例题 4.1

例题 4.1.

计算行列式 $D=\left|\begin{array}{rrrr}3& -1& 5& 4\\ 1& 2& 6& 2\\ 2& \frac{1}{2}& 0& -3\\ 1& \frac{1}{6}& 2& \frac{2}{3}\end{array}\right|.$

\begin{proof}

$$\begin{array}{rl}D& \frac{\frac{r_3÷1/2}{r_4÷1/6}}{c_4÷2}\frac{1}{6}\left|\begin{array}{rrrr}3& -1& 5& 2\\ 1& 2& 6& 1\\ 4& 1& 0& -3\\ 6& 1& 12& 2\end{array}\right|\\ & \frac{r_1\leftrightarrow r_2}{\frac{r_2-3r_1}{r_3-4r_1}}-\frac{1}{6}\left|\begin{array}{rrrr}1& 2& 6& 1\\ 0& -7& -13& -1\\ 0& -7& -24& -7\\ 0& -11& -24& -4\end{array}\right|\\ & \frac{\frac{r_i÷\left(-1\right),i=2,3,4}{c_2\leftrightarrow c_4}}{\frac{r_3-7r_2}{r_4-4r_2}}-\frac{1}{6}\left|\begin{array}{rrrr}1& 1& 6& 2\\ 0& 1& 13& 7\\ 0& 0& -67& -42\\ 0& 0& -28& -17\end{array}\right|\\ & \frac{\frac{r_4÷\left(-1\right)}{r_4-28r_3}}{r_4-28r_3}-\frac{67}{6}\left|\begin{array}{cccc}1& 1& 6& 2\\ 0& 1& 13& 7\\ 0& 0& 1& \frac{42}{67}\\ 0& 0& 0& -\frac{37}{67}\end{array}\right|\\ & =-\frac{67}{6}\times \left(-\frac{37}{67}\right)=\frac{37}{6}.\end{array}$$

\end{proof}