2024年考研数二第16题解析：矩阵的化简

二、解析

$$r\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)=2$$

$$a \neq 1$$

\begin{aligned} \left(\alpha_{2}, \alpha_{1}, \alpha_{3}\right) & = \left(\begin{array}{ccc} 1 & a & 1 \\ 1 & 1 & a \\ b & -1 & -1 \\ a & 1 & 1 \end{array}\right) \\ \\ & \rightarrow\left(\begin{array}{ccc} 1 & a & 1 \\ 0 & 1-a & a-1 \\ 0 & -1-a b & -1-b \\ 0 & 1-a^{2} & 1-a \end{array}\right) \\ \\ & \rightarrow \left(\begin{array}{ccc} 1 & a & 1 \\ 0 & 1 & -1 \\ 0 & -1-a b & -1-b \\ 0 & 1+a & 1 \end{array}\right) \end{aligned}

\begin{aligned} \frac{-1-ab}{1} & = \frac{-1-b}{-1} \Rightarrow \frac{\textcolor{springgreen}{-1-ab}}{\textcolor{orangered}{1}} = \frac{\textcolor{springgreen}{1+b}}{\textcolor{orangered}{1}}\\ \\ \frac{1+a}{1} & = \frac{1}{-1} \Rightarrow \frac{\textcolor{springgreen}{1+a}}{\textcolor{orangered}{1}} = \frac{\textcolor{springgreen}{-1}}{\textcolor{orangered}{1}} \end{aligned}

$$\begin{cases} \textcolor{springgreen}{-1-a b=1+b} \\ \textcolor{springgreen}{1+a=-1} \end{cases}$$

$$\begin{cases} a=-2 \\ b=2 \end{cases}$$