例题 1.2. 配方法化二次型为标准形

试用配方法把二次型 $f\left(x_1,x_2,x_3\right)=x_1^2+x_2^2+x_3^2+4x_1{x}_2+2x_1{x}_3+6x_2{x}_3$ 化成标准形.

解

$$\begin{array}{rl}& f\left(x_1,x_2,x_3\right)=x_1^2+2x_1\left(2x_2+x_3\right)+x_2^2+x_3^2+6x_2{x}_3\\ & ={\left(x_1+2x_2+x_3\right)}^2-{\left(2x_2+x_3\right)}^2+x_2^2+x_3^2+6x_2{x}_3\\ & ={\left(x_1+2x_2+x_3\right)}^2-3x_2^2+2x_2{x}_3\\ & ={\left(x_1+2x_2+x_3\right)}^2-3\left(x_2^2-\frac{2x_2{x}_3}{3}+\frac{x_3^2}{9}\right)+\frac{x_3^2}{3}\\ & ={\left(x_1+2x_2+x_3\right)}^2-3{\left(x_2-\frac{x_3}{3}\right)}^2+\frac{x_3^2}{3}.\end{array}$$

显然

$$\{\begin{array}{l}{y}_1=x_1+2x_2+x_3,\\ y_2=x_2-\frac{1}{3}{x}_3,\\ y_3=x_3,\end{array}$$

即

$$\begin{array}{r}\{\begin{array}{l}{x}_1=y_1-2y_2-\frac{5}{3}{y}_3,\\ x_2=y_2+\frac{1}{3}{y}_3,\\ x_3=y_3\end{array}\end{array}$$

是一个非退化的线性替换. 它把原二次型化成了标准形, 即

$$f=y_1^2-3y_2^2+\frac{1}{3}{y}_3^2$$