行列式中的“消消乐”

一、题目

难度评级：

二、解析

观察可知，行列式 $\boldsymbol{K}$ 的零元素主要集中在行列式的右上角：

$$
\begin{vmatrix}
1 & -2 & 5 & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
3 & 8 & 1 & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
5 & 0 & -3 & 2 & 1 & -1 \\
1 & 2 & 5 & 2 & 1 & -1 \\
7 & 3 & 5 & 9 & 2 & 0 \\
1 & 6 & 5 & -5 & 3 & 2 \\
\end{vmatrix}
$$

但是，该行列式的第 $3$ 行和第 $4$ 行存在一些相等的元素，刚好可以将第 $3$ 行的一些元素消为零：

$$
\begin{vmatrix}
1 & -2 & 5 & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
3 & 8 & 1 & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
5 & 0 & -3 & \textcolor{black}{\colorbox{tan}{2}} & \textcolor{black}{\colorbox{tan}{1}} & \textcolor{black}{\colorbox{tan}{-1}} \\
1 & 2 & 5 & \textcolor{black}{\colorbox{tan}{2}} & \textcolor{black}{\colorbox{tan}{1}} & \textcolor{black}{\colorbox{tan}{-1}} \\
7 & 3 & 5 & 9 & 2 & 0 \\
1 & 6 & 5 & -5 & 3 & 2 \\
\end{vmatrix}
$$

于是，在行列式 $\boldsymbol{K}$ 中，用第 $3$ 行减去第 $4$ 行，可得：

$$
\begin{aligned}
& \begin{vmatrix}
1 & -2 & 5 & 0 & 0 & 0 \\
3 & 8 & 1 & 0 & 0 & 0 \\
\textcolor{orangered}{5} & \textcolor{orangered}{0} & \textcolor{orangered}{-3} & \textcolor{orangered}{2} & \textcolor{orangered}{1} & \textcolor{orangered}{-1} \\
\textcolor{springgreen}{1} & \textcolor{springgreen}{2} & \textcolor{springgreen}{5} & \textcolor{springgreen}{2} & \textcolor{springgreen}{1} & \textcolor{springgreen}{-1} \\
7 & 3 & 5 & 9 & 2 & 0 \\
1 & 6 & 5 & -5 & 3 & 2 \\
\end{vmatrix} \\ \\
= & \begin{vmatrix}
1 & -2 & 5 & 0 & 0 & 0 \\
3 & 8 & 1 & 0 & 0 & 0 \\
\textcolor{orangered}{4} & \textcolor{orangered}{-2} & \textcolor{orangered}{-8} & \textcolor{orangered}{0} & \textcolor{orangered}{0} & \textcolor{orangered}{0} \\
\textcolor{springgreen}{1} & \textcolor{springgreen}{2} & \textcolor{springgreen}{5} & \textcolor{springgreen}{2} & \textcolor{springgreen}{1} & \textcolor{springgreen}{-1} \\
7 & 3 & 5 & 9 & 2 & 0 \\
1 & 6 & 5 & -5 & 3 & 2 \\
\end{vmatrix} \\ \\
= & \begin{vmatrix}
\textcolor{white}{\colorbox{green}{1}} & \textcolor{white}{\colorbox{green}{-2}} & \textcolor{white}{\colorbox{green}{5}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{white}{\colorbox{green}{3}} & \textcolor{white}{\colorbox{green}{8}} & \textcolor{white}{\colorbox{green}{1}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{white}{\colorbox{green}{4}} & \textcolor{white}{\colorbox{green}{-2}} & \textcolor{white}{\colorbox{green}{-8}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
1 & 2 & 5 & \textcolor{white}{\colorbox{green}{2}} & \textcolor{white}{\colorbox{green}{1}} & \textcolor{white}{\colorbox{green}{-1}} \\
7 & 3 & 5 & \textcolor{white}{\colorbox{green}{9}} & \textcolor{white}{\colorbox{green}{2}} & \textcolor{white}{\colorbox{green}{0}} \\
1 & 6 & 5 & \textcolor{white}{\colorbox{green}{-5}} & \textcolor{white}{\colorbox{green}{3}} & \textcolor{white}{\colorbox{green}{2}} \\
\end{vmatrix}
\end{aligned}
$$

于是，根据行列中的“方块乘法”可知：

$$
\begin{aligned}
& \boldsymbol{K} = \\ \\
= & \begin{vmatrix}
\textcolor{white}{\colorbox{green}{1}} & \textcolor{white}{\colorbox{green}{-2}} & \textcolor{white}{\colorbox{green}{5}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{white}{\colorbox{green}{3}} & \textcolor{white}{\colorbox{green}{8}} & \textcolor{white}{\colorbox{green}{1}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
\textcolor{white}{\colorbox{green}{4}} & \textcolor{white}{\colorbox{green}{-2}} & \textcolor{white}{\colorbox{green}{-8}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} & \textcolor{black}{\colorbox{pink}{0}} \\
1 & 2 & 5 & \textcolor{white}{\colorbox{green}{2}} & \textcolor{white}{\colorbox{green}{1}} & \textcolor{white}{\colorbox{green}{-1}} \\
7 & 3 & 5 & \textcolor{white}{\colorbox{green}{9}} & \textcolor{white}{\colorbox{green}{2}} & \textcolor{white}{\colorbox{green}{0}} \\
1 & 6 & 5 & \textcolor{white}{\colorbox{green}{-5}} & \textcolor{white}{\colorbox{green}{3}} & \textcolor{white}{\colorbox{green}{2}} \\
\end{vmatrix} \\ \\
= & \begin{vmatrix}
1 & -2 & 5 \\
3 & 8 & 1 \\
4 & -2 & -8
\end{vmatrix} \cdot \begin{vmatrix}
2 & 1 & -1 \\
9 & 2 & 0 \\
-5 & 3 & 2
\end{vmatrix} \\ \\
= & \begin{vmatrix}
0 & \textcolor{magenta}{-2} & 0 \\
7 & 0 & 0 \\
3 & 0 & \textcolor{magenta}{-22}
\end{vmatrix} \cdot \begin{vmatrix}
0 & 1 & \textcolor{tan}{-1} \\
5 & 2 & 0 \\
\textcolor{tan}{-11} & 0 & 5
\end{vmatrix} \\ \\
= & (\textcolor{magenta}{-2}) (\textcolor{magenta}{-22}) \cdot \begin{vmatrix}
0 & 1 & 0 \\
7 & 0 & 0 \\
3 & 0 & 1
\end{vmatrix} \cdot (\textcolor{tan}{-1}) (\textcolor{tan}{-11}) \begin{vmatrix}
0 & 1 & 1 \\
5 & 2 & 0 \\
1 & 0 & 5
\end{vmatrix} \\ \\
= & 44 \cdot (-7) \cdot 11 (-27) \\ \\
= & 44 \cdot 7 \cdot 11 \cdot 27 \\ \\
= & 484 \cdot 7\cdot 27 \\ \\
= & 91476
\end{aligned}
$$

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