cc}1 & 0 \ -1 & 7\end{array}ight], I=\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$ and $A^{2}=8 A+k I$, then the value of $K$ is

Aicharya · Accepted Answer

$A^{2}=A 	imes A=\left[\begin{array}{cc} 1 & 0 \ -1 & 7 \end{array}ight]\left[\begin{array}{cc} 1 & 0 \ -1 & 7 \end{array}ight]=\left[\begin{array}{cc} 1 & 0 \ -8 & 49 \end{array}ight]$ Given $A^{2}=8 A+K I$ $\begin{aligned} &=8\left[\begin{array}{cc} 1 & 0 \ -1 & 7 \end{array}ight]+\mathrm{k}\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight]=\left[\begin{array}{cc} 8 & 0 \ -8 & 56 \end{array}ight]+\left[\begin{array}{cc} \mathrm{k} & 0 \ 0 & \mathrm{k} \end{array}ight] \ 	herefore \quad\left[\begin{array}{cc} 1 & 0 \ -8 & 49 \end{array}ight] &=\left[\begin{array}{cc} 8+\mathrm{k} & 0 \ -8 & 56+\mathrm{k} \end{array}ight] \Rightarrow 8+\mathrm{k}=1 \Rightarrow \mathrm{k}=-7 \end{aligned}$

If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right], I=\left[\begin{array}{cc}1 & 0 \\ 0 &…

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