5.3 行列式 331

(g) 乘积

$$|AB| = |A| |B|$$ 这个结果很难一般性地证明，但可以通过较为繁琐的计算对 $2 \times 2$ 或 $3 \times 3$ 情况。对于 $2 \times 2$ 情形

$$\begin{aligned} |A| |B| &= (a_{11}a_{22} - a_{12}a_{21})(b_{11}b_{22} - b_{12}b_{21}) \ &= a_{11}a_{22}b_{11}b_{22} - a_{11}a_{22}b_{12}b_{21} - a_{12}a_{21}b_{11}b_{22} + a_{12}a_{21}b_{12}b_{21} \end{aligned}$$ 和

$$\begin{aligned} |AB| &= \begin{vmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{vmatrix} \ &= (a_{11}b_{11} + a_{12}b_{21})(a_{21}b_{12} + a_{22}b_{22}) - (a_{11}b_{12} + a_{12}b_{22})(a_{21}b_{11} + a_{22}b_{21}) \ &= a_{11}a_{22}b_{11}b_{22} - a_{11}a_{22}b_{12}b_{21} - a_{12}a_{21}b_{11}b_{22} + a_{12}a_{21}b_{12}b_{21} \end{aligned}$$ 例 5.16

求 $3 \times 3$ 行列式

$$(a) \begin{vmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 1 & 1 & 0 \end{vmatrix}, \quad (b) \begin{vmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ 0 & 1 & 2 \end{vmatrix}, \quad (c) \begin{vmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix}, \quad (d) \begin{vmatrix} 1 & 0 & 1 \ 0 & 2 & 4 \ 3 & 3 & 0 \end{vmatrix}$$ 解 (a) 按第一行展开：

$$\begin{vmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 1 & 1 & 0 \end{vmatrix} = 1 \begin{vmatrix} 1 & 2 \ 1 & 0 \end{vmatrix} - 0 \begin{vmatrix} 0 & 2 \ 1 & 0 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \ 1 & 1 \end{vmatrix} = -2 - 0 - 1 = -3$$ (b) 按第一列展开：

$$\begin{vmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ 0 & 1 & 2 \end{vmatrix} = 1 \begin{vmatrix} 1 & 0 \ 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 0 & 1 \ 1 & 2 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \ 1 & 0 \end{vmatrix} = 2 + 1 + 0 = 3$$ 注意 (a) 和 (b) 是同一个行列式，只是交换了两行。这一结果验证了上面刚陈述的性质 (c)。

(c) 按第三行展开：

$$\begin{vmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix} = 1 \begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix} - 0 \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} + 2 \begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = 1 - 0 + 2 = 3$$ 不