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2x1+x2-x3+x4==1
3x1-2x2+x3-3x4=4
x1+4x2-3x3+5x4==-2
解,得: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】
3x1-2x2+x3-3x4=4
x1+4x2-3x3+5x4==-2
解,得: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】
更多追问追答
追答
2x1+x2-x3+x4==1
3x1-2x2+x3-3x4=4
x1+4x2-3x3+5x4==-2
解,得:
r2-r1-r3, r1-2r3
0 -7 5 -9 5
0 -7 5 -9 5
1 4 -3 5 -2
r2-r1, r1*(-1/7), r3-4r1
0 1 -5/7 9/7 -5/7
0 0 0 0 0
1 0 -1/7 -1/7 6/7
通解为: (6/7,-5/7,0,0)'+c1(1,5,7,0)'+c2(1,-9,0,7)'
(好像有两种答案~)
这个是线性方程吧~
追问
谢谢哈
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