题目
1.(I)用非退化线性替换化下列二次型为标准形,并利用矩阵验算所得结果:(1)-4x_(1)x_(2)+2x_(1)x_(3)+2x_(2)x_(3);(2)x_(1)^2+2x_(1)x_(2)+2x_(2)^2+4x_(2)x_(3)+4x_(3)^2;(3)x_(1)^2-3x_(2)^2-2x_(1)x_(2)+2x_(1)x_(3)-6x_(2)x_(3);
1.(I)用非退化线性替换化下列二次型为标准形,并利用矩阵验算所得结果:
(1)$-4x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$;
(2)$x_{1}^{2}+2x_{1}x_{2}+2x_{2}^{2}+4x_{2}x_{3}+4x_{3}^{2}$;
(3)$x_{1}^{2}-3x_{2}^{2}-2x_{1}x_{2}+2x_{1}x_{3}-6x_{2}x_{3}$;
题目解答
答案
1. **二次型1：**
令 $x_1 = \frac{1}{2}(y_1 + y_2)$，$x_2 = \frac{1}{2}(y_1 - y_2)$，$x_3 = y_3$，则
\[
f = -y_1^2 + 4y_2^2 + y_3^2.
\]
**答案：** $-y_1^2 + 4y_2^2 + y_3^2$
2. **二次型2：**
配方得
\[
f = (x_1 + x_2)^2 + (x_2 + 2x_3)^2.
\]
**答案：** $y_1^2 + y_2^2$
3. **二次型3：**
配方得
\[
f = (x_1 - x_2 + x_3)^2 - 4(x_2 + x_3)^2.
\]
**答案：** $y_1^2 - 4y_2^2$
\[
\boxed{
\begin{array}{ll}
1) & -y_1^2 + 4y_2^2 + y_3^2 \\
2) & y_1^2 + y_2^2 \\
3) & y_1^2 - 4y_2^2 \\
\end{array}
}
\]