例題 $A = \left(\begin{array}{rrr}
1&2&3\\
2&-1&1\\
4&3&2
\end{array}\right)$ の正則性を判定し, 正則ならば逆行列を求めよう.

解答

$\displaystyle [A:I]$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrrrr}
1&2&3&1&0&0\\
2&-1&1&0&1&0\\
4&3&2&...
...y}{rrrrrr}
1&2&3&1&0&0\\
0&-5&-5&-2&1&0\\
0&-5&-10&-4&0&1
\end{array}\right )$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrrrr}
1&2&3&1&\!\!0&\!\!0\\
0&1&1&\frac{2}...
...rac{1}{5}&\!\!0\\
0&1&2&\frac{4}{5}&\!\!0&\!\!-\frac{1}{5}
\end{array}\right )$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrrrr}
1&0&0&-\frac{1}{5}&\frac{1}{5}&\frac{...
...
0&0&1&\frac{2}{5}&\frac{1}{5}&-\frac{1}{5}
\end{array}\right ) = [I:A^{-1}] .$  

よって, $A$ は正則で, $A^{-1} = \left(\begin{array}{rrr}
-\frac{1}{5}&\frac{1}{5}&\frac{1}{5}\\
0&-\frac{2}{5}&\frac{1}{5}\\
\frac{2}{5}&\frac{1}{5}&-\frac{1}{5}
\end{array}\right )$ である.