Give positive integers $m$, $n$ and example of $m \times n$ matrix $A$ with the following property: $Ax=b$ has no solutions for some $b \in \Bbb R^n$, and one solution for every other $b \in \Bbb R^n$.
Can you please explain the reasoning behind your answer?
 A: Are you sure that you’ve stated the problem correctly? There is no such example.
If $Ax=b$ has no solutions, and $\alpha\ne 0$, then $Ax=\alpha b$ has no solutions: if $Ax=\alpha b$, then $A\left(\alpha^{-1}x\right)=b$. Thus, if $b$ is the only vector in $\Bbb R^n$ for which the equation has no solutions, then $b=0$, so that $\alpha b=b$ for all $\alpha\in\Bbb R$. In other words, you’re looking for $m,n$, and an $m\times n$ matrix $A$ such that $Ax=0$ has no solution, but $Ax=b$ has one solution for each non-zero $b\in\Bbb R^n$. This, however, is impossible, since the homogeneous equation $Ax=0$ always has the solution $x=0$.
Added: If you want an example in which $Ax=b$ has solutions for some choices of $b$ but has no solution for other choices of $b$, start with a really simple example with no solutions, like
$$\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}\;;\tag{1}$$
can you see why $(1)$ has no solution? And can you see a way to change the righthand side to get a matrix equation with the same coefficient matrix that does have a solution?
A: My answer is very similar in spirit to Brian's:  If $Ax = b$ has no solution, but $Ax = c$ has a solution for every $c \ne b$, then pick $c_1 \in R^n$, $c_1 \ne b$, $c_1 \ne 0$, and find $x_1$ such that $Ax_1 = c_1$.  Now, we can assume $b \ne 0$ since $A0 = 0$; taking $b = 0$ always admits zero as a solution.  So set $c_2 = c_1 + b$.  Then since
$b \ne 0$, $c_2 \ne c_1$, and since $c_1 \ne 0$, $c_2 \ne b$.  So there is an $x_2$ with $Ax_2 = c_2$.  But then $A(x_2 - x_1) = Ax_2 - Ax_1 = c_2 - c_1 = b$, a contradiction.
So I can't give an example of such an $A$ as the OP requests, for there is none to be had.  At least knowing this will save some time in the search for such $A$!
Well, I hope this helps! Cheers!
