线性代数-北京邮电大学出版社-戴赋祥

?1?0(A?b)???0??3?1?0??0??0?1?0??0??011?12110?111000?221?r3?r2?????r4?3r1(a?3)?2b??1a?1?110?221?r4?r2?????

a?10b?1???2a?3?1?110?221??.a?10b?1??0a?10?11(i) 当a?1≠0时，R(A)=R(A)=4,方程组有惟一解.

?b?a?2??a?1?x?1???a?2b?3?x????2???a?1?. ?x3???b?1?????x4??a?1???0??(ii) 当a?1=0时,b≠?1时，方程组R(A)=2

(iii) 当a=1，b= ?1时，方程组有无穷解. 得同解方程组

?x1?x2?x3?x4?0, ?x?2x?2x?1.34?2取

?x1?x3?x4?1,?x??2x?2x?1,?234 ?x?x,33??x4?x4,?∴ 得方程组的解为

?x1??1??1???1??x???2???2??1?2???k???k?????.?x3?1?1?2?0??0?????????x0???1??0??4??112???8. 设A?224，求一秩为2的3阶方阵B使AB=0. ????336??【解】设B=(b1 b2 b3),其中bi(i=1,2,3)为列向量,

由

k1,k2?R

AB?0?A(b1?Abi?0?b1为Ax=0的解.

b2b3)?0

(i?1,2,3)b3b2?112??x1?????求224x2=0的解.由 ??????336????x3???112??112?r2?2r1?000?

A??224???????r3?3r1?????336???000??得同解方程组

?x1??x2?2x3,? ?x2?x2,?x?x,33?∴ 其解为

?x1???1???2??x??k?1??k?0?.?2?1??2??????0???1???x3??取

k1,k2?R

??1???2??0?b1??1?;b2??0?;b3??0?,

??????????0???1???0??则

??1?20?B??100?

????010??9.已知?1,?2,?3是三元非齐次线性方程组Ax=b的解，且R(A)=1及

?1??1??1??,?????1?,?????1?,

?1??2??03??2??13??????0???0???1??求方程组Ax=b的通解.

【解】Ax=b为三元非齐次线性方程组

R(A)=1?Ax=0的基础解系中含有3?R(A)=3?1=2个解向量.

?1?1??0?????1?,?1??3?(?1??2)?(?2??3)??0?1??????0????0??

?1?1??0????0?,?1??2?(?1??3)?(?2??3)??1?1??????1?0????1??由?1,?2,?3为Ax=b的解??1??3,?1??2为Ax=0的解,

且(?1??3),(?1??2)线性无关??1??3,?1??2为Ax=0的基础解系. 又

?1?1?(?1??2)?(?2??3)?(?1??3)?2?1??1??1??1??2? 1??1??1?????0?1?1??0?,2??2??2????????0???0???1???1??2?∴ 方程组Ax=b的解为

x??1?k1(?1??3)?k2(?1??2)?1??2??0??0?????0??k1??1??k2?0?.?????1???0?1????????2???2??3?????(1) ξ1=1,ξ2=0; ???????0???1??

k1,k2?R10. 求出一个齐次线性方程组，使它的基础解系由下列向量组成.

?1??2??1???2???3???2???????(2) ξ1=?0?,ξ2=?2?,ξ3=?1?.

??????35?????2??????1????3????2??【解】

??2??3???ξ=?0?设齐次线性方程组为Ax=0

(1) ξ1=1??2?????0???1??由?1,?2为Ax=0的基础解系，可知

??2??3??x1??x1???2k1?3k2????x????

x?k1?1??k2?0???xk221?????????????k2?0???1??????x3????x3???令 k1=x2 , k2=x3

?Ax=0即为x1+2x2?3x3=0.

?121???2?3?2???(2) A(?1?2?3)=0?A的行向量为方程组为(x1x2x3x4x5)?021??0的解.

??352?????1?3?2??x1?2x2?3x4?x5?0??即?2x1?3x2?2x3?5x4?3x5?0的解为 ?x?2x?x?2x?2x?02345?1?1?203?1??1?203?1?r3?r1?2?325?3?????012?1?1?

?r?2r??21?????1?212?2???001?1?1??得基础解系为?1=(?5 ?1 1 1 0)T ?2=(?1 ?1 1 0 1)T

A=?方程为

??5?1110? ???1?1101???5x1?x2?x3?x4?0, ??x?x?x?x?0.?1235