Let $A$ be a $3 \times 2$ matrix . Explain why the equation $A\vec{x} = \vec{b}$ cannot be solved for every $\vec{b}$ in $\mathbb{R}^3$. What about a $4 \times 3$ matrix?
I'm not sure how to answer this.. I know that a $3 \times 2$ matrix only has 2 pivot columns and could therefore only have 2 pivot positions maybe? I also know that the third row would have to be like maybe infinitely solutions or? I'm not really sure how to explain it
On
$A$ is a matrix and $b$ a vector. what means the equation $Ax=b$ has solutions or not?
possesses solutions means that the vector $b$ is in the sub space spanned by the column vectors of $A$. If $B$ is the matrix obtained by concatenating $A$ and $b$, then this is equivalent to $rank (A) = rank (B)$; by negation so that $Ax = b$ admits no solution iff $b$ is not in the space generated by the columns of $A$ ie $rank (B) = rank (A) +1$
In the case of a non-square matrix there is an infinity of $b$ in the space that satisfies $rank (A) = rank (B)+ 1$ and also an infinity of $b$ that satisfies $rank (A) = rank ( B)$.
Consider the linear transformation $T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{3}$ deined by $$T(X)=AX.$$ Now existence of solution for $AX=b$ means that $b\in\mathbb{R}^{3}$ has preimage under $T$ and which is not possible for every vector of $\mathbb{R}^{3}$ as $T$ can't onto. Same is true for the matrix mention by you.