If $A=\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array} ight]$, then the inverse of $\left(2 A^2+5 A ight)$ is

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If $A=\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array}ight]$, then the inverse of $\left(2 A^2+5 Aight)$ is

Aicharya · Accepted Answer

$\begin{aligned} A & =\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array}ight] \ & A^2=\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array}ight]\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array}ight] \ 	herefore \quad A^2 & =\left[\begin{array}{cc}5 & -5 \ -5 & 10\end{array}ight] \ 	ext { Now, } 2 A^2+5 A & =2\left[\begin{array}{cc}5 & -5 \ -5 & 10\end{array}ight]+5\left[\begin{array}{cc}2 & -1 \ -1 & 3\end{array}ight] \ & =\left[\begin{array}{cc}20 & -15 \ -15 & 35\end{array}ight] \ 	ext { If } A & =\left[\begin{array}{ll}a & b \ c & d\end{array}ight], \operatorname{then~} A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \ -c & a\end{array}ight] \ 	herefore \quad\left(2 A^2+5 Aight)^{-1} & =\frac{1}{475}\left[\begin{array}{cc}35 & 15 \ 15 & 20\end{array}ight] \ & =\frac{1}{95}\left[\begin{array}{ll}7 & 3 \ 3 & 4\end{array}ight]\end{aligned}$

If $A=\left[\begin{array}{cc}2 & -1 \\ -1 & 3\end{array}\right]$, then the inverse of $\left(2 A^2+5…

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