# 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?

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 $$\operatorname{Col}A = \{ \mathbf{y}\in\mathbb{R}^m \mid A\mathbf{x}=\mathbf{y} \text{ for some \mathbf{x}\in\mathbb{R}^n} \} \subset \mathbb{R}^m$$

1. 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.

2. 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, $$A = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad \dim\operatorname{Col}A=0<2, \quad \dim\operatorname{Nul}A=1>0$$

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.
• So using your guide, if $Ax = r$ has infinite solutions, then $A$ can't have a pivot in every column, but if $Ax = q$ has a unique solution, then A must have a pivot in every column, thus we come to a contradiction. There is no such matrix A that exists – Soon_to_be_code_master Mar 13 '18 at 6:31
• You have rule out the possibility of the second and third case happening simultaneously. Try to come out with an example where first case and third case happen simultaneously. – Siong Thye Goh Mar 13 '18 at 6:33
• If we rule out the possibility of the second case happening. Then such a matrix A exists if case one and case three occurs because case 3 requires A to not have a pivot in every column, and case 1 may or may not need a pivot in every column (as long as there is a zero row in RREF of A) – Soon_to_be_code_master Mar 13 '18 at 7:02
• Let $A= 0$, $p=1$, and $r=0$ . – Siong Thye Goh Mar 13 '18 at 7:06
• $A$ is the zero matrix? and $p$ is a vector of only ones? how is that possible if $p$ is supposed to be a column of $A$ – Soon_to_be_code_master Mar 13 '18 at 7:20

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.