Question

3x1+4x2-5x3+7x4=0 2x1-3x2+3x3-2x4=0 4x1+11x2-13x3+16x4=0 7x1-2x2+x3+3x4=0 解方程组，用矩阵的方法

谢谢. 紧急.

Answer 1

解: 系数矩阵 A =
3 4 -5 7
2 -3 3 -2
4 11 -13 16
7 -2 1 3

r1-r2,r3-2r2
-->
1 7 -8 9
2 -3 3 -2
0 17 -19 20
7 -2 1 3

r2-2r1,r4-7r1
-->
1 7 -8 9
0 -17 19 -20
0 17 -19 20
0 -51 57 60

r3+r2,r4-3r2,r2*(-1/17)
1 7 -8 9
0 1 -19/17 20/17
0 0 0 0
0 0 0 0

r1-7r2
1 0 -3/17 13/17
0 1 -19/17 20/17
0 0 0 0
0 0 0 0

方程组的通解为: c1(3,19,17,0)+c2(13,20,0,-17).

Answer 2

3 4 -5 7
2 -3 3 -2
4 11 -13 16
7 -2 1 3
=1 7 -8 9 =1 7 -8 9 =1 7 -8 9 =1 7 -8 9 =1 24 0 0
2 -3 3 -2 0 -17 19 -20 0 -17 19 -20 0 -17 -8 9 0 -17 -8 -9
4 11 -13 16 0 -17 19 -20 0 -17 19 -20 0 0 0 0 0 0 0 0
7 -2 1 3 0 -51 57 -60 0 0 0 0 0 0 0 0 0 0 0 0
即x1+24x2=0 所以，x1=-24x2 令，x3=c1 所以x1=(192/17)c1- (216/17)c2
-17x2-8x3+9x4=0 x2=-(8/17)x3+(9/17)x4 x4=c2 x2=(-8/17)c1+ (9/17)c2
x3=c1
x4=c2
即为此方程的解

不懂再问！！祝您学习进步！！！

Answer 3

http://zhidao.baidu.com/question/201766917.html?an=0&si=2
3 4 -5 7
2 -3 3 -2
4 11 -13 16
7 -2 1 3
=1 7 -8 9 =1 7 -8 9 =1 7 -8 9 =1 7 -8 9 =1 24 0 0
2 -3 3 -2 0 -17 19 -20 0 -17 19 -20 0 -17 -8 9 0 -17 -8 -9
4 11 -13 16 0 -17 19 -20 0 -17 19 -20 0 0 0 0 0 0 0 0
7 -2 1 3 0 -51 57 -60 0 0 0 0 0 0 0 0 0 0 0 0
即x1+24x2=0 所以，x1=-24x2 令，x3=c1 所以x1=(192/17)c1- (216/17)c2
-17x2-8x3+9x4=0 x2=-(8/17)x3+(9/17)x4 x4=c2 x2=(-8/17)c1+ (9/17)c2
x3=c1
x4=c2
即为此方程的解