Question

求2x1+x2-x3+x4=1;3x1-2x2+x3-3x4=4;x1+4x2-3x3+5x4=-2的通解 大题请给出步骤

Answer 1

2x1+x2-x3+x4=1;
3x1-2x2+x3-3x4=4;
x1+4x2-3x3+5x4=-2
2 1 -1 1
记A= 3 -2 1 -3
1 4 -3 5
2 1 -1 1 1
B=（A,b）= 3 -2 1 -3 4
1 4 -3 5 -2
r1和r3对换得
1 4 -3 5 -2
3 -2 1 -3 4
2 1 -1 1 1
r2-3r1,r3-2r1得
1 4 -3 5 -2
0 -14 10 -18 10
0 -7 5 -9 5
r3-2r1,r1+4/14 r2,-r2/14得
1 0 -1/7 -1/7 6/7
0 1 -5/7 9/7 -5/7
0 0 0 0 0
所以通解为
x1/7=6/7+x3/7+x4/7
x2=-5/7+5x3/7-9x4/7
【x3,x4为任意常量】