माना कि \(\mathrm{A}=\left[\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\)
अब \(|\mathrm{A}|=\left|\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|\)
\(=2(3-0)-0 \div(-1)(5-0)\)
\(=6-5\)
\(=1\)
\(\Rightarrow|A| \neq 0\)
अत: A invertible होगा। हम जानते हैं कि,
\(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\)
अब
\(\mathrm{C}_{11}=3, \mathrm{C}_{12}=-15, \mathrm{C}_{13}=5\)
\(\mathrm{C}_{21}=-1, \mathrm{C}_{22}=6, \mathrm{C}_{23}=-2\)
\(\mathrm{C}_{31}=1, \mathrm{C}_{72}=-5, \mathrm{C}_{33}=2\)
\(\operatorname{adj} A=\left[\begin{array}{lll}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33}\end{array}\right]^T=\left[\begin{array}{lll}C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33}\end{array}\right]\)
\(=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
\(\therefore A^{-1}=\frac{\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]}{1}=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
