If $\mathrm{A}\left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array} ight]$ then $\left(\mathrm{A}^2-5 \mathrm{~A} ight)^{-1}$ is

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If $\mathrm{A}\left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight]$ then $\left(\mathrm{A}^2-5 \mathrm{~A}ight)^{-1}$ is

Aicharya · Accepted Answer

$\begin{aligned} & A=\left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight] \ 	herefore & A^2=\left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight] 	imes\left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight]=\left[\begin{array}{cc}11 & 6 \ 42 & 23\end{array}ight] \ 	herefore & A^2-5 A=\left[\begin{array}{ll}1 & 1 \ 7 & 3\end{array}ight] \ & \ 	herefore & \ 	herefore & \left(A^2-5 A \mid=3-7=-4ight. \ & \ & =\frac{1}{4}\left[\begin{array}{cc}-3 & 1 \ 7 & -1\end{array}ight]\end{aligned}$

If $\mathrm{A}\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$ then $\left(\mathrm{A}^2-5…

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