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.