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

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