4. 原式=B?1(A?B)?1,其中
B?1??1?3?1???? 0 1 1????1 0 1????1? 1 3?2??,
???1?2 1??? 1 3?1???
???100??100???A+B=?010?, (A?B)?1??010?
?1??003??00???3??2?? 1 3??3???1?B?1(A?B)?1 =??1?2 ?
?3??1??? 1 3???3??所以
(AB?B2)?15. 由于
A1?7?0,A2??3?0, 则 A?A1A2?0所以 A1,A2,A均可逆,且有
2???7?, 3 ??7??1A1?5? 7??4????7?1A21??1 0??33?????0?1 0? ?0 0 1?????00 0 0? 0?? 0?? 1???3? 0?? 1??
A?1?5? 7???4?7???0??0??0??273 70001 030?10 06. 由题意可得 ???3x3y??x?4x?y?6??? ??????3t3u??t?u?12u?3? 则有
?3x?x?4?3y?x?y?6?, 求得x=2, y=4, t=1, u=3. ?3t?t?u?1???3u?2u?311
7. 由 XA=B可得X?BA, 其中A?1?1? 10 1??,
???21?2????31?2???则
?1?33?? 10 1???20 1???????X =4 32?21?2??85?6 ???????1?25???31?2???103?5????????18. 由AX=B及A?1?0可知X?AB,其中
?1??0?0?1A????0??0??1??0?0X????0??0??1??0?0=???0??0??110?00?110?00110?000000?110000?1100?00??000?10000?10?1??????1??00??0??0?0??,??1??1??0??123?n?1n????0??012?n?2n?1?0??001?n?3n?2????
???????000???1?12?????1??000?01??1?11?11?1?0?10?011011??1?1?? ??1??1??011?1?? 0?1 ??00???1??1????22???9. 记 A????3??1?????4??0
1??11?0?1 3????100??00??1?1 ?1 ??0011011以上矩阵中有一个三阶子式 0?11?0,且所有的四阶子式都等于零,
0
故Rank(A)=3,即秩??1,?2,?3,?4??3,且它的一个极大无关组为?1,?2,?3.
12
3?1?1 ?02210. A???03?8 ?0?07故 Rank(A) = 4.
03250?1?1 3?31??0112?0????5?3 ??0011? 2?1??43?000??11?1??0??? 3??32??11?五、求解下列各题：
1. 1)
1011A按第一行展开?0000=1+(-1)
n?1
00??1?0??0110?0011??(?1)n?1???10000?11(n?1)阶000?0000 ?1101(n?1)阶00?2,当n为奇数时, =?0,当n为偶数时?所以,当n为奇数时,A?2?0,方程组有唯一零解;当n为偶数时,A?0方程组有非零解.
?1?12) 当n = 4时,A???0??0011000111??1?00????0??0??1??0010001?0?1 ?? 11??00??x1??x4?原方程组的同解方程组为 ?x2?x4
?x??x4?3令x4= 1得原方程组的基础解系为
(?1,1,?1,1)?,
则原方程组的全部解为k(-1,1,-1,1)(k为任意常数).
??22.
?3?2?2?(??3)2(??1)
??3A??1 2??8 14
当A?0,即??1或??3时,方程组有非零解. 其中当??1时,原方程组的同解方程组为
?x1?3x2?2x3?0 ?? x2?013
求得它的一个基础解系为(2,0,?1)?,
此时原方程组的全部解为k1(2,0,?1)?(k1为任意常数);
当??3时,原方程组的同解方程组为
?x1?3x2?2x3?0 ? 2x?x?0?23求得它的一个基础解系为 此时原方程组的全部解为
(1,?1,2)?,
k2(1,?1,2)?(k2为任意常数).
1??3.
11A?111??1??2(??3) 11??
1) 当??0且???3时,A?0方程组有唯一解,其同解方程组为
?x1?x2?(1??)x3?1 ?x?x?x23?1x1?x2?x3?1. ??3 求得
?1111??1111?????2) 当??0时, A??1111???0000?
?1111??0000?????此时Rank(A)=Rank(A)=1<3,方程组有无穷多解,其同解方程组为x1?x2?x3?1,求得它的全部解为
??(1,0,0)??k1(1,0,?1)??k2(1,?1,0)?(k1,k2为任意常数,)
111??11?21???2 ?1???01?10??2 113) 当???3时, A????? ?1??1?2 1????00 01?此时Rank(A)?3?Rank(A)?2,原方程组无解. 4. 设
??k1?1?k2?2?k3?3,即
?(1??)k1?k2?k3?0??k1?(1??)k2?k3?? ?2?k1?k2?(1??)k3?? (1)
14
1??方程组(1)的系数行列式A?11111111??1?(??3)0?0??2(??3) 11??00?当??0且??3时,A?0,方程组(1)有唯一解,此时?可用?1,?2,?3线性表示;
?1110??1110?????当??0时, A??1110???0000?,Rank(A)=Rank(A)=1<3,方程组(1)有无穷多解,此时?亦可
?1110??0000?????用?1,?2,?3线性表示;
110??00 01???2 ????01?14?, Rank
?2 1?3 当???3时, A?1(A)?3?
?????1??1?2 9????11?29?Rank(A)=2,此时方程组(1)无解,即?不能用?1,?2,?3线性表示.
5.
?1?3A???0??5121411231123 1?3 6?11??1?0a????3??0??b??011001200120011?63?? 0a??0b?2?显然,Rank(A)=2,当Rank(A)?Rank(A)?2时,此方程组有解且有无穷多解,即 当a=0且b=2时,方程组有无穷多解.它的同解方程组为
?x1?x2?1?x3?x4?x5 ? x?3?2x?2x?6x?2345令x3?x4?x5?0 时,可得原方程组的一个特解为 原方程组的导出组的同解方程组为
?0?(?2,3,0,0,0)?,
?x1?x2??x3?x4?x5 ? x??2x?2x?6x?2345可求得它的一个基础解系为?1?(1,?2,1,0,0)?,?2?(1,?2,0,1,0)?,?3?(5,?6,0,0,1)?. 则原方程组的全部解为???0?k1?1?k2?2?k3?3(k1,k2,k3为任意常数).
6.
?1?2A???3??0?4 2 0?3?1?5?7 1?5 1?1?1?1??1?4 20?1??01??7??1 ?1 ?1????
?8??0000 0?????1??0000 0?15