$A=\left (\begin {array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end {array}\right)$ P 为初等方阵。若 $P A=\left (\begin {array}{ccc}{a_{11}} & {a_{12}} & {a_{13}} \\ {2 a_{11}+a_{21}} & {2 a_{12}+a_{22}} & {2 a_{13}+a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end {array}\right)$$AP$ 为?

这里极易认为只要交换行列就行了,所以答案是

$$ <br /> \left(\begin{array}{lll}<br /> {a_{11}} & {a_{12}+2a_{11}} & {a_{13}} \\<br /> {a_{21}} & {a_{22}+2a_{21}} & {a_{23}} \\<br /> {a_{31}} & {a_{32}+2a_{31}} & {a_{33}}<br /> \end{array}\right)<br /> $$

这样的做法错了.

为了避免出错,我使用如下做法.

首先,我用 $L$ 表示行,$C$ 表示列写出左乘时的表达式:

$$ <br /> L_2^{'} = L_2 + 2L_1<br /> $$

现在置换行列和下标 (倍加系数不变) 得到

$$ <br /> C_1^{'} = C_1 + 2C_2<br /> $$

这就是右乘的含义,表示第一列赋值为原第一列加上二倍原第二列.

答案:

$A P=\left(\begin{array}{lll}{2 a_{12}+a_{11}} & {a_{12}} & {a_{13}} \\ {2 a_{22}+a_{21}} & {a_{22}} & {a_{23}} \\ {2 a_{32}+a_{31}} & {a_{32}} & {a_{33}}\end{array}\right)$