( I ) 求 A.
(II) 求可逆转矩阵 $P$ 与对角矩阵 $\Lambda$ 使得 $P^{-1} A P=\Lambda$.
【答案】 (I)因为 $A\left(\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right)=\left(\begin{array}{c}x_1+x_2+x_3 \\ 2 x_1-x_2+x_3 \\ x_2-x_3\end{array}\right)=\left(\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 1 \\ 0 & 1 & -1\end{array}\right)\left(\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right)$ 对任意的 $x_1, x_2, x_3$ 均 成立, 所以 $A=\left(\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 1 \\ 0 & 1 & -1\end{array}\right)$
(II) $|\lambda E-A|=\left|\begin{array}{ccc}\lambda-1 & -1 & -1 \\ -2 & \lambda+1 & -1 \\ 0 & -1 & \lambda+1\end{array}\right|=(\lambda-1) \cdot\left|\begin{array}{cc}\lambda+1 & -1 \\ -1 & \lambda+1\end{array}\right|+2 \cdot\left|\begin{array}{cc}-1 & -1 \\ -1 & \lambda+1\end{array}\right|$

$$=(\lambda-1)\left(\lambda^2+2 \lambda\right)-2(\lambda+2)=(\lambda+2)(\lambda-2)(\lambda+1)=0 .$$

$\lambda_1=-2$ 时, $\lambda_1 E-A=\left(\begin{array}{ccc}-3 & -1 & -1 \\ -2 & -1 & -1 \\ 0 & -1 & -1\end{array}\right) \rightarrow\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right)$, 可得特征向量 $\alpha_1=(0,-1,1)^T$; $\lambda_2=2$ 时, $\lambda_2 E-A=\left(\begin{array}{ccc}1 & -1 & -1 \\ -2 & 3 & -1 \\ 0 & -1 & 3\end{array}\right) \rightarrow\left(\begin{array}{ccc}1 & 0 & -4 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{array}\right)$, 可得特征向量 $\alpha_2=(4,3,1)^T$; $\lambda_3=-1$ 时, $\lambda_3 E-A=\left(\begin{array}{ccc}-2 & -1 & -1 \\ -2 & 0 & -1 \\ 0 & -1 & 0\end{array}\right) \rightarrow\left(\begin{array}{lll}2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)$, 可得特征向荲 $\alpha_3=(1,0,-2)^T$; 令 $P=\left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\begin{array}{ccc}0 & 4 & 1 \\ -1 & 3 & 0 \\ 1 & 1 & -2\end{array}\right)$, 则 $P^{-1} A P=\left(\begin{array}{ccc}-2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1\end{array}\right)$.