Calculate a in dependence of b so that equation system is solveable?

Question

Given is following equation system:

$\begin{pmatrix} 2 & 1 & 1 \\ 3 & 2 & 3 \\ 4 & 3 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} a \\ b \\ 1 \end{pmatrix} $

Determine a in dependence of b so that system is solvable and write it as a general solution?

I tried to do elimination, hoping that I will get at the end something helpful that will lead me to the solution. But I have to many unknowns and cant get a.

$\begin{pmatrix} 2 & 1 & 1 & a\\ 0 & -1 & -3 & 3a-2b \\ 0 & 0 & 0 &a-2b+1 \end{pmatrix} $

Answer 1

The last row of your equation states that $0\times x+0\times y+0\times z=a-2b+1$ Hence the system of equations to be consistent $a-2b+1$ must be $=0$ So the dependence is $a=2b-1$

As a general solution you can write:

Since doing elimination does not change the solution of a system of equations, hence the dependency of a on b also doesnot change on performing Gaussian elimination.