Question

怎么用MATLAB解这个三元一次方程组

0.0979*x+0.225*y+0.3793*z=0.3793
0.225*x+0.3793*y+1.5*z=1.5
0.3793*x+1.5*y+5*z=2.04

Answer 1

>> [x,y,z]=solve('0.0979*x+0.225*y+0.3793*z=0.3793','0.225*x+0.3793*y+1.5*z=1.5','0.3793*x+1.5*y+5*z=2.04')

x =
6.6433242786575706312672289994801
y =

-2.1102674253251990112697519514542
z =

0.5371176478185963952929935935357

Answer 2

>> A=[0.0979 0.225 0.3793
0.225 0.3793 1.5
0.3793 1.5 5]
A =
0.0979 0.2250 0.3793
0.2250 0.3793 1.5000
0.3793 1.5000 5.0000
>> b=[0.3793;1.5;2.04]
b =
0.3793
1.5000
2.0400
>> x=A\b
x =
6.6433
-2.1103
0.5371
>>
线性方程组，直接使用左除或者用矩阵的逆求解都行。