2018年考研数二第23题解析：矩阵的秩、非齐次线性方程组、可逆矩阵

题目

$(Ⅰ)$ 求 $a$;

$(Ⅱ)$ 求满足 $AP = B$ 的可逆矩阵 $P$.

解析

第 $(Ⅰ)$ 问

$$A = \begin{bmatrix} 1 & 2 & a\\ 1 & 3 & 0\\ 2 & 7 & -a \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$A = \begin{bmatrix} 3 & 9 & 0\\ 1 & 3 & 0\\ 2 & 7 & -a \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$${\color{Red} A = \begin{bmatrix} 0 & 0 & 0\\ 1 & 3 & 0\\ 2 & 7 & -a \end{bmatrix}} \Rightarrow$$

$$r(A) = 2 \Rightarrow$$

$${\color{Red} r(B) = r(A) = 2}.$$

$$B = \begin{bmatrix} 1 & a & 2\\ 0 & 1 & 1\\ -1 & 1 & 1 \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$B = \begin{bmatrix} 1 & a & 2\\ 0 & 1 & 1\\ -1 & 1 & 1 \end{bmatrix} \Rightarrow$$

$${\color{Red} B = \begin{bmatrix} 1 & a & 2\\ 0 & 1 & 1\\ 0 & a+1 & 3 \end{bmatrix}}$$

$${\color{Red} a = 2}.$$

第 $(Ⅱ)$ 问

$$A = \begin{bmatrix} 1 & 2 & 2\\ 1 & 3 & 0\\ 2 & 7 & -2 \end{bmatrix};$$

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

$${\color{Red} \left\{\begin{matrix} A X_{1} = \beta_{1};\\ A X_{2} = \beta_{2};\\ A X_{3} = \beta_{3}. \end{matrix}\right.}$$

$$(A \vdots B) \Rightarrow$$

$$\begin{bmatrix} 1 & 2 & 2 & \vdots & 1 & 2 & 2\\ 1 & 3 & 0 & \vdots & 0 & 1 & 1\\ 2 & 7 & -2 & \vdots & -1 & 1 & 1 \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$\begin{bmatrix} 1 & 2 & 2 & \vdots & 1 & 2 & 2\\ 1 & 3 & 0 & \vdots & 0 & 1 & 1\\ 3 & 9 & 0 & \vdots & 0 & 3 & 3 \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$\begin{bmatrix} 1 & 2 & 2 & \vdots & 1 & 2 & 2\\ 1 & 3 & 0 & \vdots & 0 & 1 & 1\\ 0 & 0 & 0 & \vdots & 0 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$初等行变换 \Rightarrow$$

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

$$初等行变换，化为行最简形式 \Rightarrow$$

$${\color{Red} \begin{bmatrix} 1 & 0 & 6 & \vdots & 3 & 4 & 4\\ 0 & 1 & -2 & \vdots & -1 & -1 & -1\\ 0 & 0 & 0 & \vdots & 0 & 0 & 0 \end{bmatrix}}.$$

$$\left\{\begin{matrix} A X_{1} = \beta_{1};\\ A X_{2} = \beta_{2};\\ A X_{3} = \beta_{3}. \end{matrix}\right. \Rightarrow$$

$${\color{Red} \begin{bmatrix} 1 & 0 & 6\\ 0 & 1 & -2\\ 0 & 0 & 0 \end{bmatrix} X_{1} = \begin{bmatrix} 3\\ -1\\ 0 \end{bmatrix}};$$

$${\color{Red} \begin{bmatrix} 1 & 0 & 6\\ 0 & 1 & -2\\ 0 & 0 & 0 \end{bmatrix} X_{1} = \begin{bmatrix} 4\\ -1\\ 0 \end{bmatrix}};$$

$${\color{Red} \begin{bmatrix} 1 & 0 & 6\\ 0 & 1 & -2\\ 0 & 0 & 0 \end{bmatrix} X_{1} = \begin{bmatrix} 4\\ -1\\ 0 \end{bmatrix}}.$$

$${\color{Red} X_{1} = k_{1} \begin{bmatrix} -6\\ 2\\ 1 \end{bmatrix} + \begin{bmatrix} 3\\ -1\\ 0 \end{bmatrix}};$$

$${\color{Red} X_{2} = k_{2} \begin{bmatrix} -6\\ 2\\ 1 \end{bmatrix} + \begin{bmatrix} 4\\ -1\\ 0 \end{bmatrix}};$$

$${\color{Red} X_{3} = k_{3} \begin{bmatrix} -6\\ 2\\ 1 \end{bmatrix} + \begin{bmatrix} 4\\ -1\\ 0 \end{bmatrix}}.$$

$$P = (X_{1}, X_{2}, X_{3}) \Rightarrow$$

$${\color{Red} \begin{bmatrix} -6k_{1} + 3 & -6k_{2} + 4 & -6k_{3} + 4\\ 2 k_{1} – 1 & 2k_{2} – 1 & 2k_{3} – 1\\ k_{1} & k_{2} & k_{3} \end{bmatrix}}.$$

$${\color{Red} |P| \neq 0} \Rightarrow$$

$$\begin{vmatrix} -6k_{1} + 3 & -6k_{2} + 4 & -6k_{3} + 4\\ 2 k_{1} – 1 & 2k_{2} – 1 & 2k_{3} – 1\\ k_{1} & k_{2} & k_{3} \end{vmatrix} \neq 0 \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$\begin{vmatrix} -6k_{1} + 3 & -6k_{2} + 4 & -6k_{3} + 4\\ -1 & – 1 & – 1\\ k_{1} & k_{2} & k_{3} \end{vmatrix} \neq 0 \Rightarrow$$

$$初等行变换 \Rightarrow$$

$$\begin{vmatrix} 3 & 4 & 4\\ -1 & – 1 & – 1\\ k_{1} & k_{2} & k_{3} \end{vmatrix} \neq 0 \Rightarrow$$

$$初等行变换 \Rightarrow$$

$${\color{Red} \begin{vmatrix} -1 & 0 & 0\\ -1 & – 1 & – 1\\ k_{1} & k_{2} & k_{3} \end{vmatrix} \neq 0} \Rightarrow$$

$$k_{3} – k_{2} \neq 0 \Rightarrow$$

$${\color{Red} k_{3} \neq k_{2}}.$$

$${\color{Red} P = \begin{bmatrix} -6k_{1} + 3 & -6k_{2} + 4 & -6k_{3} + 4\\ 2 k_{1} – 1 & 2k_{2} – 1 & 2k_{3} – 1\\ k_{1} & k_{2} & k_{3} \end{bmatrix}}.$$