Finding all left inverses of a matrix I have to find all left inverses of a matrix 
$$A = \begin{bmatrix}
 2&-1 \\ 
5 & 3\\ 
 -2& 1
\end{bmatrix}$$
I created a matrix to the left of $A$, 
$$\begin{bmatrix}
a &b  &c \\ 
d &e &f 
\end{bmatrix}  \begin{bmatrix}
 2&-1 \\ 
5 & 3\\ 
 -2& 1
\end{bmatrix} = \begin{bmatrix}
 1&0 \\ 
 0&1 
\end{bmatrix}$$
and I got the following system of equations: 
\begin{array} {lcl} 2a+5b-2c & = & 1 \\-a+3b+c & = & 0 \\
2d+5e-2f & = & 0 \\ -d+3e+f & = & 1 \end{array}
After this step, I am unsure how to continue or form those equations into a solvable matrix, and create a left inverse matrix from the answers of these equations. 
 A: Hint:
solve the first two equations using $c$ as a parameter and find 
$b=\frac1{11}$ , $a=c+\frac3{11}$
do the same in last two equation and you have the result.
A: Okay. I got the following matrix as my final answer as the general form of all left inverses of A:
\begin{bmatrix}
a & \frac1{11}  &a-\frac3{11} \\ 
d & \frac2{11} &d+\frac5{11}
\end{bmatrix}  
A: Alternative less advanced solution. Multiply it with its own transpose:
$A^TA$ is a $2\times2$ matrix:
$\left(
\begin{array}{cc}
 33 & 11 \\
 11 & 11 \\
\end{array}
\right)$
This has a trivial inverse by swapping the diagonals and finding the determinant:
$\frac{1}{22}\left(
\begin{array}{cc}
 1 & -1 \\
 -1 & 3 \\
\end{array}
\right)$
If you multiply this with the original matrix:
$(A^TA)^{-1}A^TA=I_2$
And note that this is your inverse:
$(A^TA)^{-1}A^T$
Which gives:
$\begin{pmatrix}\frac{3}{22}&\frac{1}{11}&-\frac{3}{22}\\ -\frac{5}{22}&\frac{2}{11}&\frac{5}{22}\end{pmatrix}$

Even less advanced, your system of linear equations work too. What I think you're not realising is, due to the very fact that there are less equations than variables, that there are an infinite number of inverses to matrices. Due to that fact you can take the liberty of just letting two variables be 0, and solving the rest. If you choose $c=f=0$, you can quite easily solve $b=\frac{1}{11}, a=\frac{3}{11}, e=\frac{2}{11}, d=-\frac{5}{11}$, which gives the solution:
$\begin{pmatrix}
\frac{3}{11}&\frac{1}{11}&0\\
-\frac{5}{11}&\frac{2}{11}&0
\end{pmatrix}$
Whilst this method is superior from a linear algebra perspective (because unlike the first solution above I can find multiple by choosing different values for a pair of unknowns), it's less practical if all you want is to find another inverse.

A third idea I have is to use block matrices. Generally speaking I can write your matrix as a joining of the vector $\mathbf{a}=\begin{pmatrix}-2&1\end{pmatrix}$ with the matrix $\begin{pmatrix}
 2&-1 \\ 
5 & 3
\end{pmatrix}$
Written as a block matrix:
\begin{bmatrix}
\mathbf{a}\\M_a
\end{bmatrix}
Writing the inverse, $B$, as a block matrix of a similar form (albeit with a
vetrical vector), we get:
$\begin{bmatrix}
\mathbf{b}&M_b
\end{bmatrix}
\begin{bmatrix}
\mathbf{a}\\M_a
\end{bmatrix}
=I_2$
By the property of block matrices, this gives us:
$I_2=\mathbb{a}\mathbb{b}+M_aM_b$
Hence for convenience, if $\mathbf{b}$ is a $0$ vector (the equivalent of setting $a=d=0$ in your linear equations), then all you need to do is invert $M_a$:
$M_b=\begin{pmatrix}
5&3\\
-2&1
\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{11}&-\frac{3}{11}\\ \frac{2}{11}&\frac{5}{11}\end{pmatrix}$
Hence your answer would be:
$B=\begin{pmatrix}0&\frac{1}{11}&-\frac{3}{11}\\ 0& \frac{2}{11}&\frac{5}{11}\end{pmatrix}$
For explanatory purposes I wrote everything out. In general, just invert the largest square you can fit (in any position) in the matrix you wish to invert, and just bung it in a $0$ matrix with reversed dimensions (in the same position along the height or width)
