武汉大学真题 第20652题
📝 题目
1．（10 分）计算行列式
$$
\left|\begin{array}{llll}
x_{1} & a_{2} & a_{3} & a_{4} \\
a_{1} & x_{2} & a_{3} & a_{4} \\
a_{1} & a_{2} & x_{3} & a_{4} \\
a_{1} & a_{2} & a_{3} & x_{4}
\end{array}\right| .
$$
💡 答案解析
1．解 记原行列式为 $\Delta_{4}$ ，令 $x_{4}=x_{4}-a_{4}+a_{4}$ ．
$$
\begin{aligned}
\Delta_{4}=\left|\begin{array}{llll}
x_{1} & a_{2} & a_{3} & a_{4} \\
a_{1} & x_{2} & a_{3} & a_{4} \\
a_{1} & a_{2} & x_{3} & a_{4} \\
a_{1} & a_{2} & a_{3} & x_{4}
\end{array}\right| & =\left|\begin{array}{cccc}
x_{1} & a_{2} & a_{3} & a_{4} \\
a_{1} & x_{2} & a_{3} & a_{4} \\
a_{1} & a_{2} & x_{3} & a_{4} \\
a_{1} & a_{2} & a_{3} & a_{4}
\end{array}\right|+\left|\begin{array}{cccc}
x_{1} & a_{2} & a_{3} & 0 \\
a_{1} & x_{2} & a_{3} & 0 \\
a_{1} & a_{2} & x_{3} & 0 \\
a_{1} & a_{2} & a_{3} & x_{4}-a_{4}
\end{array}\right| \\
& =a_{4}\left|\begin{array}{cccc}
x_{1} & a_{2} & a_{3} & 1 \\
a_{1} & x_{2} & a_{3} & 1 \\
a_{1} & a_{2} & x_{3} & 1 \\
a_{1} & a_{2} & a_{3} & 1
\end{array}\right|+\left(x_{4}-a_{4}\right) \Delta_{3} \\
& =a_{4}\left|\begin{array}{ccc}
x_{1}-a_{1} & 1 \\
x_{2}-a_{2} & x_{3}-a_{3} & 1 \\
1
\end{array}\right|+\left(x_{4}-a_{4}\right) \Delta_{3} \\
& =a_{4}\left(x_{1}-a_{1}\right)\left(x_{2}-a_{2}\right)\left(x_{3}-a_{3}\right)+\left(x_{4}-a_{4}\right) \Delta_{3}
\end{aligned}
$$
同样，$\quad \Delta_{3}=a_{3}\left(x_{1}-a_{1}\right)\left(x_{2}-a_{2}\right)+\left(x_{3}-a_{3}\right) \Delta_{2}$ ，
$$
\Delta_{2}=a_{2}\left(x_{1}-a_{1}\right)+\left(x_{2}-a_{2}\right) x_{1}
$$
📋 详细解题步骤
步骤 1/1
目标：将第四列拆分为两列，利用行列式按列线性性质递推降阶，得递推公式。
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