Does there exist a matrix $A$ with vectors $p,q,r$ such that.... Does there exist a matrix $A$ with vectors $p, q, r$ such that $Ax = p$ has no solution, $Ax = q$ has exactly one solution, and $Ax = r$ has infinitely many solutions?  Justify your answer.  If such a matrix does not exist, are there matrices for which two out of the three cases can hold?  What are they?
I believe that there is such a matrix $A$ because if $Ax = q$ and $Ax = r$ has at least one solution, then that means that $q$ and $r$ are in the span of the vectors in the matrix $A$, which implies that $q$ and $r$ are multiples of the vectors in the matrix.
This also implies that the columns of $A$ are linearly dependent and therefore NOT invertible.
Do you think this is a proper justification? did i make any errors somewhere?
 A: Guide:


*

*Consider reducing the system to reduced row echelon form. Let the RREF of $A$ be $R$

*It has exactly one solution if and only if every column of $R$ is a pivot column and there is no zero rows with the right hand side being non-zero.

*Consider what happens if $R$ has non-pivot columns.

A: The answer is easy if you know what $\operatorname{Nul}A$ and $\operatorname{Col}A$ mean.
Let $A$ be an $m\times n$ matrix. The null space of $A$ is defined by
$$
\operatorname{Nul}A = \{ \mathbf{x}\in\mathbb{R}^n \mid A\mathbf{x}=\mathbf{0} \} \subset \mathbb{R}^n
$$
and the column space of $A$ is defined by
$$
\operatorname{Col}A = \{ \mathbf{y}\in\mathbb{R}^m \mid A\mathbf{x}=\mathbf{y} \text{ for some $\mathbf{x}\in\mathbb{R}^n$} \} \subset \mathbb{R}^m
$$

  
*
  
*If $\dim\operatorname{Col}A=m$, then $A\mathbf{x}=\mathbf{y}$ has a solution for all $\mathbf{y}\in\mathbb{R}^m$. Otherwise, there exists $\mathbf{y}\in\mathbb{R}^m$ such that $A\mathbf{x}=\mathbf{y}$ does not have a solution.
  
*If $\dim\operatorname{Nul}A=0$, then $A\mathbf{x}=\mathbf{y}$ has a unique solution when any solution exists. Otherwise, $A\mathbf{x}=\mathbf{y}$ has infinitely many solutions when any solution exists.

Therefore, there is no $A$ satisfying all three cases because of (2). But there are matrices for which two out of the three cases can hold.

(Case 1) $\dim\operatorname{Col}A<m$ and $\dim\operatorname{Nul}A=0$: Either $A\mathbf{x}=\mathbf{y}$ has a unique solution or no solution. For example,
  $$
A = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \dim\operatorname{Col}A=1<2, \quad \dim\operatorname{Nul}A=0
$$
(Case 2) $\dim\operatorname{Col}A<m$ and $\dim\operatorname{Nul}A>0$: Either $A\mathbf{x}=\mathbf{y}$ has infinitely many solutions or no solution. For example,
  $$
A = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad \dim\operatorname{Col}A=0<2, \quad \dim\operatorname{Nul}A=1>0
$$

A: Suppose that $Ax=r$ has infinitely many solutions and let $x_1,x_2$ solutions of this equation with $x_1 \ne x_2$.
If $x_0$ is  a solution of $Ax=q$ and $z:=x_0+x_1-x_2$, then $z \ne x_0$ and $Az=q+r-r=q$. Hence the equation $Ax=q$ has more then one solution.
