Hi there i can't remember exactly how look like my exercise but i came here to ask if i did good or wrong the way i solve it.I will write similar equations try to understand my logic how i tryed solving it and tell me if it is good.
The exercise ask me to investigate a system 2x2.
What i did was the exercise was similar like this x(λ-1)+3y-5λ=-1 and -x(λ-1)+4y+12=1.What i did was to add first equation x(λ-1)+3y-5λ=-1 with the second equation -x(λ-1)+4y+12=1.As a result the variable x will be gone.
Step 1: I find a relation i name it A which relation hasn't the variable x.
Step 2: Next i solve after y=... and example i had y=λ-1 (lets say)
Step 3: I took y and i go to the relation A(is the sum of first equation with the second) and where was y i put the result (which was lets say λ-1).
Step 4: By this way I found the λ and I remember it was -1.
Step 5: I take the λ and I go to y=λ-1 and put λ=-1 then I found y.
Step 6: I take the y and I go to first equation and i solve and i find x My way is right?
Thanks in advance.
Upon eliminating $x$ you find $y(\lambda)$. You substitute that into one of the original equations (doesn't matter which) to find $x(y,\lambda)$. Replacing $y$ by the $y(\lambda)$ you previously found gives you $x(\lambda)$. You should not "find" $\lambda$ at any stage of this process: the solution depends on $\lambda$, but it does not include $\lambda$.
The main interesting question, which I don't see in your description of your method of solution, is how the number of solutions depends on $\lambda$. Typically these problems are written in such a way that there is in fact such a dependence (i.e. the number of solutions is not the same for all values of $\lambda$). In your example here for instance, the number of solutions is different for $\lambda \neq 1$ and $\lambda=1$. There is some deep linear algebra explanation for this, but if you don't know that you can just look at the system itself to see what's happening.