$A = \begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix}$
$A^{-1} = \begin{pmatrix}x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}$
因为 $A \cdot A^{-1} = I = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$
所以 $\begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix}x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$
$\left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array} \right) -> \left( \begin{array}{cc|cc} 1 & 0 & x_{11} & x_{12} \\ 0 & 1 & x_{21} & x_{22} \end{array} \right)$
原矩阵高斯乔丹消元法如果能得出单位矩阵，则右侧的矩阵就是逆矩阵
$\left( \begin{array}{c|c} A & I \end{array} \right) -> \left( \begin{array}{c|c} I & A^{-1} \end{array} \right)$
解法：$\left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array} \right) -> \left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array} \right) -> \left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & 1.5 & -0.5 \end{array} \right)$
$ ->
\left( \begin{array}{cc|cc} 1 & 0 & -2 & 1 \\ 0 & 1 & 1.5 & -0.5 \end{array} \right)$
因此：$A^{-1} = \begin{pmatrix}-2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$
系数矩阵化为行最简形式时有零行，则无解。