System of linear equations with parameter - strange result, does this make sense

Question

I have a system of linear equations $Ax = b$ where $A \in \mathbb R^{3\times 3}$, and $x,b = \in \mathbb R^{3 \times 1}$.

$A$ has some parameter $\alpha$ in its entries.

I was asked to find for which values of $\alpha$ does this system have a unique solution, no solutions and infinite many solutions.

I found two different problematic values for $\alpha$, but my problem is that it seems like in both these values, I have infinite solutions.

There seems to be no case in which we have no solution, 2 different cases where we get infinite many solutions, and the rest ofcourse is unique.

Does this make sense? this is the first question I encountered where there is no value for no solutions and 2 different values for infinite solutions. so far its been one value of alpha for each, but I checked myself 5 times and I've yet to find a mistake.

I don't want to trouble you with the details of the problem, just the final question - does this situation make sense or did I make a mistake for sure?

Answer 1

Here is an example concerning your question ""Can a linear system of equations with one parameter have two different values of the parameter for which the system has infinite solutions". Consider the three linear equations $$ x=0, y=0, \alpha(\alpha-1)z=0. $$ For $\alpha=1$ and $\alpha=0$ there are infinitely many solutions $(0,0,z)\in K^3$, provided the field $K$ is infinite. Otherwise there is only the trivial solution $(0,0,0)$.

Answer 2

Lets us have a look at this simple system: $$ \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ It will have either unique solutions ($\alpha \ne 0$) or infinite many solutions. But no case of no solution. So such systems exist.