Find a matrix of $A$ such that $Ax=b$ has a solution for infinitely many $b\in\mathbb{R}^3$ Find a matrix of $A$ such that $Ax=b$ has a solution for infinitely many $b\in\mathbb{R}^3$ but $Ax=b$ does not have a solution for some $b\in\mathbb{R}^3$.
The matrix is supposed to be $3\times4$. I'm shaky on finding a matrix that has infinitely many let alone finding that also doesn't have a solution for some $b$. I appreciate the help.

Consider for example
$$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}.$$
Its kernel is spanned by $e_3\equiv(0,0,1)^T$, while its range is spanned by $e_1$ and $e_2$.
It follows that $Ax=b$ has solution for all $b\in\mathrm{span}(\{e_1,e_2\})$, but has no solution for all $b\in\mathbb F e_3$.
More explicitly, given any $b=c_1 e_1+c_2 e_2$, the solution of the linear system is $x=c_1 e_2+c_2 e_2$, while for any $b$ of the form $b=c_3 e_3$ there is no$x$ such that $Ax=b$.
More generally, $Ax=b$ has a solution for $x$ if and only if $b\in\mathrm{Range}(A)$, so any matrix whose range is both nontrivial and smaller than the full space will satisfy your constraint.
If you want the matrix to be $3\times 4$, you can trivially add a zero column to the above matrix $A$.
