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

2 visualizzazioni (ultimi 30 giorni)
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,:))

Risposte (1)

Davide Masiello
Davide Masiello il 25 Set 2022
Modificato: Davide Masiello il 25 Set 2022
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 Commento
SAM MATHWORKS
SAM MATHWORKS il 26 Set 2022
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')

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