线性代数,x1+x2-3x3-x4=0 3x1-x2-3x3+4x4=0 x1+5x2-9x3-8x4=0

2024-12-01 17:17:19
推荐回答(1个)
回答1:

解: 系数矩阵 =
1 1 -3 -1
3 -1 -3 4
1 5 -9 -8

r2-3r1, r3-r1
1 1 -3 -1
0 -4 6 7
0 4 -6 -7

r3+r2
1 1 -3 -1
0 -4 6 7
0 0 0 0

r2*(-1/4)
1 1 -3 -1
0 1 -3/2 -7/4
0 0 0 0 0

r1-r2
1 0 -3/2 3/4
0 1 -3/2 -7/4
0 0 0 0

通解为: c1(3,3,-2,0)^T + c2(3,-7,0,-4)^T, c1,c2为任意常数.