# 在实际的考试中，我们没必要把矩阵化简得这么“彻底”再去求未知数

## 二、解析

$$\boldsymbol { A } = \left( \boldsymbol { \alpha } _ { 1 } , \boldsymbol { \alpha } _ { 2 } , \boldsymbol { \alpha } _ { 3 } \right) = \begin{bmatrix} 1 & 1 & – 2 \\ 1 & – 2 & 1 \\ a & b & c \end{bmatrix}$$

*“$\boldsymbol { \beta }$ 可由 $\boldsymbol { \alpha } _ { 1 }$, $\boldsymbol { \alpha } _ { 2 }$, $\boldsymbol { \alpha } _ { 3 }$ 线性表示”，就是说方程组 $\boldsymbol { A } x = \boldsymbol { \beta }$ 有解，即 $r ( \boldsymbol { A } )$ $=$ $r ( \boldsymbol { A } , \boldsymbol { \beta } )$
**“且表示方法不唯一”，就是说方程组 $\boldsymbol { A } x = \boldsymbol { \beta }$ 有无穷多解。

$$r ( \boldsymbol { A } ) = r ( \boldsymbol { A } , \boldsymbol { \beta } ) < 3$$

$$\left| \begin{array} { c c } 1 & 1 \\ 1 & – 2 \end{array} \right| \neq 0$$

$$r ( \boldsymbol { A } ) \geqslant 2$$

$$3 > r ( \boldsymbol { A } ) \geqslant 2$$

$$r ( \boldsymbol { A } ) = \textcolor{springgreen}{a = 2}$$

\begin{aligned} ( \boldsymbol { A } , \boldsymbol { \beta } ) \\ \\ & = \begin{bmatrix} 1 & 1 & – 2 & \textcolor{yellow}{\vdots} & 1 \\ 1 & – 2 & 1 & \textcolor{yellow}{\vdots} & 2 \\ 2 & b & c & \textcolor{yellow}{\vdots} & 0 \end{bmatrix} \\ \\ & \rightarrow \begin{bmatrix} \textcolor{orangered}{1} & \textcolor{orangered}{0} & – 1 & \textcolor{yellow}{\vdots} & \frac { 4 } { 3 } \\ \\ \textcolor{orangered}{0} & \textcolor{orangered}{1} & – 1 & \textcolor{yellow}{\vdots} & \frac { -1 } { 3 } \\ \\ 0 & 0 & b + c + 2 & \textcolor{yellow}{\vdots} & \frac { 1 } { 3 } b – \frac { 8 } { 3 } \end{bmatrix} \\ \\ & \rightarrow \end{aligned}

$$\begin{cases} b + c + 2 = 0 , \\ \frac { 1 } { 3 } b – \frac { 8 } { 3 } = 0 \end{cases}$$

$$\textcolor{springgreen}{ \begin{cases} b = 8 \\ c = – 1 0 \end{cases} }$$

### 实战经验

\begin{aligned} & ( \boldsymbol { A } , \boldsymbol { \beta } ) = \begin{bmatrix} 1 & 1 & – 2 & \textcolor{yellow}{\vdots} & 1 \\ 1 & – 2 & 1 & \textcolor{yellow}{\vdots} & 2 \\ 2 & b & c & \textcolor{yellow}{\vdots} & 0 \end{bmatrix} \\ \\ & \underrightarrow{\text{ 第2行减第1行 } \ } \begin{bmatrix} 1 & 1 & – 2 & \textcolor{yellow}{\vdots} & 1 \\ 0 & – 3 & 3 & \textcolor{yellow}{\vdots} & 1 \\ 2 & b & c & \textcolor{yellow}{\vdots} & 0 \end{bmatrix} \\ \\ & \underrightarrow{\text{ 第3行减第1行的二倍 } \ } \begin{bmatrix} 1 & 1 & – 2 & \textcolor{yellow}{\vdots} & 1 \\ 0 & – 3 & 3 & \textcolor{yellow}{\vdots} & 1 \\ 0 & b-2 & c+4 & \textcolor{yellow}{\vdots} & -2 \end{bmatrix} \end{aligned}

$$\begin{cases} \frac{b-2}{-3} = \frac{-2}{1} \\ \\ \frac{c+4}{3} = \frac{-2}{1} \end{cases} \Rightarrow$$

$$\begin{cases} b-2 = 6 \\ c+4 = -6 \end{cases} \Rightarrow$$

$$\textcolor{springgreen}{ \begin{cases} b = 8 \\ c = -10 \end{cases} }$$