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