3 10-1 0 26 01226 00 3200 00 00000 00 00 00000 runk(A)=2<5,所以原方程组有非零解 同解方程组为 1=x3+x4+5xs 2 ixa-2x-6x 3 4 法1. 3 11[11「5 取 4 0 0得 6 5            → 0 0 0 0 0 0 0 0 0 0 0 1 2 2 6 1 2 3 3 7             − − − → 0 0 0 0 0 0 0 0 0 0 0 1 2 2 6 1 0 1 1 5 rank(A) = 2  5, 所以原方程组有非零解. 同解方程组为    = − − − = + + 2 3 4 5 1 3 4 5 2 2 6 5 x x x x x x x x 法1. , 0 0 1 5 4 3           =           x x x 取 , 0 1 0                     1 0 0 , 2 1 2 1       − =      x x 得 , 2 1       −       − 6 5