Question

设线性方程组x1+x2-2x3=0,3x1+2x3+ax3=1,x1-x2-6x3=2b,讨论ab为？

Answer 1

原方程的增广矩阵为
1 1 -2 0
3 2 a 1
1 -1 -6 2b
=>
1 1 -2 0
0 -1 a+6 1
0 -2 -4 2b
=>
1 1 -2 0
0 -1 a+6 1
0 1 2 -b
=>
1 0 -4 b
0 0 a+8 1-b
0 1 2 -b
当a=-8时,若b不等于1,无解
若b=1,则x1=1+4x3,x2=-1-2x3,无穷多解
当a不等于-8时,x3=(1-b)/(a+8)
x1=b+4(1-b)/(a+8),x2=-b-2(1-b)/(a+8),有唯一解,1,设线性方程组x1+x2-2x3=0,3x1+2x3+ax3=1,x1-x2-6x3=2b,讨论ab为
方程无解,有解,当有无穷多借时,求出通解.