\(A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1\end{array}\right]\) and \(A^{-1}=\frac{1}{2}\left[\begin{array}{ccc}1 & -1 & 1 \\ -8 & 6 & 2 c \\ 5 & -3 & 1\end{array}\right]\), then values of \(a\) and \(\mathrm{c}\) are respectively

We know that \(\mathrm{AA}^{-1}=1\)
\(
\begin{aligned}
& \therefore\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & \mathrm{a} & 1
\end{array}\right]\left[\begin{array}{ccc}
\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\
-4 & 3 & \mathrm{c} \\
\frac{5}{2} & \frac{-3}{2} & \frac{1}{2}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \therefore\left[\begin{array}{ccc}
1 & 0 & \mathrm{c}+1 \\
0 & 1 & 2+2 \mathrm{c} \\
4-4 \mathrm{a} & 3 \mathrm{a}-3 & 2+\mathrm{ac}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
\)
Thus \(\mathrm{c}+1 \Rightarrow \mathrm{c}=-1\) and \(4-4 \mathrm{a}=0 \Rightarrow \mathrm{a}=1\)