高阶行列式的计算思路：降阶或者找规律

一、题目

(A) $a _{ 1 } a _{ 2 } a _{ 3 } a _{ 4 }$ $-$ $b _{ 1 } b _{ 2 } b _{ 3 } b _{ 4 }$

(B) $a _{ 1 } a _{ 2 } a _{ 3 } a _{ 4 }$ $+$ $b _{ 1 } b _{ 2 } b _{ 3 } b _{ 4 }$

(C) $\left( a _{ 2 } a _{ 3 } – b _{ 2 } b _{ 3 } \right)$ $\left( a _{ 1 } a _{ 4 } – b _{ 1 } b _{ 4 } \right)$

(D) $\left( a _{ 1 } a _{ 2 } – b _{ 1 } b _{ 2 } \right)$ $\left( a _{ 3 } a _{ 4 } – b _{ 3 } b _{ 4 } \right)$

难度评级：

二、解析

解法一

Tip

该解法就是利用行列式的性质“降阶”，之后再按照低阶行列式的计算公式计算。

zhaokaifeng.com

$$
\begin{aligned}
\left| \begin{array} { c c c c } a _{ 1 } & 0 & 0 & b _{ 1 } \\ 0 & a _{ 2 } & b _{ 2 } & 0 \\ 0 & b _{ 3 } & a _{ 3 } & 0 \\ b _{ 4 } & 0 & 0 & a _{ 4 } \end{array} \right| \\ \\
& \underrightarrow{\quad \textcolor{blue}{降为 \ 3 \ 阶行列式} \quad} \\ \\
& = a _{ 1 } \times ( – 1 ) ^ { 1 + 1 } \left| \begin{array} { c c c } a _{ 2 } & b _{ 2 } & 0 \\ b _{ 3 } & a _{ 3 } & 0 \\ 0 & 0 & a _{ 4 } \end{array} \right| \\ \\
& + b _{ 1 } \times ( – 1 ) ^ { 1 + 4 } \left| \begin{array} { c c c } 0 & a _{ 2 } & b _{ 2 } \\ 0 & b _{ 3 } & a _{ 3 } \\ b _{ 4 } & 0 & 0 \end{array} \right| \\ \\
& \underrightarrow{\quad \textcolor{blue}{降为 \ 2 \ 阶行列式} \quad} \\ \\
& = a _{ 1 } a _{ 4 } \left| \begin{array} { l l } a _{ 2 } & b _{ 2 } \\ b _{ 3 } & a _{ 3 } \end{array} \right| – b _{ 1 } b _{ 4 } \left| \begin{array} { l l } a _{ 2 } & b _{ 2 } \\ b _{ 3 } & a _{ 3 } \end{array} \right|\\ \\
& = \textcolor{springgreen}{\boldsymbol{\left( a _{ 2 } a _{ 3 } – b _{ 2 } b _{ 3 } \right) \left( a _{ 1 } a _{ 4 } – b _{ 1 } b _{ 4 } \right)}}
\end{aligned}
$$

因此，本题应选 C.

解法二

Tip

该解法是先对行列式进行变形，形成位于对角线上的分块行列式，之后再利用分块行列式的计算公式计算。

zhaokaifeng.com

$$
\begin{aligned}
& \left| \begin{array} { c c c c } a _{ 1 } & 0 & 0 & b _{ 1 } \\ 0 & a _{ 2 } & b _{ 2 } & 0 \\ 0 & b _{ 3 } & a _{ 3 } & 0 \\ b _{ 4 } & 0 & 0 & a _{ 4 } \end{array} \right|
\underrightarrow{\quad 交换第 \ 2 \ 列和第 \ 4 \ 列 \quad } \\ \\
& \left| \begin{array} { c c c c } a _{ 1 } & b_{1} & 0 & 0 \\ 0 & 0 & b _{ 2 } & a_{2} \\ 0 & 0 & a _{ 3 } & b_{3}\\ b _{ 4 } & a_{4} & 0 & 0 \end{array} \right| \underrightarrow{\quad 交换第 \ 2 \ 行和第 \ 4 \ 行 \quad } \\ \\
& \left| \begin{array} { c c c c } \textcolor{springgreen}{a _{ 1 }} & \textcolor{springgreen}{b_{1}} & 0 & 0 \\ \textcolor{springgreen}{b _{ 4 }} & \textcolor{springgreen}{a_{4}} & 0 & 0 \\ 0 & 0 & \textcolor{orangered}{a _{ 3 }} & \textcolor{orangered}{b_{3}} \\ 0 & 0 & \textcolor{orangered}{b _{ 2 }} & \textcolor{orangered}{a_{2}} \end{array} \right| \\ \\
& = \begin{vmatrix}
\textcolor{springgreen}{a_{1}} & \textcolor{springgreen}{b_{1}} \\
\textcolor{springgreen}{b_{4}} & \textcolor{springgreen}{a_{4}}
\end{vmatrix} \cdot \begin{vmatrix}
\textcolor{orangered}{a_{3}} & \textcolor{orangered}{b_{3}} \\
\textcolor{orangered}{b_{2}} & \textcolor{orangered}{a_{2}}
\end{vmatrix} \\ \\
& = (a_{1} a_{4} – b_{1} b_{4}) (a_{2} a_{3} – b_{2} b_{3}) \\ \\
& = \textcolor{springgreen}{\boldsymbol{(a_{2} a_{3} – b_{2} b_{3}) (a_{1} a_{4} – b_{1} b_{4})}}
\end{aligned}
$$

综上可知，本 题 应 选 C

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