https://ccjou.wordpress.com/2013/02/22/%E9%AB%98%E6%96%AF%E2%94%80%E7%B4%84%E7%95%B6%E6%B3%95/
高斯─約當法
在解線性方程組的應用上，高斯─約當法[1] (Gauss-Jordan method) 是高斯消去法的延伸 (見“高斯消去法”)，其目的要得到最簡約的列等價方程組。高斯消去法產生梯形矩陣後，我們可以繼續執行取代運算將軸元 (pivot) 上方的元悉數消去，並使用伸縮運算迫使軸元為 
。高斯─約當法產生的矩陣稱為簡約列梯形式 (reduced row echelon form)，由下列四個條件定義 (前兩個條件即為梯形矩陣的性質)：
。高斯─約當法產生的矩陣稱為簡約列梯形式 (reduced row echelon form)，由下列四個條件定義 (前兩個條件即為梯形矩陣的性質)：- 零列置於矩陣最底下。
- 每列軸元的位置都位於其上方各列軸元的右側。
- 軸元等於
- 軸元其上方與下方的元皆為零。
下面列舉兩個簡約列梯形式。數字 表示軸元，每一軸元上方和下方的元皆為零，其他各元 (以 
表示) 可以是任意數：
表示軸元，每一軸元上方和下方的元皆為零，其他各元 (以 表示) 可以是任意數：![\left[\!\begin{array}{ccccc} 1 & \square & 0 & 0 & \square \\ 0 & 0 & 1 & 0 & \square \\ 0 & 0 & 0 & 1 & \square \\ 0 & 0 & 0 & 0 & 0 \end{array}\!\right],~~ \left[\!\begin{array}{ccccccc} 0 & 1 & 0 & \square & 0 & 0 & \square \\ 0 & 0 & 1 & \square & 0 & 0 & \square \\ 0 & 0 & 0 & 0 & 1 & 0 & \square \\ 0 & 0 & 0 & 0 & 0 & 1 & \square \end{array}\!\right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Bccccc%7D++1+%26+%5Csquare+%26+0+%26+0+%26+%5Csquare+%5C%5C++0+%26+0+%26+1+%26+0+%26+%5Csquare+%5C%5C++0+%26+0+%26+0+%26+1+%26+%5Csquare+%5C%5C++0+%26+0+%26+0+%26+0+%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D%2C%7E%7E++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Bccccccc%7D++0+%26+1+%26+0+%26+%5Csquare+%26+0+%26+0+%26+%5Csquare+%5C%5C++0+%26+0+%26+1+%26+%5Csquare+%26+0+%26+0+%26+%5Csquare+%5C%5C++0+%26+0+%26+0+%26+0+%26+1+%26+0+%26+%5Csquare+%5C%5C++0+%26+0+%26+0+%26+0+%26+0+%26+1+%26+%5Csquare++%5Cend%7Barray%7D%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "\left[\!\begin{array}{ccccc} 1 & \square & 0 & 0 & \square \\ 0 & 0 & 1 & 0 & \square \\ 0 & 0 & 0 & 1 & \square \\ 0 & 0 & 0 & 0 & 0 \end{array}\!\right],~~ \left[\!\begin{array}{ccccccc} 0 & 1 & 0 & \square & 0 & 0 & \square \\ 0 & 0 & 1 & \square & 0 & 0 & \square \\ 0 & 0 & 0 & 0 & 1 & 0 & \square \\ 0 & 0 & 0 & 0 & 0 & 1 & \square \end{array}\!\right]")
。
Carl Friedrich Gauss (1777-1855) From Wikimedia
我們用一個例子說明高斯─約當法的計算過程。先使用高斯消去法將給定的增廣矩陣 (代表線性方程組) 化簡至梯形矩陣 (運算細節省略)：
![\left[\!\begin{array}{rrrrcr} 0 & 2 & -4 & 1 &\vline& -6 \\ -1 & 0 & -5 & 1 &\vline& -4 \\ 1 & 2 & 1 & 0 &\vline& -2 \\ 2 & -4 & 18 & -7 &\vline& 14 \end{array}\!\right]\to\left[\!\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & -3 &\vline& -6 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++0+%26+2+%26+-4+%26+1+%26%5Cvline%26+-6+%5C%5C++-1+%26+0+%26+-5+%26+1+%26%5Cvline%26+-4+%5C%5C++1+%26+2+%26+1+%26+0+%26%5Cvline%26+-2+%5C%5C++2+%26+-4+%26+18+%26+-7+%26%5Cvline%26+14++%5Cend%7Barray%7D%5C%21%5Cright%5D%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+1+%26%5Cvline%26+-4+%5C%5C++0+%26+2+%26+-4+%26+1+%26%5Cvline%26+-6+%5C%5C++0+%26+0+%26+0+%26+-3+%26%5Cvline%26+-6+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "\left[\!\begin{array}{rrrrcr} 0 & 2 & -4 & 1 &\vline& -6 \\ -1 & 0 & -5 & 1 &\vline& -4 \\ 1 & 2 & 1 & 0 &\vline& -2 \\ 2 & -4 & 18 & -7 &\vline& 14 \end{array}\!\right]\to\left[\!\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & -3 &\vline& -6 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\!\right]")
接著繼續對梯形矩陣執行基本列運算。以伸縮運算使最底下的軸元為 ，再消去該軸元以上的所有元，沿用此方式由下而上直到產生簡約列梯形式，如下：
，再消去該軸元以上的所有元，沿用此方式由下而上直到產生簡約列梯形式，如下：![\begin{array}{llll} &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & -3 &\vline& -6 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \\ &&\\ \rightarrow &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 0 &\vline& -8 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 0 &\vline& -6 \\ 0 & 2 & -4 & 0 &\vline& -8 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \\ &&\\ \rightarrow &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 0 &\vline& -6 \\ 0 & 1 & -2 & 0 &\vline& -4 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{ccrccr} 1 & 0 & 5 & 0 &\vline& 6 \\ 0 & 1 & -2 & 0 &\vline& -4 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bllll%7D++%26%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+1+%26%5Cvline%26+-4+%5C%5C++0+%26+2+%26+-4+%26+1+%26%5Cvline%26+-6+%5C%5C++0+%26+0+%26+0+%26+-3+%26%5Cvline%26+-6+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D%26%5Crightarrow+%26++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+1+%26%5Cvline%26+-4+%5C%5C++0+%26+2+%26+-4+%26+1+%26%5Cvline%26+-6+%5C%5C++0+%26+0+%26+0+%26+1+%26%5Cvline%26+2+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D+%5C%5C++%26%26%5C%5C++%5Crightarrow++%26%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+1+%26%5Cvline%26+-4+%5C%5C++0+%26+2+%26+-4+%26+0+%26%5Cvline%26+-8+%5C%5C++0+%26+0+%26+0+%26+1+%26%5Cvline%26+2+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D%26%5Crightarrow+%26++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+0+%26%5Cvline%26+-6+%5C%5C++0+%26+2+%26+-4+%26+0+%26%5Cvline%26+-8+%5C%5C++0+%26+0+%26+0+%26+1+%26%5Cvline%26+2+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D+%5C%5C++%26%26%5C%5C++%5Crightarrow++%26%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrrcr%7D++-1+%26+0+%26+-5+%26+0+%26%5Cvline%26+-6+%5C%5C++0+%26+1+%26+-2+%26+0+%26%5Cvline%26+-4+%5C%5C++0+%26+0+%26+0+%26+1+%26%5Cvline%26+2+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D%26%5Crightarrow+%26++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Bccrccr%7D++1+%26+0+%26+5+%26+0+%26%5Cvline%26+6+%5C%5C++0+%26+1+%26+-2+%26+0+%26%5Cvline%26+-4+%5C%5C++0+%26+0+%26+0+%26+1+%26%5Cvline%26+2+%5C%5C++0+%26+0+%26+0+%26+0+%26%5Cvline%26+0++%5Cend%7Barray%7D%5C%21%5Cright%5D++%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{llll} &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & -3 &\vline& -6 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 1 &\vline& -6 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \\ &&\\ \rightarrow &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 1 &\vline& -4 \\ 0 & 2 & -4 & 0 &\vline& -8 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 0 &\vline& -6 \\ 0 & 2 & -4 & 0 &\vline& -8 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \\ &&\\ \rightarrow &\left[\!\begin{array}{rrrrcr} -1 & 0 & -5 & 0 &\vline& -6 \\ 0 & 1 & -2 & 0 &\vline& -4 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{ccrccr} 1 & 0 & 5 & 0 &\vline& 6 \\ 0 & 1 & -2 & 0 &\vline& -4 \\ 0 & 0 & 0 & 1 &\vline& 2 \\ 0 & 0 & 0 & 0 &\vline& 0 \end{array}\!\right] \end{array}")
與高斯消去法得到的梯形矩陣比較，簡約列梯形式的主要價值在於不需要使用反向代入法，當下便可決定線性方程組的解。上例中，從簡約列梯形式具有完整的零列可斷定此方程組是一致的，第一、二、四行為軸行，故
是軸變數，
是自由變數。寫出對應簡約列梯形式的線性方程組
令 
，移項後立刻得到通解
，移項後立刻得到通解
Wilhelm Jordan (1842-1899) From http://upload.wikimedia.org/wikipedia/commons/7/70/Wilhelm_Jordan.png
對於任意給定矩陣，雖然存在許多不同的列等價梯形矩陣，但是萬法歸一，不論基本列運算的執行方式或次序為何，簡約列梯形式總是唯一的，這是因為簡約列梯形式與方程組的解具有一對一的關係 (證明見“簡約列梯形式的唯一性”)。就這層意義來說，列等價的簡約列梯形式是一種典型的矩陣形式。提醒讀者，如果我們的問題僅須解線性方程組，在一般情況下，使用高斯消去法再以反向代入求解的計算效率略優於高斯—約當法直接得到簡約列梯形式。雖然如此，高斯—約當法自有一個特別的應用──計算一個
階方陣的逆矩陣。令
![A=[a_{ij}]](https://s0.wp.com/latex.php?latex=A%3D%5Ba_%7Bij%7D%5D&bg=ffffff&fg=000000&s=0 "A=[a_{ij}]")
為 
階矩陣，![b=[b_i]](https://s0.wp.com/latex.php?latex=b%3D%5Bb_i%5D&bg=ffffff&fg=000000&s=0 "b=[b_i]")
為 維向量，寫出增廣矩陣![\begin{bmatrix} A\vert\mathbf{b} \end{bmatrix}=\left[\!\!\begin{array}{cccccc} a_{11}&a_{12}&\cdots&a_{1n}&\vline&b_1\\ a_{21}&a_{22}&\cdots&a_{2n}&\vline&b_2\\ \vdots&\vdots&\ddots&\vdots&\vline&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}&\vline&b_n \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D++A%5Cvert%5Cmathbf%7Bb%7D++%5Cend%7Bbmatrix%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcccccc%7D++a_%7B11%7D%26a_%7B12%7D%26%5Ccdots%26a_%7B1n%7D%26%5Cvline%26b_1%5C%5C++a_%7B21%7D%26a_%7B22%7D%26%5Ccdots%26a_%7B2n%7D%26%5Cvline%26b_2%5C%5C++%5Cvdots%26%5Cvdots%26%5Cddots%26%5Cvdots%26%5Cvline%26%5Cvdots%5C%5C++a_%7Bn1%7D%26a_%7Bn2%7D%26%5Ccdots%26a_%7Bnn%7D%26%5Cvline%26b_n++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "\begin{bmatrix} A\vert\mathbf{b} \end{bmatrix}=\left[\!\!\begin{array}{cccccc} a_{11}&a_{12}&\cdots&a_{1n}&\vline&b_1\\ a_{21}&a_{22}&\cdots&a_{2n}&\vline&b_2\\ \vdots&\vdots&\ddots&\vdots&\vline&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}&\vline&b_n \end{array}\!\!\right]")
。
若 
是可逆的，則 
恆有唯一解，基本列運算最終可將增廣矩陣化簡為
是可逆的，則 恆有唯一解，基本列運算最終可將增廣矩陣化簡為![\begin{bmatrix} I\vert\mathbf{c} \end{bmatrix}=\left[\!\!\begin{array}{cccccc} 1&0&\cdots&0&\vline&c_1\\ 0&1&\cdots&0&\vline&c_2\\ \vdots&\vdots&\ddots&\vdots&\vline&\vdots\\ 0&0&\cdots&1&\vline&c_n \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D++I%5Cvert%5Cmathbf%7Bc%7D++%5Cend%7Bbmatrix%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcccccc%7D++1%260%26%5Ccdots%260%26%5Cvline%26c_1%5C%5C++0%261%26%5Ccdots%260%26%5Cvline%26c_2%5C%5C++%5Cvdots%26%5Cvdots%26%5Cddots%26%5Cvdots%26%5Cvline%26%5Cvdots%5C%5C++0%260%26%5Ccdots%261%26%5Cvline%26c_n++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "\begin{bmatrix} I\vert\mathbf{c} \end{bmatrix}=\left[\!\!\begin{array}{cccccc} 1&0&\cdots&0&\vline&c_1\\ 0&1&\cdots&0&\vline&c_2\\ \vdots&\vdots&\ddots&\vdots&\vline&\vdots\\ 0&0&\cdots&1&\vline&c_n \end{array}\!\!\right]")
。
線性方程組 的解即為最右行，
，
。若 的簡約列梯形式不等於單位矩陣 
，便可判定 是不可逆的。令 
為 階矩陣使得 
，也就是說，是 的逆矩陣 
。將 和 以行向量表示：
且 
，其中 
代表第 
元等於 的單位向量，則有
的解即為最右行，，。若 的簡約列梯形式不等於單位矩陣 ，便可判定 是不可逆的。令 為 階矩陣使得 ，也就是說，是 的逆矩陣 。將 和 以行向量表示： 且 ，其中 代表第 元等於 的單位向量，則有
。
因此，矩陣方程 可分解成 個線性方程 
。分別對 個增廣矩陣 
執行高斯─約當法，可得
可分解成 個線性方程 。分別對 個增廣矩陣 執行高斯─約當法，可得![\begin{aligned} \begin{bmatrix} A\vert\mathbf{e}_1 \end{bmatrix}&\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\vert\mathbf{x}_1 \end{bmatrix}\\ &\vdots\\ \begin{bmatrix} A\vert\mathbf{e}_n \end{bmatrix}&\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\vert\mathbf{x}_n \end{bmatrix}.\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D++%5Cbegin%7Bbmatrix%7D++A%5Cvert%5Cmathbf%7Be%7D_1++%5Cend%7Bbmatrix%7D%26%5Cxrightarrow%5B%5D%7B%7E%7E%5Cmathrm%7BGauss-Jordan%7D%7E%7E%7D%5Cbegin%7Bbmatrix%7D++I%5Cvert%5Cmathbf%7Bx%7D_1++%5Cend%7Bbmatrix%7D%5C%5C++%26%5Cvdots%5C%5C++%5Cbegin%7Bbmatrix%7D++A%5Cvert%5Cmathbf%7Be%7D_n++%5Cend%7Bbmatrix%7D%26%5Cxrightarrow%5B%5D%7B%7E%7E%5Cmathrm%7BGauss-Jordan%7D%7E%7E%7D%5Cbegin%7Bbmatrix%7D++I%5Cvert%5Cmathbf%7Bx%7D_n++%5Cend%7Bbmatrix%7D.%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned} \begin{bmatrix} A\vert\mathbf{e}_1 \end{bmatrix}&\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\vert\mathbf{x}_1 \end{bmatrix}\\ &\vdots\\ \begin{bmatrix} A\vert\mathbf{e}_n \end{bmatrix}&\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\vert\mathbf{x}_n \end{bmatrix}.\end{aligned}")
但因為這 個增廣矩陣 擁有相同的係數矩陣 ，我們可以將它們合併成一個大增廣矩陣 
。直接以高斯─約當法化簡大增廣矩陣，即得
個增廣矩陣 擁有相同的係數矩陣 ，我們可以將它們合併成一個大增廣矩陣 。直接以高斯─約當法化簡大增廣矩陣，即得![\begin{bmatrix} A\!\!&\vert&\!\! I \end{bmatrix}\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\!\!&\vert&\!\! X \end{bmatrix}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D++A%5C%21%5C%21%26%5Cvert%26%5C%21%5C%21+I++%5Cend%7Bbmatrix%7D%5Cxrightarrow%5B%5D%7B%7E%7E%5Cmathrm%7BGauss-Jordan%7D%7E%7E%7D%5Cbegin%7Bbmatrix%7D++I%5C%21%5C%21%26%5Cvert%26%5C%21%5C%21+X++%5Cend%7Bbmatrix%7D&bg=ffffff&fg=000000&s=0 "\begin{bmatrix} A\!\!&\vert&\!\! I \end{bmatrix}\xrightarrow[]{~~\mathrm{Gauss-Jordan}~~}\begin{bmatrix} I\!\!&\vert&\!\! X \end{bmatrix}")
。最後舉一例展示高斯—約當法於計算逆矩陣的應用。考慮
![A=\left[\!\!\begin{array}{ccc} 1&1&0\\ 2&3&3\\ 2&2&1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=A%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccc%7D++1%261%260%5C%5C++2%263%263%5C%5C++2%262%261++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "A=\left[\!\!\begin{array}{ccc} 1&1&0\\ 2&3&3\\ 2&2&1 \end{array}\!\!\right]")
，
高斯─約當法計算逆矩陣 的過程如下：
的過程如下：![\begin{array}{rlll} \begin{bmatrix} A\!\!&\vert&\!\! I \end{bmatrix}=&\left[\!\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 2 & 3 & 3 &\vline& 0 &1& 0 \\ 2 & 2 & 1 &\vline& 0 &0& 1 \end{array}\!\!\right]&\rightarrow & \left[\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 0 & 1 & 3 &\vline& -2 &1& 0 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right] \\ &&&\\ \rightarrow &\left[\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 0 & 1 & 0 &\vline& 4 &1& -3 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrcrrr} 1 & 0 & 0 &\vline& -3 &-1& 3 \\ 0 & 1 & 0 &\vline& 4 &1& -3 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right]\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brlll%7D++%5Cbegin%7Bbmatrix%7D++A%5C%21%5C%21%26%5Cvert%26%5C%21%5C%21+I++%5Cend%7Bbmatrix%7D%3D%26%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrcrrr%7D++1+%26+1+%26+0+%26%5Cvline%26+1+%260%26+0+%5C%5C++2+%26+3+%26+3+%26%5Cvline%26+0+%261%26+0+%5C%5C++2+%26+2+%26+1+%26%5Cvline%26+0+%260%26+1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%26%5Crightarrow+%26++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrcrrr%7D++1+%26+1+%26+0+%26%5Cvline%26+1+%260%26+0+%5C%5C++0+%26+1+%26+3+%26%5Cvline%26+-2+%261%26+0+%5C%5C++0+%26+0+%26+1+%26%5Cvline%26+-2+%260%26+1++%5Cend%7Barray%7D%5C%21%5Cright%5D+%5C%5C++%26%26%26%5C%5C++%5Crightarrow++%26%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrcrrr%7D++1+%26+1+%26+0+%26%5Cvline%26+1+%260%26+0+%5C%5C++0+%26+1+%26+0+%26%5Cvline%26+4+%261%26+-3+%5C%5C++0+%26+0+%26+1+%26%5Cvline%26+-2+%260%26+1++%5Cend%7Barray%7D%5C%21%5Cright%5D%26%5Crightarrow+%26++%5Cleft%5B%5C%21%5Cbegin%7Barray%7D%7Brrrcrrr%7D++1+%26+0+%26+0+%26%5Cvline%26+-3+%26-1%26+3+%5C%5C++0+%26+1+%26+0+%26%5Cvline%26+4+%261%26+-3+%5C%5C++0+%26+0+%26+1+%26%5Cvline%26+-2+%260%26+1++%5Cend%7Barray%7D%5C%21%5Cright%5D%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{rlll} \begin{bmatrix} A\!\!&\vert&\!\! I \end{bmatrix}=&\left[\!\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 2 & 3 & 3 &\vline& 0 &1& 0 \\ 2 & 2 & 1 &\vline& 0 &0& 1 \end{array}\!\!\right]&\rightarrow & \left[\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 0 & 1 & 3 &\vline& -2 &1& 0 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right] \\ &&&\\ \rightarrow &\left[\!\begin{array}{rrrcrrr} 1 & 1 & 0 &\vline& 1 &0& 0 \\ 0 & 1 & 0 &\vline& 4 &1& -3 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right]&\rightarrow & \left[\!\begin{array}{rrrcrrr} 1 & 0 & 0 &\vline& -3 &-1& 3 \\ 0 & 1 & 0 &\vline& 4 &1& -3 \\ 0 & 0 & 1 &\vline& -2 &0& 1 \end{array}\!\right]\end{array}")
從最後一個矩陣可以判斷 是可逆的，
是可逆的，![A^{-1}=\left[\!\!\begin{array}{rrr} -3&-1&3\\ 4&1&-3\\ -2&0&1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrr%7D++-3%26-1%263%5C%5C++4%261%26-3%5C%5C++-2%260%261++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0 "A^{-1}=\left[\!\!\begin{array}{rrr} -3&-1&3\\ 4&1&-3\\ -2&0&1 \end{array}\!\!\right]")
。註解
[1] 高斯─約當法的約當常被人誤認為是法國數學家 Marie Ennemond Camille Jordan(1838-1922)，矩陣分析理論中 Jordan 典型形式即因他而名。事實上，高斯─約當法是由德國測地學家 Wihelm Jordan (1842-1899) 於公元1888年提出。
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