How to plot the linear system solutions with multiple solutions?

[gruba]

Homework Statement

Solve the linear system of equations:
ax+by+z=1
x+aby+z=b
x+by+az=1
for a,b\in\mathbb R
and plot equations and solutions in cases where the system is consistent.

Homework Equations

-Cramer's rule
-Kronecker-Capelli's theorem

The Attempt at a Solution

Using Cramer's rule, we find the determinant of the system and determinant for each variable:
D=b(a-1)^2(a+2)
D_x=b(a-b)(a-1)
D_y=(a-1)(ab+b-2)
D_z=b(a-1)(a-b)

For b\neq 0 \land a\neq 1\land a\neq -2\Rightarrow D\neq 0 system has unique solution:
(x,y,z)=\left(\frac{a-b}{(a-1)(a+2)},\frac{ab+b-2}{b(a-1)},\frac{a-b}{(a-1)(a+2)}\right).

How to plot the equations with intersection (point) in this case?

Second case, a=1.
Solvind the system using Kronecker-Capelli's theorem gives:
b=1\Rightarrow infinitely many solutions.
b\neq 1\Rightarrow the system is inconsistent.
This gives (x,y,z)=(1-y-z,y,z).

How to plot the equations with intersection (line) in this case?

Third case, a=b=-2\Rightarrow infinitely many solutions.
(x,y,z)=\left(z,\frac{-z-1}{2},z\right).

How to plot the equations with intersection (line) in this case?

 

[geoffrey159]

I didn't check the correctness of your result, but assuming all is right :

(x,y,z)=\left(\frac{a-b}{(a-1)(a+2)},\frac{ab+b-2}{b(a-1)},\frac{a-b}{(a-1)(a+2)}\right).

If it has a unique solution, so it's a point.

This gives (x,y,z)=(1-y-z,y,z). How to plot the equations with intersection (line) in this case?

That means that the solution verifies $x+ y + z - 1 = 0$ when $x,y,z$ describe $\mathbb{R}^3$.
It's a plane passing through (1,0,0), (0,1,0), and (0,0,1).

Third case, a=b=-2\Rightarrow infinitely many solutions.
(x,y,z)=\left(z,\frac{-z-1}{2},z\right).

How to plot the equations with intersection (line) in this case?

We already discussed that in length yesterday