2026年考研数二第09题解析：线性方程组有解的条件

一、题目

»A« $a = -1$, $b = -1$

»B« $a = 2$, $b = 2$

»C« $a = -1$, $b = 2$

»D« $a = 2$, $b = -1$

二、解析

解法 1

首先，设：

$$
\begin{aligned}
\boldsymbol{A} & = (\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}) \\ \\
\boldsymbol{C} & = (\boldsymbol{r}_{1}, \boldsymbol{r}_{2})
\end{aligned}
$$

于是，由 $\boldsymbol{AB} = \boldsymbol{C}$ 可知，矩阵 $\boldsymbol{B}$ 的列向量就是 $\boldsymbol{A} \boldsymbol{x}_{1} = \boldsymbol{r}_{1}$ 和 $\boldsymbol{A} \boldsymbol{x}_{2} = \boldsymbol{r}_{2}$ 的解 $\boldsymbol{x}_{1}$ 和 $\boldsymbol{x}_{2}$, 也就是说，$\boldsymbol{A} \boldsymbol{x}_{1} = \boldsymbol{r}_{1}$ 和 $\boldsymbol{A} \boldsymbol{x}_{2} = \boldsymbol{r}_{2}$ 都是有解的线性方程组.

接着，由 $\boldsymbol{A} \boldsymbol{x}_{1} = \boldsymbol{r}_{1}$ 有解，得 $\boldsymbol{r}(\boldsymbol{A}) = \boldsymbol{r}(\boldsymbol{A}, \boldsymbol{r}_{1})$, 又因为：

$$
\begin{aligned}
(\boldsymbol{A}, \boldsymbol{r}_{1}) & = \begin{pmatrix}
1 & 0 & 1 & 2 \\
0 & 0 & 1 & 1 \\
1 & 1 & 3 & 1 \\
1 & 1 & 1 & a
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 \\
0 & 0 & 1 & 1 \\
0 & 1 & 2 & -1 \\
0 & 1 & 0 & a-2
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 \\
0 & 1 & 2 & -1 \\
0 & 0 & 1 & 1 \\
0 & 0 & -2 & a-1
\end{pmatrix}
\end{aligned}
$$

因此，矩阵 $\begin{pmatrix}
1 & 0 & 1 & 2 \\
0 & 1 & 2 & -1 \\
0 & 0 & 1 & 1 \\
0 & 0 & -2 & a-1
\end{pmatrix}$ 的最后两行要成比例，即：

$$
\begin{aligned}
& \ \frac{1}{-2} = \frac{1}{a-1} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ a = -1
\end{aligned}
$$

类似地，由 $\boldsymbol{A} \boldsymbol{x}_{2} = \boldsymbol{r}_{2}$ 有解，得 $\boldsymbol{r}(\boldsymbol{A}) = \boldsymbol{r}(\boldsymbol{A}, \boldsymbol{r}_{2})$, 又因为：

$$
\begin{aligned}
(\boldsymbol{A}, \boldsymbol{r}_{2}) & = \begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 \\
1 & 1 & 3 & 1 \\
1 & 1 & 1 & b
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 \\
0 & 1 & 2 & 1 \\
0 & 1 & 0 & b
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 \\
0 & 1 & 2 & 1 \\
0 & 0 & -2 & b-1
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 1 & 2 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & -2 & b-1
\end{pmatrix} \\ \\
\end{aligned}
$$

因此，矩阵 $\begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 1 & 2 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & -2 & b-1
\end{pmatrix}$ 的最后两行要成比例，即：

$$
\begin{aligned}
& \ \frac{1}{-2} = \frac{1}{b-1} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ b = -1
\end{aligned}
$$

综上可知，本 题 应 选 A

解法 2

首先，由 $\boldsymbol{C}$ $=$ $\begin{pmatrix} 2 & 0 \\ 1 & 1 \\ 1 & 1 \\ a & b \end{pmatrix}$ $\textcolor{lightgreen}{ \leadsto }$ $\begin{pmatrix} 2 & 0 \\ 1 & 1 \\ 0 & 0 \\ a & b \end{pmatrix}$ 可知：

$$
\boldsymbol{r} (\boldsymbol{A}) \leqslant 3
$$

又由 $\boldsymbol{A}$ $=$ $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 3 \\ 1 & 1 & 1 \end{pmatrix}$ $\textcolor{lightgreen}{ \leadsto }$ $\begin{vmatrix}
1 & 0 & 1 \\
0 & 0 & 1 \\
1 & 1 & 3
\end{vmatrix}$ $=$ $-1$ $\neq$ $0$ 可知：

$$
\boldsymbol{r} (\boldsymbol{B}) \geqslant 3
$$

于是，结合题目条件可知：

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

又因为：

$$
\begin{aligned}
(\boldsymbol{A}, \boldsymbol{C}) & = \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
1 & 1 & 3 & 1 & 1 \\
1 & 1 & 1 & a & b
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
0 & 1 & 2 & -1 & 1 \\
0 & 1 & 0 & a-2 & b
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
0 & 1 & 2 & -1 & 1 \\
0 & 0 & -2 & a-1 & b-1
\end{pmatrix}
\end{aligned}
$$

由于 $\boldsymbol{r} (\boldsymbol{A}, \boldsymbol{C})$ $=$ $3$, 所以：

$$
\begin{aligned}
& \frac{1}{-2} = \frac{1}{a-1} = \frac{1}{b-1} \\ \\
\textcolor{lightgreen}{ \leadsto } & \begin{cases}
a=-1 \\
b=-1
\end{cases}
\end{aligned}
$$

综上可知，本 题 应 选 A

解法 3

首先：

$$
\begin{aligned}
(\boldsymbol{A}, \boldsymbol{C}) & = \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
1 & 1 & 3 & 1 & 1 \\
1 & 1 & 1 & a & b
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
0 & 1 & 2 & -1 & 1 \\
0 & 0 & -2 & a-1 & b-1
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 0 & 1 & 1 & 1 \\
0 & 1 & 2 & -1 & 1 \\
0 & 0 & 0 & a+1 & b+1
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & 1 & 2 & 0 \\
0 & 1 & 2 & -1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & a+1 & b+1
\end{pmatrix}
\end{aligned}
$$

又因为，若要使 $\boldsymbol{AB} = \boldsymbol{C}$ 成立，就必须有：

$$
\boldsymbol{r} (\boldsymbol{A}, \boldsymbol{C}) = \boldsymbol{r}(\boldsymbol{A})
$$

于是：

$$
\begin{aligned}
& \begin{cases}
a+1=0 \\
b+1=0
\end{cases} \\ \\
\textcolor{lightgreen}{ \leadsto } & \begin{cases}
a=-1 \\
b=-1
\end{cases}
\end{aligned}
$$

综上可知，本 题 应 选 A

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