2017年考研数二第23题解析：二次型、标准型、特征值与特征向量

解析

$$A = \begin{bmatrix} 2 & 1 & -4\\ 1 & -1 & 1\\ -4 & 1 & a \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$A = \begin{bmatrix} 3 & 0 & -3\\ 1 & -1 & 1\\ -3 & 0 & a+1 \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$A = \begin{bmatrix} 3 & 0 & -3\\ 1 & -1 & 1\\ 0 & 0 & a-2 \end{bmatrix}.$$

$$r(A) < 3 \Rightarrow$$

$$|A| = 0 \Rightarrow$$

$$a -2 = 0 \Rightarrow$$

$$a = 2.$$

$$A = \begin{bmatrix} 2 & 1 & -4\\ 1 & -1 & 1\\ -4 & 1 & 2 \end{bmatrix}.$$

$$\begin{vmatrix} \lambda – 2 & -1 & 4\\ -1 & \lambda + 1 & -1\\ 4 & -1 & \lambda – 2 \end{vmatrix} = 0 \Rightarrow$$

$$\begin{vmatrix} \lambda – 6 & 0 & 6 – \lambda\\ -1 & \lambda + 1 & -1\\ 0 & 4 \lambda + 3 & \lambda – 6 \end{vmatrix} = 0 \Rightarrow$$

$$(\lambda – 6)^{2}(\lambda + 1) + 2 (4 \lambda + 3)(\lambda – 6) \Rightarrow$$

$$(\lambda – 6)[(\lambda – 6)(\lambda + 1) + 2(4\lambda + 3)] \Rightarrow$$

$$(\lambda – 6)[\lambda ^{2} – 5 \lambda – 6 + 8 \lambda + 6] \Rightarrow$$

$$(\lambda – 6) (\lambda + 3) \lambda = 0 \Rightarrow$$

$$\lambda_{1} = 6, \lambda_{2} = -3, \lambda_{3} = 0.$$

$$(\lambda_{1} E – A) \Rightarrow$$

$$(6E – A) \Rightarrow$$

$$\begin{bmatrix} 4 & -1 & 4\\ -1 & 7 & -1\\ 4 & -1 & 4 \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 4 & -1 & 4\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}.$$

$\lambda_{1} = 6$ 对应的特征向量为：

$$\alpha_{1} =\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}.$$

$$(\lambda_{2} E – A) \Rightarrow$$

$$(-3E – A) \Rightarrow$$

$$\begin{bmatrix} -5 & -1 & 4\\ -1 & -2 & -1\\ 4 & -1 & -5 \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}.$$

$\lambda_{2} = -3$ 对应的特征向量为：

$$\alpha_{2} =\begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}.$$

$$(\lambda_{3} E – A) \Rightarrow$$

$$(0E – A) \Rightarrow$$

$$\begin{bmatrix} -2 & -1 & 4\\ -1 & 1 & -1\\ 4 & -1 & – 2 \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -2\\ 0 & 0 & 0 \end{bmatrix}.$$

$\lambda_{3} = 0$ 对应的特征向量为：

$$\alpha_{3} = \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}.$$

$$\gamma_{1} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix} = \begin{bmatrix} \frac{-1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}} \end{bmatrix}.$$

$$\gamma_{2} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{3}}\\ \frac{-1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} \end{bmatrix}.$$

$$\gamma_{3} = \frac{1}{\sqrt{6}} \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{6}}\\ \frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}} \end{bmatrix}$$

$$Q = \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}}\\ 0 & \frac{-1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{bmatrix}.$$

$$6 y_{1} – 3 y_{2}.$$