# 2011年考研数二第22题解析：线性相关、线性表示、秩、可逆矩阵

## 题目

$(Ⅰ)$ 求 $a$ 的值；

$(Ⅱ)$ 将 $\beta_{1}$, $\beta_{2}$, $\beta_{3}$ 用 $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$ 线性表示。

## 解析

### 第 $(Ⅰ)$ 问

$$r(\beta_{1}, \beta_{2}, \beta_{3}) = r(\beta_{1}, \beta_{2}, \beta_{3}, \alpha_{1}) \Rightarrow$$

$$r(\beta_{1}, \beta_{2}, \beta_{3}) = r(\beta_{1}, \beta_{2}, \alpha_{1}, \beta_{3})$$

$$|\beta_{1}, \beta_{2}, \alpha_{1}| =$$

$$\begin{vmatrix} 1 & 1 & 1\\ 1 & 2 & 0\\ 1 & 3 & 1 \end{vmatrix} =2 \neq 0$$

$$r(\beta_{1}, \beta_{2}, \beta_{3}) = r(\beta_{1}, \beta_{2}, \alpha_{1}, \beta_{3}) = 3.$$

$$r(\beta_{1}, \beta_{2}, \beta_{3}) \neq 3.$$

$$r(\beta_{1}, \beta_{2}, \beta_{3}) \leqslant 3.$$

$$r(\beta_{1}, \beta_{2}, \beta_{3}) < 3.$$

$$|\beta_{1}, \beta_{2}, \beta_{3}| = 0 \Rightarrow$$

$$\begin{vmatrix} 1 & 1 & 3\\ 1 & 2 & 4\\ 1 & 3 & a \end{vmatrix} =0 \Rightarrow$$

$$a=5.$$

### 第 $(Ⅱ)$ 问

$$(\alpha_{1}, \alpha_{2}, \alpha_{3})\begin{bmatrix} a\\ b\\ c \end{bmatrix}=\beta_{1} \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 3\\ 1 & 1 & 5 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix} =\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}.$$

$$\beta_{1} = a \alpha_{1} + b \alpha_{2} + c \alpha_{3}. ①$$

$$\beta_{2} = d \alpha_{1} + e \alpha_{2} + f \alpha_{3}. ②$$

$$\beta_{3} = g \alpha_{1} + h \alpha_{2} + i \alpha_{3}. ③$$

$$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 3\\ 1 & 1 & 5 \end{bmatrix} \begin{bmatrix} a & d & h\\ b & e & i\\ c & f & j \end{bmatrix} =\begin{bmatrix} 1 & 1 & 3\\ 1 & 2 & 4\\ 1 & 3 & 5 \end{bmatrix}.$$

$$A = \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 3\\ 1 & 1 & 5 \end{bmatrix};$$

$$X = \begin{bmatrix} a & d & h\\ b & e & i\\ c & f & j \end{bmatrix};$$

$$B = \begin{bmatrix} 1 & 1 & 3\\ 1 & 2 & 4\\ 1 & 3 & 5 \end{bmatrix}.$$

$$AX=B \Rightarrow$$

$$X = A^{-1}B.$$

$$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 & 0\\ 0 & 1 & 3 & 0 & 1 & 0\\ 1 & 1 & 5 & 0 & 0 & 1 \end{bmatrix} \Rightarrow 初等行变换 \Rightarrow$$

$$\begin{bmatrix} 1 & 0 & 0 & 2 & 1 & -1\\ 0 & 1 & 0 & 3 & 4 & -3\\ 0 & 0 & 1 & -1 & -1 & 1 \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} 2 & 1 & -1\\ 3 & 4 & -3\\ -1 & -1 & 1 \end{bmatrix}$$

$$X = A^{-1}B =$$

$$\begin{bmatrix} 2 & 1 & -1\\ 3 & 4 & -3\\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3\\ 1 & 2 & 4\\ 1 & 3 & 5 \end{bmatrix} =\begin{bmatrix} 2 & 1 & 5\\ 4 & 2 & 10\\ -1 & 0 & -2 \end{bmatrix}$$

$$\begin{bmatrix} a & d & h\\ b & e & i\\ c & f & j \end{bmatrix} =\begin{bmatrix} 2 & 1 & 5\\ 4 & 2 & 10\\ -1 & 0 & -2 \end{bmatrix}$$

$$\beta_{1} = 2 \alpha_{1} + 4 \alpha_{2} – \alpha_{3};$$

$$\beta_{2} = 1 \alpha_{1} + 2 \alpha_{2} + 0 \alpha_{3};$$

$$\beta_{3} = 5 \alpha_{1} + 10 \alpha_{2} – 2 \alpha_{3}.$$