Left Inverses of a Matrix I am confused about left inverses after examining two particular problems. From what I understand in order to find the left inverse of an $m\times n$ matrix I must first transpose it then augment it with the identity matrix that would result after multiplying $A$ by an unknown $X$. After this I should perform RREF as far as I can and whatever I'm left with (barring any inconsistent rows) will be the columns of $X$ (the inverse of $A$). At that point it should be true that $XA = I$, however I do not get that for the first problem. I checked to make sure the first matrix does not have a right inverse so I don't know why $AX$ produces the identity matrix when it was the left inverse that was being solved for.
$A = \begin{bmatrix}1 & 2 \\ 2 & 5 \\ 3 & 7\end{bmatrix}$
$\begin{bmatrix}1 & 2 & 3 & | & 1 & 0\\ 2 & 5 & 7 & | & 0 & 1\\\end{bmatrix} => RREF => \begin{bmatrix}1 & 0 & 1 & | & 5 & | & -2\\0 & 1 & 1 & | & -2 & | & 1\end{bmatrix}$
Therefore,
$X = \begin{bmatrix}5-r & -2-t \\ -2-r & 1-t \\ r & t\end{bmatrix}$ (See correction)
However $XA$ does not produce the identity matrix while $AX$ does. Is this because I transposed $A$ in order to augment it and find $X$? As far as I know, $X$ should be the left inverse, so $XA$ should work. What really confuses me is that for this second matrix, $B$, I attempt the exact same thing and find that $XB = I$ as expected.
$B = \begin{bmatrix}2 & -1 \\ 4 & -1 \\ 2 & 2\end{bmatrix}$
$\begin{bmatrix}2 & 4 & 2 & | & 1 & 0\\ -1 & -1 & 2 & | & 0 & 1\end{bmatrix} => RREF => \begin{bmatrix}2 & 4 & 2 &| & 1 & 0 \\ 0 & 1 & 3 & | & 1/2 & 1\end{bmatrix}$
Therefore,
$X = \begin{bmatrix}-1/2+5s & 1/2-3s & s \\ -2+5t & 1-3t & t\end{bmatrix}$ 
Could someone please explain what is happening? I strongly suspect it has to do with taking the transpose, but why then does it work for one and not the other? The problems are both "cooked" so I know there are definitely left inverses for both.
**** Correction / Solved ****
$X = \begin{bmatrix}5-r & -2-r & r\\ -2-t & 1-t & t\\\end{bmatrix}$

The matrix and Moore-Penrose pseudoinverse are
$$
 \mathbf{A} =
%
\left(
\begin{array}{cc}
 1 & 2 \\
 2 & 5 \\
 3 & 7 \\
\end{array}
\right), \qquad
%
 \mathbf{A}^{\dagger} =
\frac{1}{3}
\left(
\begin{array}{ccc}
 12 & -9 & 3 \\
 -5 & 4 & -1 \\
\end{array}
\right)
%
$$
The reduced row echelon form is 
$$
\mathbf{E}_{\mathbf{A}} =
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
 0 & 0 \\
\end{array}
\right)
$$
The matrix $\mathbf{A}$ has rank $\rho = 2$. Because the number of columns $n=2$ matches the rank, $\mathbf{A}$ has full column rank $n=\rho$. Therefore the nullspace $\mathcal{N}\left( \mathbf{A} \right)$ is trivial. Therefore the pseudoinverse is a left inverse:
$$
\mathbf{A}^{\dagger} = \mathbf{A}^{L}.
$$
That is,
$$
\begin{align}
 \mathbf{A}^{L} \mathbf{A} &= \mathbf{I}_{2} \\
%
\frac{1}{3}
\left(
\begin{array}{rrr}
 12 & -9 & 3 \\
 -5 & 4 & -1 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
 1 & 2 \\
 2 & 5 \\
 3 & 7 \\
\end{array}
\right)
%
&=
%
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)
%
\end{align}
$$
The matrix $\mathbf{A}$ has a row rank deficiency so the nullspace is nontrivial. 
$$
  \mathcal{N}\left( \mathbf{A}^{*} \right) = \text{span }
\left\{\,
\left(
\begin{array}{r}
 -1  \\
 -1 \\
  1
\end{array}
\right)
\, \right\}
$$
The matrix product
$$
\mathbf{A} \mathbf{A}^{\dagger} =
\left(
\begin{array}{rrr}
 2 & -1 & 1 \\
 -1 & 2 & 1 \\
 1 & 1 & 2 \\
\end{array}
\right)
$$
is not an identity matrix.
