I would like to plot row 1 and row 3 of B as (x vs y plot graph).

1 次查看(过去 30 天)
I would like to plot row 1 and row 3 of B as (y vs x plot graph). However, I could only plot using the last iteration which is 10. What can i do to get all the iteration so i can plot y against x plot?
syms x y z
for i=1:10
eqn1 = 2.*x + 2.*i.*y + z == 2;
eqn2 = -x + y - i.*z == 3;
eqn3 = x + 2.*y + 3.*z == 10.*i;
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3], [x, y, z])
end
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
A = 
B = 
figure(1)
plot(B(1,:),B(3,:))

回答(1 个)

Davide Masiello
Davide Masiello 2022-9-25
编辑:Davide Masiello 2022-9-25
I suspect you want something like this
syms x y z
for i=1:10
eqn1 = 2.*x + 2.*i.*y + z == 2;
eqn2 = -x + y - i.*z == 3;
eqn3 = x + 2.*y + 3.*z == 10.*i;
[A(:,:,i),B(:,1,i)] = equationsToMatrix([eqn1, eqn2, eqn3], [x, y, z]);
end
figure(1)
plot(B(:,1,1),B(:,1,3),'-o')
  1 个评论
SAM MATHWORKS
SAM MATHWORKS 2022-9-26
Dear Mr. Masiello, it was not what I wanted, however your code had given me an idea to find the plot that I was looking for.
syms x y z
for i=1:10
eqn1 = 2.*x + 2.*i.*y + z == 2;
eqn2 = -x + y - i.*z == 3;
eqn3 = x + 2.*y + 3.*z == 10.*i;
[A(:,:,i),B(:,1,i)] = equationsToMatrix([eqn1, eqn2, eqn3], [x, y, z]);
end
figure(1)
plot(B(1,:),B(3,:),'-o')

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品


版本

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by