A rank=r=n recatangular matrix has a left inverse (B) and Ax=b has solutions as Bb; then can't I solve Ax=b for any b? Let's take a rectangular matrix A with rank=r=n
$$A=\begin{bmatrix}4&0\\0&5\\0&0\end{bmatrix}$$
Infinitely many left inverses B exist of the form
$$B=\begin{bmatrix}1/4&0&b13\\0&1/5&b23\end{bmatrix}$$
Coming to the system of linear equations
$$Ax=\begin{bmatrix}b1\\b2\\b3\end{bmatrix}$$
This can have solutions only if b3=0.
But if I take $$b=\begin{bmatrix}4\\5\\5\end{bmatrix}$$
And try to solve Ax=b by multiplying by B (letting b12=b23=0) on both sides I get
$$BAx=Bb$$
$$x=Bb$$
$$x=\begin{bmatrix}1\\1\end{bmatrix}$$
Subsitituting this x back in Ax=b gives $$Ax=\begin{bmatrix}4\\5\\0\end{bmatrix}=\begin{bmatrix}4\\5\\5\end{bmatrix}$$
This implies $$0=5$$
I am just trying to understand why I am getting such a contradiction. I know that for $$b=\begin{bmatrix}4\\5\\5\end{bmatrix}$$ Ax=b is unsolvable, but what went wrong in the algebraic manipulations I did to show 0=5. Was it like a divide by zero thingy?
 A: You're getting a contradiction (absurdity) because you start with an absurdity.
$$A\mathbf x=\begin{bmatrix}4&0\\0&5\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4x\\5y\\0\end{bmatrix}=\mathbf b$$
ie, $b_3$ must be $0$ for a solution to exist. Since you take a $\mathbf b$ not of that form, you get an absurdity implying that no solution $\mathbf x$ exists for that choice of $\mathbf b$

Consider the following analogy: let us consider the map $f\colon x\mapsto x^2$ on $\Bbb R$ and find a real solution $x$ to the equation $f(x)=-1$, ie, $x^2+1=0$ which is an absurdity for real $x$, ergo no real solution $x$ exists. Indeed, solutions exists if the map was on $\Bbb C$, namely $x=\pm i$
One more example: Let us try to find a solution to the equation $x+1=x$ where $x\in\Bbb C$. But this gives us $x+1-x=0$, ie, $1=0$, an absurdity, ie, no solution exists on $\Bbb C$. However, if $x\in R$ where $R=(\{0\},+,\cdot)$ is the trivial (zero) ring where $1$ and $0$ mean the same element, it has a solution, namely the only element of $R$
