对没有平方项的二次项使用拉格朗日配方法：有时候直接反解方程组比求解逆矩阵更简单

一、题目

使用拉格朗日配方法将 $f$ $=$ $2 x_{1} x_{2}+2 x_{1} x_{3}-6 x_{2} x_{3}$ 化为标准形，并求出对应的线性变换矩阵 $C$.

难度评级：

二、解析

Tips:

关于拉格朗日配方法的完整讲解，可以参阅《将二次型化为标准型（规范型）的方法之：拉格朗日配方法》这篇文章。

已知：

$$
f=2 x_{1} x_{2}+2 x_{1} x_{3}-6 x_{2} x_{3}
$$

令：

$$
\tag{1} \left\{\begin{array}{l}x_{1}=y_{1}+y_{2} \\ x_{2}=y_{1}-y_{2} \\ x_{3}=y_{3} .\end{array} \right. \Rightarrow
$$

$$
\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right]\
$$

$$
(1) \Rightarrow f \Rightarrow
$$

$$
f = 2\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)+2\left(y_{1}+y_{2}\right) y_{3}-
$$

$$
6\left(y_{1}-y_{2}\right) y_{3} \Rightarrow
$$

$$
f = 2 y_{1}^{2}-2 y_{2}^{2}+2 y_{1} y_{3}+2 y_{2} y_{3}-
$$

$$
6 y_{1} y_{3}+6 y_{2} y_{3} \Rightarrow
$$

$$
f = 2 y_{1}^{2}-2 y_{2}^{2}+8 y_{2} y_{3}-4 y_{1} y_{3} \Rightarrow
$$

$$
f=2\left(y_{1}^{2}-2 y_{1} y_{3}\right)-2 y_{2}^{2}+8 y_{2} y_{3} \Rightarrow
$$

$$
f=2\left(y_{1}-y_{3}\right)^{2}-2 y_{2}^{2}-2 y_{3}^{2}+8 y_{2} y_{3} \Rightarrow
$$

$$
f=2\left(y_{1}-y_{3}\right)^{2}-2\left(y_{2}^{2}-4 y_{2} y_{3}\right)-2 y_{3}^{2} \Rightarrow
$$

$$
f=2\left(y_{1}-y_{3}\right)^{2}-2\left(y_{2}-2 y_{3}\right)^{2}+6 y_{3}^{2}
$$

于是：

$$
\left\{\begin{array}{l}z_{1}=y_{1}-y_{3} \\ z_{2}=y_{2}-2 y_{3} \\ z_{3}=y_{3}\end{array} \right. \Rightarrow
$$

$$
\left\{\begin{array}{l}z_{1}=y_{1}-z_{3} \\ z_{2}=y_{2}-2 z_{3} \\ y_{3}=z_{3}\end{array} \Rightarrow\right.
$$

$$
\left\{\begin{array}{l}y_{1}=z_{1}+z_{3} \\ y_{2}=z_{2}+2 z_{3} \\ y_{3}=z_{3}\end{array} \right. \Rightarrow
$$

$$
\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}z_{1} \\ z_{2} \\ z_{3}\end{array}\right]
$$

于是可知，二次型 $f$ 的标准型为：

$$
f=2 z_{1}^{2}-2 z_{2}^{2}+6 z_{3}^{2}
$$

对应的变换矩阵为：

$$
C=\left[\begin{array}{ccc}1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 3 \\ 1 & -1 & -1 \\ 0 & 0 & 1\end{array}\right]
$$

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