# Plotting the linear system

Tags:
1. Mar 8, 2016

### gruba

1. The problem statement, all variables and given/known data
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.
2. Relevant equations
-Cramer's rule
-Kronecker-Capelli's theorem

3. 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?

2. Mar 9, 2016

### geoffrey159

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

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

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).

We already discussed that in length yesterday