求矩阵A=｛x 1 1,1 x 1,1 1 x}的秩

｛x 1 1, 1 x 1, 1 1 x}→r2+(-r3)→｛x 1 1, 0 x-1 1-x, 1 1 x}→r3+(-r1/x)→｛x 1 1, 0 x-1 1-x, 0 1-1/x x-1/x}→r1/x→｛1 1/x 1/x, 0 x-1 1-x, 0 1-1/x x-1/x }→r1+r3→｛1 1 x, 0 x-1 1-x, 0 1-1/x x-1/x }→xr3→｛1 1 x, 0 x-1 1-x, 0 x-1 x^2-1}→r3+(-r2)→ ｛1 1 x, 0 x-1 1-x, 0 0 x^2+x-2}→r2/(x-1)→｛1 1 x, 0 1 -1, 0 0 x^2+x-2}→r1+(-r2)→｛1 0 x+1, 0 1 -1, 0 0 x^2+x-2},所以非零行为0秩为3