使用初等行變換的方法解題寫出方程對應的增廣矩陣為1 1 1 1 1 73 2 1 1 -3 -20 1 2 2 6 235 4 -3 3 -1 12 r2-3r1，r4-5r1~1 1 1 1 1 70 -1 -2 -2 -6 -230 1 2 2 6 230 -1 -8 -2 -6 -23 r1+r2，r2+r3，r4+r3~1 1 1 1 1 70 0 0 0 0 00 1 2 2 6 230 0 -6 0 0 0 r1-r3，r4/-6，r1+r4，r3-2r4，交換行次序~1 0 0 -1 -5 -160 1 0 2 6 230 0 1 0 0 00 0 0 0 0 0得到通解為(-16,23,0,0,0)^T+c1(1,-2,0,1,0)^T+c2(5,-6,0,0,1)^T，c1c2為常數

解: 增廣矩陣 =

1 3 5 -4 0 1

1 3 2 -2 1 -1

1 -2 1 -1 -1 3

1 -4 1 1 -1 3

1 2 1 -1 1 -1

ri-r1, i=2,3,4,5

1 3 5 -4 0 1

0 0 -3 2 1 -2

0 -5 -4 3 -1 2

0 -7 -4 5 -1 2

0 -1 -4 3 1 -2

r1+3r5,r3-5r5,r4-7r5,r5*(-1)

1 0 7 5 3 -5

0 0 -3 2 1 -2

0 0 16 -12 -6 12

0 0 24 -16 -8 16

0 1 4 -3 -1 2

r1-3r2,r3+6r2,r4+8r2,r5+r2 (這樣可避免分數運算)

1 0 -2 -1 0 1

0 0 -3 2 1 -2

0 0 -2 0 0 0

0 0 0 0 0 0

0 1 1 -1 0 0

r3*(-1/2),r1+2r3,r2+3r3,r5-r3

1 0 0 -1 0 1

0 0 0 2 1 -2

0 0 1 0 0 0

0 0 0 0 0 0

0 1 0 -1 0 0

所以方程組的全部解為: (1,0,0,0,-2)^T+c(1,1,0,1,-2)^T.