例 2.29

已知 ${\mathbf{A}}^{\ast }=\left(\begin{array}{ccc}1& -2& 1\\ 0& 2& -2\\ -1& 2& 1\end{array}\right)$ ,求 $\mathbf{A}$ .

解 计算 ${\mathbf{A}}^{\ast }$ 的行列式可得 $\left|{\mathbf{A}}^{\ast }\right|=4$ . 因为 $\left|{\mathbf{A}}^{\ast }\right|={\left|\mathbf{A}\right|}^{n-1}$ ,得 ${\left|\mathbf{A}\right|}^2=4$ , $\left|\mathbf{A}\right|=\pm 2$ .

若 $\left|\mathbf{A}\right|=2$ ,由 ${\mathbf{A}}^{\ast }=\left|\mathbf{A}\right|{\mathbf{A}}^{-1}$ 得

$${\mathbf{A}}^{-1}=\frac{1}{2}{\mathbf{A}}^{\ast }=\left(\begin{array}{ccc}\frac{1}{2}& -1& \frac{1}{2}\\ 0& 1& -1\\ -\frac{1}{2}& 1& \frac{1}{2}\end{array}\right)$$

于是

$$\mathbf{A}={\left({\mathbf{A}}^{-1}\right)}^{-1}=\left(\begin{array}{lll}3& 2& 1\\ 1& 1& 1\\ 1& 0& 1\end{array}\right)$$

若 $\left|\mathbf{A}\right|=-2$ ,则 ${\mathbf{A}}^{-1}=-\frac{1}{2}{\mathbf{A}}^{\ast }$ ,故

$$\mathbf{A}=\left(\begin{array}{ccc}-3& -2& -1\\ -1& -1& -1\\ -1& 0& -1\end{array}\right).\square \text{\hspace{1em}(口)}$$