一、题目
$$
\begin{vmatrix} \boldsymbol{A} \end{vmatrix} = \begin{vmatrix}
a & 0 & b & 0 \\
0 & c & 0 & d \\
e & 0 & f & 0 \\
0 & g & 0 & h
\end{vmatrix} = ?
$$
难度评级：
二、解析 
§2.1 不一般的解法
首先，我们先来看看行列式 $\begin{vmatrix} \boldsymbol{A} \end{vmatrix}$ 中有哪些可能非零的元素：
$$
\begin{vmatrix} \boldsymbol{A} \end{vmatrix} = \begin{vmatrix}
\mathbf{\textcolor{orangered}{a}} & \textit{0} & \mathbf{\textcolor{orangered}{b}} & \textit{0} \\
\textit{0} & \mathbf{\textcolor{orangered}{c}} & \textit{0} & \mathbf{\textcolor{orangered}{d}} \\
\mathbf{\textcolor{orangered}{e}} & \textit{0} & \mathbf{\textcolor{orangered}{f}} & \textit{0} \\
\textit{0} & \mathbf{\textcolor{orangered}{g}} & \textit{0} & \mathbf{\textcolor{orangered}{h}}
\end{vmatrix}
$$
Note
题目并没有说行列式 $\begin{vmatrix} \boldsymbol{A} \end{vmatrix}$ 中的元素 $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ 这八个元素都一定是非零元素，但是，由题目已知条件可知，除此之外的其他元素一定是零元素——所有零元素在行列式展开式中一定都是没有作用的，所以不用考虑。
zhaokaifeng.com
由于行列式 $\begin{vmatrix} \boldsymbol{A} \end{vmatrix}$ 是一个四阶行列式，对于标准的四阶行列式，我们可以表示如下：
$$
\begin{vmatrix}
\mathbf{\textcolor{orangered}{a_{11}}} & a_{12} & \mathbf{\textcolor{orangered}{a_{13}}} & a_{14} \\
a_{21} & \mathbf{\textcolor{orangered}{a_{22}}} & a_{23} & \mathbf{\textcolor{orangered}{a_{24}}} \\
\mathbf{\textcolor{orangered}{a_{31}}} & a_{32} & \mathbf{\textcolor{orangered}{a_{33}}} & a_{34} \\
a_{41} & \mathbf{\textcolor{orangered}{a_{42}}} & a_{43} & \mathbf{\textcolor{orangered}{a_{44}}} \\
\end{vmatrix}
$$
所以，我们实际要在该行列式的展开式中考虑的元素就是：
$$
\begin{aligned}
\textcolor{springgreen}{a} & \Leftrightarrow \textcolor{springgreen}{a_{11}} \\
\textcolor{springgreen}{b} & \Leftrightarrow \textcolor{springgreen}{a_{13}} \\
\textcolor{yellow}{c} & \Leftrightarrow \textcolor{yellow}{a_{22}} \\
\textcolor{yellow}{d} & \Leftrightarrow \textcolor{yellow}{a_{24}} \\
\textcolor{orangered}{e} & \Leftrightarrow \textcolor{orangered}{a_{31}} \\
\textcolor{orangered}{f} & \Leftrightarrow \textcolor{orangered}{a_{33}} \\
\textcolor{pink}{g} & \Leftrightarrow \textcolor{pink}{a_{42}} \\
\textcolor{pink}{h} & \Leftrightarrow \textcolor{pink}{a_{44}}
\end{aligned}
$$
我们知道，行列式展开式每个展开项中的元素都来自行列式的不同行不同列，但是，我们应该怎么在展开项中排列这些元素的先后顺序呢？
其实，根据荒原之梦考研数学的《如何确定行列式展开计算公式中每一项的正负？- zhaokaifeng.com》这篇文章可知，“先将展开项中的元素按照行下标 $1$ 到 $n$ 顺序排列并穷举对应的列下标”，与“行下标和列下标都穷举”，效果是一样的。
因此，如果我们对 $a$ 到 $h$ 这八个可能非零的元素的行下标先按照 $1$, $2$, $3$, $4$ 排列好，那么：
第 $1$ 行元素可以出的列为：$1$ 或 $3$;
第 $2$ 行元素可以出的列为：$2$ 或 $4$;
第 $3$ 行元素可以出的列为：$1$ 或 $3$;
第 $4$ 行元素可以出的列为：$2$ 或 $4$;
于是，在行下标按顺序排列的情况下，我们可以组合出来的全部列下标为（注意列下标不能重复）：
$$
\begin{aligned}
\begin{pmatrix}
\textcolor{#161615}{\colorbox{#99F944}{1}} & & 3 & \\
& 2 & & \textcolor{#161615}{\colorbox{#99F944}{4}} \\
1 & & \textcolor{#161615}{\colorbox{#99F944}{3}} & \\
& \textcolor{#161615}{\colorbox{#99F944}{2}} & & 4
\end{pmatrix} & \Rightarrow \textcolor{#161615}{\colorbox{#99F944}{1432}} \\ \\
\begin{pmatrix}
\textcolor{#161615}{\colorbox{#99F944}{1}} & & 3 & \\
& \textcolor{#161615}{\colorbox{#99F944}{2}} & & 4 \\
1 & & \textcolor{#161615}{\colorbox{#99F944}{3}} & \\
& 2 & & \textcolor{#161615}{\colorbox{#99F944}{4}}
\end{pmatrix} & \Rightarrow \textcolor{#161615}{\colorbox{#99F944}{1234}} \\ \\
\begin{pmatrix}
1 & & \textcolor{#161615}{\colorbox{#99F944}{3}} & \\
& \textcolor{#161615}{\colorbox{#99F944}{2}} & & 4 \\
\textcolor{#161615}{\colorbox{#99F944}{1}} & & 3 & \\
& 2 & & \textcolor{#161615}{\colorbox{#99F944}{4}}
\end{pmatrix} & \Rightarrow \textcolor{#161615}{\colorbox{#99F944}{3214}} \\ \\
\begin{pmatrix}
1 & & \textcolor{#161615}{\colorbox{#99F944}{3}} & \\
& 2 & & \textcolor{#161615}{\colorbox{#99F944}{4}} \\
\textcolor{#161615}{\colorbox{#99F944}{1}} & & 3 & \\
& \textcolor{#161615}{\colorbox{#99F944}{2}} & & 4
\end{pmatrix} & \Rightarrow \textcolor{#161615}{\colorbox{#99F944}{3412}} \\ \\
\end{aligned}
$$
于是，根据行列式的标准展开计算公式 – zhaokaifeng.com，我们有：
$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{A} \end{vmatrix} \\ \\
& = (-1)^{\tau \left( \textcolor{#161615}{\colorbox{#99F944}{1432}} \right)} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{4}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{2}}} \\
& + (-1)^{\tau \left( \textcolor{#161615}{\colorbox{#99F944}{1234}} \right)} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{2}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{4}}} \\
& + (-1)^{\tau \left( \textcolor{#161615}{\colorbox{#99F944}{3214}} \right)} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{2}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{4}}} \\
& + (-1)^{\tau \left( \textcolor{#161615}{\colorbox{#99F944}{3412}} \right)} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{4}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{2}}} \\ \\
& = (-1)^{3} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{4}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{2}}} \\
& + (-1)^{0} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{2}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{4}}} \\
& + (-1)^{3} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{2}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{4}}} \\
& + (-1)^{4} \cdot a_{1 \textcolor{#161615}{\colorbox{#99F944}{3}}} \cdot a_{2 \textcolor{#161615}{\colorbox{#99F944}{4}}} \cdot a_{3 \textcolor{#161615}{\colorbox{#99F944}{1}}} \cdot a_{4 \textcolor{#161615}{\colorbox{#99F944}{2}}} \\ \\
& = -a_{1 1} \cdot a_{2 4} \cdot a_{3 3} \cdot a_{4 2} \\
& + a_{1 1} \cdot a_{2 2} \cdot a_{3 3} \cdot a_{4 4} \\
& – a_{1 3} \cdot a_{2 2} \cdot a_{3 1} \cdot a_{4 4} \\
& + a_{1 3} \cdot a_{2 4} \cdot a_{3 1} \cdot a_{4 2} \\ \\
& = a_{11} a_{33} \left( a_{22} a_{44} – a_{24} a_{42} \right) \\
& + a_{13} a_{31} \left( a_{24} a_{42} – a_{22} a_{44} \right) \\ \\
& = \left( a_{22} a_{44} – a_{24} a_{42} \right) \left( a_{11} a_{33} – a_{13} a_{31} \right) \\ \\
& = \textcolor{springgreen}{\boldsymbol{ \left( ch – dg \right) \left( af – be \right) }}
\end{aligned}
$$
附上前文中制作的对照表：
$\begin{aligned}
\textcolor{springgreen}{a} & \Leftrightarrow \textcolor{springgreen}{a_{11}} &
\textcolor{springgreen}{b} & \Leftrightarrow \textcolor{springgreen}{a_{13}} \\
\textcolor{yellow}{c} & \Leftrightarrow \textcolor{yellow}{a_{22}} & \textcolor{yellow}{d} & \Leftrightarrow \textcolor{yellow}{a_{24}} \\
\textcolor{orangered}{e} & \Leftrightarrow \textcolor{orangered}{a_{31}} & \textcolor{orangered}{f} & \Leftrightarrow \textcolor{orangered}{a_{33}} \\
\textcolor{pink}{g} & \Leftrightarrow \textcolor{pink}{a_{42}} & \textcolor{pink}{h} & \Leftrightarrow \textcolor{pink}{a_{44}}
\end{aligned}$
§2.2 一般的解法
由于行列式 $\begin{vmatrix} \boldsymbol{A} \end{vmatrix}$ 中的任意一行或者一列都只含有 $2$ 个两个非零元素，所以，我们可以随便挑选其中的一行或者一列，使用按行或者按列展开的方式将四阶行列式降为 $2$ 个三阶行列式，之后就可以继续降阶，或者直接利用三阶行列式的计算公式进行计算即可。
按第一行展开并直接按照三阶行列式计算：
$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{A} \end{vmatrix} \\ \\
& = \begin{vmatrix}
a & 0 & b & 0 \\
0 & c & 0 & d \\
e & 0 & f & 0 \\
0 & g & 0 & h
\end{vmatrix} \\ \\
& = \begin{vmatrix}
\textcolor{black}{\colorbox{pink}{a}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{b}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{black}{\colorbox{pink}{0}} & c & 0 & d \\
\textcolor{black}{\colorbox{pink}{e}} & 0 & f & 0 \\
\textcolor{black}{\colorbox{pink}{0}} & g & 0 & h
\end{vmatrix} \\ \\
& = \begin{vmatrix}
\textcolor{black}{\colorbox{pink}{a}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{b}} & \textcolor{black}{\colorbox{pink}{0}} \\
0 & c & \textcolor{black}{\colorbox{pink}{0}} & d \\
e & 0 & \textcolor{black}{\colorbox{pink}{f}} & 0 \\
0 & g & \textcolor{black}{\colorbox{pink}{0}} & h
\end{vmatrix} \\ \\
& = (-1) ^{1+1} \cdot a \begin{vmatrix}
c & 0 & d \\
0 & f & 0 \\
g & 0 & h
\end{vmatrix} + (-1) ^{1+3} \cdot b \begin{vmatrix}
0 & c & d \\
e & 0 & 0 \\
0 & g & h
\end{vmatrix} \\ \\
& = a \left[ cfh – dfg \right] + b \left[ dge – ceh \right] \\ \\
& = acfg – adfg + bdge – bceh \\ \\
& = afch – bech + bedg – afdg \\ \\
& = ch \left( af – be \right) + dg \left( be – af \right) \\ \\
& = \textcolor{springgreen}{\boldsymbol{ \left( ch – dg \right) \left( af – be \right) }}
\end{aligned}
$$
Next 
按第一行展开并降阶至二阶行列式计算：
$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{A} \end{vmatrix} \\ \\
& = \begin{vmatrix}
a & 0 & b & 0 \\
0 & c & 0 & d \\
e & 0 & f & 0 \\
0 & g & 0 & h
\end{vmatrix} \\ \\
& = \begin{vmatrix}
\textcolor{black}{\colorbox{pink}{a}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{b}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{black}{\colorbox{pink}{0}} & c & 0 & d \\
\textcolor{black}{\colorbox{pink}{e}} & 0 & f & 0 \\
\textcolor{black}{\colorbox{pink}{0}} & g & 0 & h
\end{vmatrix} \\ \\
& = \begin{vmatrix}
\textcolor{black}{\colorbox{pink}{a}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{b}} & \textcolor{black}{\colorbox{pink}{0}} \\
0 & c & \textcolor{black}{\colorbox{pink}{0}} & d \\
e & 0 & \textcolor{black}{\colorbox{pink}{f}} & 0 \\
0 & g & \textcolor{black}{\colorbox{pink}{0}} & h
\end{vmatrix} \\ \\
& = (-1) ^{1+1} \cdot a \begin{vmatrix}
c & 0 & d \\
0 & f & 0 \\
g & 0 & h
\end{vmatrix} + (-1) ^{1+3} \cdot b \begin{vmatrix}
0 & c & d \\
e & 0 & 0 \\
0 & g & h
\end{vmatrix} \\ \\
& = a \begin{vmatrix}
c & 0 & d \\
0 & f & 0 \\
g & 0 & h
\end{vmatrix} + b \begin{vmatrix}
0 & c & d \\
e & 0 & 0 \\
0 & g & h
\end{vmatrix} \\ \\
& = a \begin{vmatrix}
\textcolor{white}{\colorbox{brown}{c}} & \textcolor{white}{\colorbox{brown}{0}} & \textcolor{white}{\colorbox{brown}{d}} \\
\textcolor{white}{\colorbox{brown}{0}} & f & 0 \\
\textcolor{white}{\colorbox{brown}{g}} & 0 & h
\end{vmatrix} + b \begin{vmatrix}
\textcolor{white}{\colorbox{brown}{0}} & \textcolor{white}{\colorbox{brown}{c}} & \textcolor{white}{\colorbox{brown}{d}} \\
e & \textcolor{white}{\colorbox{brown}{0}} & 0 \\
0 & \textcolor{white}{\colorbox{brown}{g}} & h
\end{vmatrix} \\ \\
& = a \begin{vmatrix}
\textcolor{white}{\colorbox{brown}{c}} & \textcolor{white}{\colorbox{brown}{0}} & \textcolor{white}{\colorbox{brown}{d}} \\
0 & f & \textcolor{white}{\colorbox{brown}{0}} \\
g & 0 & \textcolor{white}{\colorbox{brown}{h}}
\end{vmatrix} + b \begin{vmatrix}
\textcolor{white}{\colorbox{brown}{0}} & \textcolor{white}{\colorbox{brown}{c}} & \textcolor{white}{\colorbox{brown}{d}} \\
e & 0 & \textcolor{white}{\colorbox{brown}{0}} \\
0 & g & \textcolor{white}{\colorbox{brown}{h}}
\end{vmatrix} \\ \\
& = (-1) ^{1+1} ac \begin{vmatrix}
f & 0 \\
0 & h
\end{vmatrix} + (-1) ^{1+3} ad \begin{vmatrix}
0 & f \\
g & 0
\end{vmatrix} \\
& + (-1)^{1+2} bc \begin{vmatrix}
e & 0 \\
0 & h
\end{vmatrix} + (-1)^{1+3} bd \begin{vmatrix}
e & 0 \\
0 & g
\end{vmatrix} \\ \\
& = acfh – adfg – bceh + bdeg \\ \\
& = afch – bech + bedg – afdg \\ \\
& = ch \left( af – be \right) + dg \left( be – af \right) \\ \\
& = \textcolor{springgreen}{\boldsymbol{ \left( ch – dg \right) \left( af – be \right) }}
\end{aligned}
$$
高等数学
涵盖高等数学基础概念、解题技巧等内容，图文并茂，计算过程清晰严谨。
线性代数
以独特的视角解析线性代数，让繁复的知识变得直观明了。
特别专题
通过专题的形式对数学知识结构做必要的补充，使所学知识更加连贯坚实。
让考场上没有难做的数学题！