Here, $A$ has $5$ columns w/ $4$ rows. $b$ is a given $4\times1$ vector.
Is it possible for $b$ not to belong to $\mathrm{col}(A)$? I am confused because I know that $\mathrm{col}(A)=\mathrm{span}\{\text{columns of} A\}$ and b/c there are more columns than rows, columns of A are linearly dependent. Wouldn't this mean that $\mathrm{col}(A)$ spans the entirety of $\mathbb{R}^4$? If so, how can it be that after solving $Ax=b$ I ended up with something that cannot be (e.g. 1=0). Because there are more columns than rows, does $\mathrm{col}(A)$ no longer cover all of $\mathbb{R}^4$? Thanks in advance.
$A=\pmatrix{1&0&2&0&-1\\0&1&3&0&2\\0&0&0&3&9\\1&0&2&0&-1}$
$%\pmatrix{1&0&0&1\\0&1&0&0\\2&3&0&2\\0&0&3&0\\-1&2&9&-1}^T$
$b\pmatrix{1\\1\\1\\2}$
Best Answer
$\dim \mathrm{col} (A)=\dim \mathrm{row} (A)=\mathrm{rank}(A)$
Since in your case the first row of $A$ equal to the last one you got $\mathrm{rank}(A)\le3 <4$ and therefore $\mathrm{col} (A)$ doesn't span $\mathrm{R}^4$, i.e. $ \mathrm{col}(A)\subsetneq \mathrm{R}^4$. Thus there is $b\in \mathrm{R}^4$ such that $b\not\in \mathrm{col}(A)$.