Consider this problem:
Given are the matrix $A$ and the vector $\mathbf{b}$ with $ \alpha $ and $\beta \in \Bbb R $;
$$
State for which alpha and beta the linear system of equations $$ A \cdot x = \mathbf b $$ has none, one or infinite number of solutins.
Now what I tried to do is set up a system of equations. That looks like this
$$ \begin{matrix} 0 & 1 & 2 | 5\\ 1 & 3 & 4 | 6 \\ 2 & 4 & a | b \end{matrix} $$
Now the goal is to get $0 \ 0 \ a$ in the last line so what I did i took the $-2$ second line and added it to the 3rd. So we get
$$ \begin{matrix} 0 & 1 & 2 | 5\\ 1 & 3 & 4 | 6 \\ 0 & 2 & -8+a |-12+ b \end{matrix} $$
Than I had to get rid of the so I took $-2 \cdot 1 + 3$ and I got.
$$ \begin{matrix} 0 & 1 & 2 | 5\\ 1 & 3 & 4 | 6 \\ 0 & 0 & -4+a |-22+ b \end{matrix} $$
Now I've look at the cases when when the system has no solutions. That is when $\operatorname{rank} (A) < \operatorname{rank}(A|b)$. Now I'm not sure how I can confirm the rank of a matrix, I would assume the rank is 3 because we have 3 lines. So if the rank of $A$ is 3 than the rank of $(A|b)$ has to be 2 or lower. Now if I want to get a rank 2, $(A|b)$ do I have to set $\alpha=0$?
Because than we would have all $0$ in the last line thus the rank would be $2$. Some insight would be great. Thank you!
PS: If the solution to my system is false, please let me know.
Suppose that $\alpha=4$. Then you have the matrix$$\left[\begin{array}{ccc|c}0&1&2&5\\1&3&4&6\\0&0&0&b-22\end{array}\right].\tag1$$The matrix to the left of the vertical bar has rank $2$ (the rank is the number of linearly independent lines, whereas the rank of $(1)$ depends upon $b$: it is $3$ if $b\ne22$, but it is $2$ if $b=22$. So, if $b\ne22$, the system has no solutions, but it has solutions if $b=22$.
And if $\alpha\ne4$ both matrices have rank $3$, and therefor the system has solutions.