已知 3 阶方阵 $A$ 的逆矩阵为 $A^{-1}=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3\end{array}\right)$, 试求伴随矩阵 $A^*$ 的逆矩阵.
【答案】 解: 因为 $\left(A^*\right)^{-1}=\left(|A| \cdot A^{-1}\right)^{-1}=\frac{1}{|A|} \cdot A=\left|A^{-1}\right| \cdot A$.
$$
\left|A^{-1}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 3
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}\right|=2,
$$
$$
\left(\begin{array}{ll}
A^{-1} & E
\end{array}\right)=\left(\begin{array}{llllll}
1 & 1 & 1 & 1 & 0 & 0 \\
1 & 2 & 1 & 0 & 1 & 0 \\
1 & 1 & 3 & 0 & 0 & 1
\end{array}\right) \sim\left(\begin{array}{cccccc}
1 & 0 & 0 & \frac{5}{2} & -1 & -\frac{1}{2} \\
0 & 1 & 0 & -1 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2}
\end{array}\right),
$$
所以 $A=\left(A^{-1}\right)^{-1}=\left(\begin{array}{ccc}\frac{5}{2} & -1 & -\frac{1}{2} \\ -1 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2}\end{array}\right)$.
$$
\left(A^*\right)^{-1}=\left|A^{-1}\right| A=2 A=2\left(\begin{array}{ccc}
\frac{5}{2} & -1 & -\frac{1}{2} \\
-1 & 1 & 0 \\
-\frac{1}{2} & 0 & \frac{1}{2}
\end{array}\right)=\left(\begin{array}{ccc}
5 & -2 & -1 \\
-2 & 2 & 0 \\
-1 & 0 & 1
\end{array}\right) \text {. }
$$


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