$f\left(x_1, x_2, x_3\right)=4 x_2^2-3 x_3^2+4 x_1 x_2-4 x_1 x_3+8 x_2 x_3$.
(1)写出二次型 $f$ 的矩阵表达式 ；
(2)用正交变换把二次型 $f$ 化为标准型，并写出相应正交矩阵.
【答案】 【参考解析】 (1) $f$ 的矩阵表达式为
$$f\left(x_1, x_2, x_3\right)=\left(x_1, x_2, x_3\right)\left(\begin{array}{ccc} 0 & 2 & -2 \\ 2 & 4 & 4 \\ -2 & 4 & -3 \end{array}\right)\left(\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right)$$
(2) 二次型矩阵为 $A=\left(\begin{array}{ccc}0 & 2 & -2 \\ 2 & 4 & 4 \\ -2 & 4 & -3\end{array}\right) ， A$ 的特征方程为
$$|\lambda E-A|=\left|\begin{array}{ccc} \lambda & -2 & 2 \\ -2 & \lambda-4 & -4 \\ 2 & -4 & \lambda+3 \end{array}\right|=(\lambda-1)\left(\lambda^2-36\right)=0$$

$$\alpha_1=\left(\begin{array}{c} 2 \\ 0 \\ -1 \end{array}\right), \alpha_2=\left(\begin{array}{l} 1 \\ 5 \\ 2 \end{array}\right), \alpha_3=\left(\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right) .$$

$$\alpha_1^{\circ}=\left(\begin{array}{c} \frac{2}{\sqrt{5}} \\ 0 \\ \frac{-1}{\sqrt{5}} \end{array}\right), \alpha_2^{\circ}=\left(\begin{array}{c} \frac{1}{\sqrt{30}} \\ \frac{5}{\sqrt{30}} \\ \frac{2}{\sqrt{30}} \end{array}\right), \alpha_3^{\circ}=\left(\begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{-1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \end{array}\right) \text {. }$$

$$P=\left(\begin{array}{ccc} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{30}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{5}{\sqrt{30}} & \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{5}} & \frac{2}{\sqrt{30}} & \frac{2}{\sqrt{6}} \end{array}\right)$$

$$f\left(x_1, x_2, x_3\right)=y_1^2+6 y_2^2-6 y_3^2$$