求解方程组：
{3x1+4x2-6x3=4
{x1-x2+4x3=1
{-x1+3x2-10x3=1
【正确答案】：方程组的增广矩阵为 (A，β)= (3 4 -6 ┆ 4 1 -1 4 ┆ 1 -1 3 -10 ┆ 1) 对增广矩阵作初等变换得 (A，β)= (3 4 -6 ┆ 4 1 -1 4 ┆ 1 -1 3 -10 ┆ 1) → (1 -1 4 ┆ 1 3 4 -6 ┆ 4 -1 3 -10 ┆ 1) → (1 -1 4 ┆ 1 0 7 -18 ┆ 1 0 2 -6 ┆ 2) → (1 -1 4 ┆ 1 0 1 -3 ┆ 1 0 7 -18┆ 1) → (1 -1 4 ┆ 1 0 1 -3 ┆ 1 0 0 3 ┆ -6) → (1 0 0 ┆ 4 0 1 0 ┆ -5 0 0 1 ┆ -2) 因此方程组的解为x1=4，x2=-5，x3=-2．
