Question

试用克拉默法则求下列线性方程组的解 x1+x3=1;2x1+2x2+3x3=3;x2+x3=-1？

Answer 1

根据克拉默法则,该线性方程组的系数行列式为
| 1 0 1 |
D = | 2 2 3 | = 1 x (2 - 3) + 1 x (2 - 0) = 1
| 0 1 1 |
而各个未知数对应的行列式分别为
| 1 0 1 |
D1 = | 3 2 3 | = 1 x (2 - 3) + 1 x (3 + 2) = 4
| -1 1 1 |
| 1 1 1 |
D2 = | 2 3 3 | = 1 x (3 + 3) - 1 x (2 - 0) + 1 x (-2 - 0) = 2
| 0 -1 1 |
| 1 0 1 |
D3 = | 2 2 3 | = 1 x (-2 - 3) + 1 x (2 - 0) = -3
| 0 1 -1 |
根据克拉默法则x1 = D1/D = 4,x2 = D2/D = 2,x3 = D3/D = -3.,2,试用克拉默法则求下列线性方程组的解 x1+x3=1;2x1+2x2+3x3=3;x2+x3=-1
x1+x3=1
2x1+2x2+3x3=3
x2+x3=-1