例题 2.3.3

求矩阵 $\mathbf{A}=\left(\begin{array}{rrrr}0& 2& -1& 3\\ 1& 4& -2& 0\\ 2& 3& 5& 2\\ 0& 0& 13& 19\end{array}\right)$ 的秩. 解 对矩阵作如下初等变换：

$$\begin{array}{rl}& \left(\begin{array}{rrrc}0& 2& -1& 3\\ 1& 4& -2& 0\\ 2& 3& 5& 2\\ 0& 0& 13& 19\end{array}\right)\\ & \stackrel{r_1\leftrightarrow r_2}{\to }\left(\begin{array}{rrrc}1& 4& -2& 0\\ 0& 2& -1& 3\\ 2& 3& 5& 2\\ 0& 0& 13& 19\end{array}\right)\\ & \stackrel{-2r_1+r_3}{\to }\left(\begin{array}{rrrc}1& 4& -2& 0\\ 0& 2& -1& 3\\ 0& -5& 9& 2\\ 0& 0& 13& 19\end{array}\right)\\ & \stackrel{\frac{5}{2}{r}_2+r_3}{\to }\left(\begin{array}{cccc}1& 4& -2& 0\\ 0& 2& -1& 3\\ 0& 0& \frac{13}{2}& \frac{19}{2}\\ 0& 0& 13& 19\end{array}\right)\\ & \stackrel{-2r_3+r_4}{\to }\left(\begin{array}{cccc}1& 4& -2& 0\\ 0& 2& -1& 3\\ 0& 0& \frac{13}{2}& \frac{19}{2}\\ 0& 0& 0& 0\end{array}\right)\end{array}$$

由此可知 $r(\mathbf{A})=3$ 。