有的行列式可能越化简计算步骤越复杂

一、题目

$$
\left|\begin{array}{llll}1 & a & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1\end{array}\right|=?
$$

难度评级：

二、解析

方法 1：直接计算（计算步骤较简单）

$$
1 \cdot\left|\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right|+a \cdot(-1)^{4+1}\left|\begin{array}{lll}a & 0 & 0 \\ 1 & a & 0 \\ 0 & 1 & a\end{array}\right|=
$$

$$
1 -a\left(a^{3}\right)=1-a^{4}.
$$

方法 2：先化简再计算（计算步骤较复杂）

$$
\left|\begin{array}{llll}1 & a & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1\end{array}\right|=
$$

$$
\left|\begin{array}{cccc}a+1 & a+1 & a+1 & a+1 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1\end{array}\right|=
$$

$$
(a+1)\left|\begin{array}{llll}1 & 1 & 1 & 1 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1\end{array}\right|=
$$

$$
(a+1) \times 1 \times\left|\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right| +
$$

$$
(a+1) \times a \times(-1)^{4+1} \times\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & a & 0 \\ 0 & 1 & a\end{array}\right| =
$$

$$
(a+1)-a(a+1)\left(a^{2}+1-a\right)=
$$

$$
(a+1)\left[1-a\left(a^{2}+1-a\right)\right]=
$$

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

$$
a-a^{4}-a^{2}+a^{3}+1-a^{3}-a+a^{2}=
$$

$$
1-a^{4}.
$$

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