If \(A^{-1}=\frac{-1}{2}\left[\begin{array}{cc}5 & 8 \\ -1 & 2\end{array}\right]\), then \(2 A+I_2=\), where \(I_2\) is a unit matrix of order 2

\(
\begin{aligned}
& \mathrm{A}^{-1}=\frac{-1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{cc}
\frac{-1}{2} & 2 \\
\frac{1}{2} & -1
\end{array}\right] \\
& \mathrm{AA}^{-1}=\mathrm{I} \Rightarrow \mathrm{A}\left[\begin{array}{cc}
\frac{-1}{2} & 2 \\
\frac{1}{2} & -1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{R}_2 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \\
& \mathrm{~A}\left[\begin{array}{ll}
\frac{-1}{2} & 2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& \mathrm{R}_1 \rightarrow \mathrm{R}_1+2 \mathrm{R}_2 \\
& \mathrm{~A}\left[\begin{array}{cc}
\frac{-1}{2} & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\
& \mathrm{R}_1 \rightarrow-2 \mathrm{R}_1 \\
& \mathrm{~A}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
1 & 1
\end{array}\right] \Rightarrow \mathrm{A}=\left[\begin{array}{ll}
2 & 4 \\
1 & 1
\end{array}\right] \\
& \therefore 2 \mathrm{~A}+\mathrm{I}_2=\left[\begin{array}{ll}
4 & 8 \\
2 & 2
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
5 & 8 \\
2 & 3
\end{array}\right]
\end{aligned}
\)