2021年考研数二第08题解析：二次型、正负惯性指数、二次型的配方

一、题目

»A« $2, 0$

»B« $1, 1$

»C« $2, 1$

»D« $1, 2$

难度评级：

二、解析

解法 1

首先：

$$
\begin{aligned}
f\left(x_{1},x_{2},x_{3}\right) & = \left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}-\left(x_{3}-x_{1}\right)^{2} \\ \\
& = \textcolor{gray}{x_{1}^{2}} + x_{2}^{2} + 2x_{1} x_{2} + x_{2}^{2} + \textcolor{gray}{ x_{3}^{2} } + 2x_{2} x_{3} – \left( \textcolor{gray}{ x_{3}^{2} } + \textcolor{gray}{ x_{1}^{2} } – 2x_{1} x_{3} \right) \\ \\
& = 2x_{2}^{2}+2x_{1}x_{2}+2x_{2}x_{3}+2x_{1}x_{3}
\end{aligned}
$$

所以，根据二次型的定义可知，该二次型对应的矩阵为：

$$
\boldsymbol{A} = \begin{pmatrix}
0 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
\end{pmatrix}
$$

于是，特征多项式为：

$$
\begin{aligned}
\begin{vmatrix}
\lambda \boldsymbol{E}-\boldsymbol{A}
\end{vmatrix} & = \begin{vmatrix}
\lambda & -1 & -1 \\
-1 & \lambda – 2 & -1 \\
-1 & -1 & \lambda
\end{vmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\text{第一行减去第三行}} \\ \\
& = \begin{vmatrix}
\lambda + 1 & 0 & – \left( \lambda + 1 \right) \\
-1 & \lambda – 2 & -1 \\
-1 & -1 & \lambda
\end{vmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\text{第一列加到第三列}} \\ \\
& = \begin{vmatrix}
\lambda + 1 & 0 & 0 \\
-1 & \lambda – 2 & -2 \\
-1 & -1 & \lambda – 1
\end{vmatrix} \\ \\
& = \left( \lambda + 1 \right) \begin{vmatrix}
1 & 0 & 0 \\
-1 & \lambda – 2 & -2 \\
-1 & -1 & \lambda – 1
\end{vmatrix} \\ \\
& = \left(\lambda + 1\right) \left(\lambda – 3\right) \lambda
\end{aligned}
$$

令：

$$
\left(\lambda + 1\right) \left(\lambda – 3\right) \lambda = 0
$$

则可得二次型矩阵 $\boldsymbol{A}$ 的特征值为：

$$
-1, \ 3, \ 0
$$

综上，该二次型的正惯性指数为 $1$, 负惯性指数为 $1$.

综上可知，本 题 应 选 »B«

解法 2

根据拉格朗日配方法，得：

$$
\begin{aligned}
f\left(x_{1}, x_{2}, x_{3}\right) & = 2 x_{2}^{2} + 2 x_{1} x_{2} + 2 x_{2} x_{3} + 2 x_{1} x_{3} \\ \\
& = 2 \left(x_{2} + \dfrac{x_{1} + x_{3}}{2}\right)^{2} – \dfrac{\left(x_{1} + x_{3}\right)^{2}}{2} + 2 x_{1} x_{3} \\ \\
& = 2 \left(x_{2} + \dfrac{x_{1} + x_{3}}{2}\right)^{2} – \dfrac{1}{2} x_{1}^{2} – \dfrac{1}{2} x_{3}^{2} + x_{1} x_{3} \\ \\
& = 2 \left(x_{2} + \dfrac{1}{2} x_{1} + \dfrac{1}{2} x_{3}\right)^{2} – \dfrac{1}{2} \left(x_{1} – x_{3}\right)^{2}
\end{aligned}
$$

综上，该二次型的正惯性指数为 $1$, 负惯性指数为 $1$.

综上可知，本 题 应 选 »B«

解法 3

根据偏导数配方法可知，由于给定的二次型 $f\left(x_{1},x_{2},x_{3}\right) 2x_{2}^{2}+2x_{1}x_{2}+2x_{2}x_{3}+2x_{1}x_{3}$ 含有平方项 $2x_{2}^{2}$, 所以先对 $x_{2}$ 求偏导，得：

$$
\frac{\partial f}{\partial x_{2}} = 4x_{2}+2x_{1}+2x_{3} = 4\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)
$$

这说明应先配出：

$$
2\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2}
$$

于是：

$$
\begin{aligned}
f & = 2\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2}
+\left[
f-2\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2}
\right] \\ \\
& = 2\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2}
-\frac{\left(x_{1}+x_{3}\right)^{2}}{2}
+2x_{1}x_{3} \\ \\
& = 2\left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2} – \frac{1}{2}x_{1}^{2} + x_{1}x_{3} – \frac{1}{2}x_{3}^{2} \\ \\
& = 2 \left(x_{2}+\frac{x_{1}+x_{3}}{2}\right)^{2} – \frac{1}{2}\left(x_{1}-x_{3}\right)^{2} \\ \\
& = 2 \left(x_{2}+\frac{1}{2}x_{1}+\frac{1}{2}x_{3}\right)^{2} – \frac{1}{2}\left(x_{1}-x_{3}\right)^{2}
\end{aligned}
$$

综上，该二次型的正惯性指数为 $1$, 负惯性指数为 $1$.

综上可知，本 题 应 选 »B«

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