How I should discuss the solution of the following system of equations depending on the parameter of Lambda
$x-y+z=\lambda \ , \ \ \lambda \in \mathbb{R} \\ 2x-3y+4z=0\\ 3x-4y+5z=1$
The only way that I have discovered that the function has one solution on 1 and none solution if lambda is different from 1, was testing the equation. Exist a way to conclude the same but without testing random numbers? thanks.
Big hint: every system of linear equations has either 0 solutions, exactly 1 solution, or infinitely many solutions (this is a foundational result in linear algebra). So, which case is it for these equations?
This particular problem is easy to simplify. Subtracting the second equation from the last equation gives $$ x - y + z = 1. $$ So in order for the set of equations to be consistent, $\lambda$ must equal 1, or the first equation would be false. That is, if $\lambda \neq 1$, you know there are 0 solutions. But now you have to decide: in the $\lambda = 1$ case, do you have a single solution, or infinitely many? Remember, if you can find two solutions, then you can find infinitely many solutions.
More hint: what happens when $x = 1$, $y = 2$, and $z = 1$?