矩阵经初等行变换化为1123x12x3x4x3x4是自由未知量）．A0102则方程组的一般解为x22x4000059当1
xx21时，方程组1有无穷多解．x1x21
5
f11221．设矩阵A235B0324解：利用初等行变换得112235324
15，且有AXB，求X．11121001121012301210012101511020107211511
100101000010112100110112100100151100110922100107210100151100
201即A721511由矩阵乘法和转置运算得20120111XAB721111135115162110200121B050，求A1B．2．设矩阵A223005
1
解：利用初等行变换得110100111210100122300104100110110011110010001641001
010011103201100531641
100431010531001641
即
431A15316414312008155AB5310501015564100512205
1
由矩阵乘法得
6
f2311233．设矩阵A011B112，求：（1）AB；（2）A1．001012231解：（1）因为A0112001
12301111B112112112012012
所以
ABAB2．
（2）因为
231100AI011010001001
230101100123210100110100110010010010011232r