线性无关的向量经运算之后变相关，则背后隐藏的矩阵一定线性相关

一、题目

已知 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 线性无关，若 $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, a \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}$ 线性相关，则 $a=?$

难度评级：

二、解析

$$
\left(\alpha_{1}+\alpha_{2}, a \alpha_{2}-\alpha_{3}, \alpha_{1}-\alpha_{2}+\alpha_{3}\right)=
$$

$$
\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)\left[\begin{array}{ccc}1 & 0 & 1 \\ 1 & a & -1 \\ 0 & -1 & 1\end{array}\right] \Rightarrow
$$

$$
\left|\begin{array}{ccc}1 & 0 & 1 \\ 1 & a & -1 \\ 0 & -1 & 1\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}1 & 0 & 1 \\ 2 & a & 0 \\ 0 & -1 & 1\end{array}\right|=0 \Rightarrow
$$

$$
a-2=0 \Rightarrow a=2.
$$

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