plot solution of linear systems (linsolve)

Hello.

I am looking for a code that will help me plot a graph of solution of these two linear systems. My codes are the following, however, how can I get a solution into one plot with lines representing solution for each equation?

? +? +? = 2

? +2? −? = 4

−? +2? +? = 2

% LS for A
A = [1 1 1; 1 2 -1; -1 2 1];
a = [2; 4; 2];
linsolve(A,a)
plot ????

[Matt J]

The result of your example code is 3 values.

>> linsolve(A,a)

ans =

    0.7500
    1.5000
   -0.2500
   

Please elaborate how these 3 values would form a plot

[M03]

Right? I was also wondering. But I need a plot, which contains three lines (one for every equation) describing the relationship of linear system. It should look like the one on the picture

[M03]

I am actually looking for ther intersection if they have a solution

[Matt J]

But your original example is 3x3 system (with 3 unknowns), while your figure illustration is a 3x2 system (2 unknowns).

[M03]

That's the point. I should make one for the system above.

And if I have a 3x2 system like this, how will the code look like?

+2? +? =

7 ? −2? = 1

−? +5? = 4

[Matt J]

An equation in 3 unknowns does not define a line, but rather a plane.

Answer 1

[madhan ravi]

doc syms
doc fplot
doc equationsToMatrix
doc linsolve
doc plot % see how to plot a single point especially markers
doc legend
doc xlim

Answer 2

[Star Strider]

This was something of a challenge in the format presented.

% % % Equations: -2*x + y = 1;  3*x + 2*y = 10;  -15*x + 2*y = -80
X = [-5 1; 10 1];                                           % Create: [X 1] Matrix
A = [2 1; -3/2 10/2; 15/2 -80/2];                           % ‘Solve’ The Equations In ‘y = B1*x + B2’ Form
Y = X * A';                                                 % Solve For ‘Y’
CoefVct = linsolve([A(:,1), ones(size(A(:,1)))], A(:,2));

figure
plot(X(:,1), Y)
Ax = gca;
Ax.YAxisLocation = 'origin';
ylim([0 20])

producing:

CoefVct =
         -5.21862348178138
          2.58299595141701

and the plot.

The plot appears to be as requested.