I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.

1 次查看（过去 30 天）
Oddur Bjarnason
Oddur Bjarnason 2016-10-12
回答： Oddur Bjarnason 2016-10-14
3*x/5+y/5=x
2*x/5+0.8*y=y
x+y=1

请先登录，再回答此问题。

采纳的回答

Oddur Bjarnason
Oddur Bjarnason 2016-10-14
I found the answer to my question:
3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1
3*x/5-5/5*x+y/5=0 2*x/5+4/5*y-5/5*y=0 x+y=1
-2/5*x+y/5=0 2/5*x-1/5*y=0 x+y=1
A=[-2/5,1/5;2/5,-1/5;1,1] B=[0;0;1] linsolve(A,B)
ans = 0.3333 0.6667

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Linear Algebra 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by