例题 4.3 求逆矩阵

例题 4.3.

求矩阵

$$\mathbf{A}=\left(\begin{array}{rrr}2& 1& -1\\ 1& 4& 2\\ 5& -3& 1\end{array}\right)$$

的逆矩阵.

解 首先， $|\mathbf{A}|=52$ 。

并且

$$\begin{array}{rl}& A_{11}=\left|\begin{array}{rr}4& 2\\ -3& 1\end{array}\right|=10,A_{12}=-\left|\begin{array}{ll}1& 2\\ 5& 1\end{array}\right|=9,A_{13}=\left|\begin{array}{rr}1& 4\\ 5& -3\end{array}\right|=-23,\\ & A_{21}=-\left|\begin{array}{rr}1& -1\\ -3& 1\end{array}\right|=2,A_{22}=\left|\begin{array}{rr}2& -1\\ 5& 1\end{array}\right|=7,A_{23}=-\left|\begin{array}{rr}2& 1\\ 5& -3\end{array}\right|=11,\\ & A_{31}=\left|\begin{array}{rr}1& -1\\ 4& 2\end{array}\right|=6,A_{32}=-\left|\begin{array}{rr}2& -1\\ 1& 2\end{array}\right|=-5,A_{33}=\left|\begin{array}{ll}2& 1\\ 1& 4\end{array}\right|=7,\end{array}$$

于是

$$A^{-1}=\frac{1}{52}\left(\begin{array}{rrr}10& 2& 6\\ 9& 7& -5\\ -23& 11& 7\end{array}\right)$$

直接验算可知确有 $\mathbf{A}{\mathbf{A}}^{-1}={\mathbf{A}}^{-1}\mathbf{A}=\mathbf{E}$.