r0??r??1?/1?2)?)?r?10101/201/22?(4??r?2?(1??0?20?2?1010???00?2?2?1100???00?2200?11??
r?(??100?101/21/20?3?r1)?r4??3?1?/2??r?1??0?20?2?1010???00?2?2?1100???00041?1?11??
rr4?1/4?r1?10001/41/41/41/4?4?1/2??r??1?/2??r?r2?43??0?200?1/2?1/21/21/2???00?20?1/21/2?1/21/2???00041?1?11??rr2?(??11//22))??10001/41/41/41/4?3?(?r?4?1?/4????01001/41/4?1/4?1/4???00101/4?1/41/4?1/4???00011/4?1/4?1/41/4??因此有
??1/41/41/41/4?A1??1/41/4?1/4?1/4????1/4?1/41/4?1/4???1A3)
?1/4?1/4?1/41/4?4?
??0a10?0100?0?a2?0010?0?[A|I]??00???????????????000?a0?n?1000????an00?0000?1?? r00?000?01??rn??rnr?1?ann?1n?2?a10?010?00??r?1??r2????0??00a?2?001?00????????????????000?an?100?10??r100?000?0?r?1//aan?12?11?10?01/a10?0?r0?n?1?/a?n??1????001?001/a2?0????????????000?100?1/an?1因此, 最后得
11
/an?0?0?????0?? 1
0?0?1/a0?1A?1??01/a2?????0?0????000??1/an?1
#. 解下列矩阵方程, 求出未知矩阵X.
1/an?0??0????0??
?25??4?13?X??2??1) ??25?A????13?解: 令
21??0???123?X?2?13?6????2?31??????3341??? 2) ?
?4?6?B????21?, 则要解的方程为AX=B
将方程两边左乘上A的逆A-1, 可得A-1AX=A-1B, 即 X=A-1B 下面求A-1:
03?5?1?12?? ?3?5?A?1?????12? 因此有
?3?5??4?6??2?23?X?A?1B????21???0??128?????? 因此
21??0?A??2?13?????33?4??2) 令
-1
?25[A|I]???13r2?3?r1r2?(?1)?1???????010?r1?r2?1301?r1?(?2)?r2?1301???????????????0?11?2?01?2510?????
?123?B????2?31?-1
则矩阵方程为XA=B
设A的逆存在为A, 则方程两边右乘A, 得XAA-1=BA-1, 即
X=BA-1 下面求A-1:
21100?r1?r2?1?1/23/201/20??01/2???r?1??A|I???2?13010????021100??????3?4001???33?4001????3? r1?3?r3?1?1/23/201/20?r2?1/2???????011/21/200????03/21/203/21??
r2?1/2?r11/41/20??107/4r2?(?3/2)?r3?????????011/21/200????00?1/4?3/43/21??
12
?107/41/41/20?r3?(?4)???????011/21/200???3?6?4??001?
r3?(?1/2)?r2?100?5117?r3?(?7/4)?r1?????????010?132????0013?6?4??
??5117??A?1???132????3?6?4?? 因此,
??5117??123?????2?1?1?X?BA?1???132??????472?314???3?6?4?????最后得
#. 求矩阵X满足AX=A+2X, 其中
解: 将方程两边减去2X, 得AX-2X=A
因2X=2IX, 因此上面的方程可以从右边提取公因子X, 得 (A-2I)X=A
假设A-2I可逆, 则方程两边同时左乘(A-2I)-1, 得(A-2I)-1(A-2I)X=(A-2I)-1A, 即X=(A-2I)-1A
?301??A??110????014??
?301??200??101????020???1?10?B??110????????014????002????012?? 设B=A-2I, 则X=B-1A, 而
下面用行初等变换求B的逆B-1:
?10?B|I????1?1??01r2?1?r3?1r2?(?1)???????0??0B?1则
1100??10r1?(?1)?r2?0010?????????0?1?2001???0101100?r3?(?1)?r2?1r3?(?1)?r1?111?10?????????0?01?111???0?2?1?1????2?2?1???1???11?
00??1?110??2001??
002?1?1?102?2?1??01?111??
11?2?1?1??301??5?2?2???110???4?3?2?X?B?1A??2?2?1???????1?3???11???014?????22? 最后得
13
验算:
?301????4?4???4?3?A?2X???110?108?6?4??139?5?4???014????????446????????4510??? ?301???2???3?AX???110??5?24?3?2??13?49?5?4???014???????223????????4510???
#. 利用分块的方法, 求下列矩阵的乘积:
?0c0???a000??1?20??0a??100?010c?????01?01110??0b0????00d0??1) ??102???????01??; 2) ?1?010b????000d??
解:
??1?20??01??011????10????A|B???C??I?1) 将乘积分块为??102????01??2?
?1?A???0???20?,B??11?,C??0其中?????1??1???02??
?1???20??01???2?A|B???C????????0?????2?I??AC?BI2?2??0???01???11???00???11???1?1????02????01????02????02) 将乘积分块为
??a000??0c0??0a00??1010c???10b0???0d0?????aI2O2??I???I2cI2??2bI2??OdI2??010b??0???000d??
??a0ac0????aI2acI?2??I(c?bd)I???0a0ac?22???10c?bd0??010c?bd??
0?407?0?7?0?(?4)?0.
#. 计算下列三阶行列式:
14
1?1?3???
abc?124bca031?142; 2) cab 1)
解: 1) 将行列式按第一列展开
?1242)
031??1??1???2?10?84231?142 abc3124
#. 计算下列行列式:
bca?abc?abc?abc?a3?b3?c3?3abc?a3?b3?c3cab
a1a2a3a41)
b1b2b3b4b5c1c2000d1d2000e1e2000;
2)
xy0?000xy?00Dn???????0y0000??x0yx;
a50a000ef
bc000d3) 000解: 1) 将行列式按第一列展开后, 得到的各子式再按第二列展开, 这样展开后的后三列构成的任何三阶子式都至少包括一行0, 因此后三列任何三阶子式均为0, 整个行列式的值D=0. 2) 将行列式按第一列展开得
3) 先对第一列展开, 然后对第二列展开, 得
xy?00y0?00?????xy?00Dn?x?(?1)n?1y?xn?(?1)n?1yn00?xy?????00?0x00?xy
a00defD??b0d00e??ba0f??badf??abdf
#. 利用行列式的性质计算下列行列式
21413?12111) 5
15
20
3262;
?abacbd?cdbfcf2) aede?ef;