I'm trying to figure out all matrices that satisfy this condition: For all $x$ there exists a $b$ such that $Ax = b$.
$A$ is an $n \, \text{x} \, n$ matrix. Also both $x,b$ are $n \, \text{x} \, 1$ column vectors.
I know that if a matrix is invertible then $Ax = b \Longrightarrow x = A^{-1}b$.
What I'm uncertain about is whether there are non invertible matrices that also satisfy this condition. I was wondering whether such a matrix exists or not?
On
If the question is really finding all matrices such that:
For all $x$ there exists a $b$ such that $Ax=b.$
Then you do not have to go too far to find an answer.
Pick any matrix $A$ that you like and multiply it by $x$. The answer is $b=Ax$
That was too easy. I guess you wanted to ask:
Given $x$ and $b$, is it possible to find a non invertible matrix such that $Ax=b?$
It depend on $x$ and $b$. For example if $x = \begin{bmatrix} 1 \\1 \end{bmatrix}$ and $b = \begin{bmatrix} 2 \\2 \end{bmatrix}$
then the matrix $ A=\begin{bmatrix} 1&1 \\1&1 \end{bmatrix} $ satisfies $Ax=b.$
It is true if and only if $A=0$ and $b=0$.
Indeed
$$A(1,0,...,0)=A(2,0,...,0)\iff (A)_{i,1}=0 \quad b(1)=0$$
and so on.