QUESTION: Answer as a series of four letters $(Y$ for Yes and $N$ for No$)$ in order.
Is it possible to find a $2 × 2$ matrix $M$ for which the equation $M\vec{x} = \vec{p}$ has:
$(a)$ no solutions for some but not all $\vec{p}$; exactly one solution for all other $\vec{p} ?$
$(b)$ exactly one solution for some but not all $\vec{p}$; more than one solution for all other $\vec{p} ?$
$(c)$ no solutions for some but not all $\vec{p}$; more than one solution for all other $\vec{p} ?$
$(d)$ no solutions for some $\vec{p}$, exactly one solution for some $\vec{p}$ and more than one solution for some $\vec{p} ?$
MY ANSWER: Looking at the equation we can deduce some basic information as follows-
case 1: if det$(M)$ is non zero Then the system has a unique solution that is $\vec{x}=M^{-1}\vec{p}$.
case 2: if det$(M)$ is zero
$$ case \space 2.a$$
If $\vec{p}$ is also zero then clearly $\vec{x}=\frac{0}0$ has infinite number of solutions..
$$case \space 2.b$$
If $\vec{p}$ is not zero then $\vec{x}$ is of the form $\frac{non-zero}{zero}$ which is inconsistent..
Following this knowledge let us examine the options one by one-
(a)
For us to have no solution, $M$ must have a zero determinant and $\vec{p}$ may take any non-zero value. But in that case, we cannot have exactly one solution for any $\vec{p}$. Therefore, $(a)$ is false $(N)$.
(b)
For us to have exactly one solution, $det(M)≠0$. Then, we cannot have more than one solution for any $\vec{p}$. So, this too is false $(N)$.
(c)
As said before, no solution occurs only when $det(M)=0$ and $\vec{p}≠0$. And if we choose $\vec{p}=0$ here then we have more than one (actually infinite) number of solutions. So this is possible.. $(Y)$.
(d)
Clearly, the three conditions given cannot hold simultaneously for some given $M$. So it is false $(N)$.
Are my reasoning correct? Is there a better way to solve this equation?
Also, it would be much help if someone can solve it geometrically.. Since, I am keen to know that approach.
Thank you so much in advance :)