Question

求方程组：x1+x2-x3+2x4=3, 2x1+x2-3x4 =1, -2x1-2x3+10x4=4

(2) X1-4x2+2x3-3x4=11
4x1+3x2+6x3-x4=-1
2x1+4x2+2x3+x4=-6

Answer 1

(1) 解: 增广矩阵 =
1 1 -1 2 3
2 1 0 -3 1
-2 0 -2 10 4

r3+r2, r2-2r1
1 1 -1 2 3
0 -1 2 -7 -5
0 1 -2 7 5

r1+r2,r3+r2,r2*(-1)
1 0 1 -5 -2
0 1 -2 7 5
0 0 0 0 0

方程组的通解为: (-2,5,0,0)^T+c1(-1,2,1,0)^T+c2(5,-7,0,1)^T.

(2)
解: 增广矩阵 =
1 -4 2 -3 11
4 3 6 -1 -1
2 4 2 1 -6

r2-4r1,r3-2r1
1 -4 2 -3 11
0 19 -2 11 -45
0 12 -2 7 -28

r2-r3
1 -4 2 -3 11
0 7 0 4 -17
0 12 -2 7 -28

r2*(1/7),r1+4r2,r3-12r2
1 0 2 -5/7 9/7
0 1 0 4/7 -17/7
0 0 -2 1/7 8/7

r1+r3,r3*(-1/2)
1 0 0 -4/7 17/7
0 1 0 4/7 -17/7
0 0 1 -1/14 -4/7

方程组的通解为: (17/7,-17/7,-4/7,0)^T + c(8,-8,1,14)^T.

Answer 2

(1)
x1 = -x3 + 5x4 - 2
x2 = 2x3 - 7x4 + 5
(x3,x4为自由变量)

(2)
x1 = 17 / 7 + 4x4 / 7
x2 = -17 / 7 - 4x4 / 7
x3 = -4 / 7 + x4 /14

追问

谢谢了，能否把解题步骤写详细一点啊，我要的是解题过程

追答

学过矩阵没？用矩阵的初等行变换求

追问

没有学过啊，帮帮忙了

追答

那就只能用消元法了