$ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$
矩阵的某一行乘以一个常数:$ \begin{pmatrix}k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}k \cdot a & k \cdot b & k \cdot c \\ d & e & f \\ g & h & i \end{pmatrix}$
矩阵的一行加(减)另一行:$ \begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}a-g & b-h & c-i \\ d & e & f \\ g & h & i \end{pmatrix}$
矩阵的一行加(减)另一行的若干倍:$ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}a & b & c \\ d-pg & e-ph & f-pi \\ g & h & i \end{pmatrix}$
交换矩阵的两行:$ \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}a & b & c \\ g & h & i \\ d & e & f \end{pmatrix}$ (交换单位矩阵的两行即可)
因此 Gauss-Jordan 消元法把矩阵化为行最简形式的过程,寻找一系列初等矩阵E使得:
$$ E_p \cdot ... \cdot E_3 \cdot E_2 \cdot E_1 \cdot A = rref(A) = I$$
初等矩阵对单位矩阵进行一次初等变换得到
因为初等变换是可逆的,所以初等矩阵是可逆的
$E_1 = \begin{pmatrix}k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} or \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{pmatrix} or \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$
$E_2 = \begin{pmatrix}\frac{1}{k} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} or \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{pmatrix} or \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$
$E_1 \cdot E_2 = I 而且 E_2 \cdot E_1 = I$
根据上面 Gauss-Jordan 消元法: $$ E_p \cdot ... \cdot E_3 \cdot E_2 \cdot E_1 \cdot A = rref(A) = I$$ $$ E_p \cdot ... \cdot E_3 \cdot E_2 \cdot E_1 \cdot A \cdot A^{-1} = I \cdot A^{-1}$$ $$ E_p \cdot ... \cdot E_3 \cdot E_2 \cdot E_1 \cdot I = A^{-1}$$ $$\left( \begin{array}{c|c} A & I \end{array} \right) \longrightarrow \left( \begin{array}{c|c} I & A^{-1} \end{array} \right)$$