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．
求解方程组： {3x1+4x2-6x3=4 {x1-x2+4x3=1 {-x1+3x2-10x3=1
- 2024-11-07 03:14:08
- 线性代数（工）(13175)