Left inverses for matrix Consider $A=\pmatrix{1&2\\1&-2\\0&1}$.
I'm trying to see how I can find 2 different left inverses for this $3\times2$ matrix. If it has a left inverse, then it is injective. 
Also, for the Right inverse, how come it doesn't have any? 
 A: You can find a left inverse of the form $\pmatrix{a&0&b\\c&0&d}$ simply by ignoring the middle row of $A$ and inverting the rest. Repeat that to find a left inverse of the form $\pmatrix{0&p&q\\0&r&s}$.
As for a right inverse, assume that $AB=I_{3\times 3}$. Then, in particular $AB\mathbf e_1=\mathbf e_1$, $AB\mathbf e_2=\mathbf e_2$, and $AB\mathbf e_3=\mathbf e_3$. But $\{B\mathbf e_1, B\mathbf e_2, B\mathbf e_3\}$ are three vectors in $\mathbb R^2$ and therefore they must be linearly dependent. Multiplying with $A$ cannot then stop them being linearly dependent. Therefore, by contradiction, $AB\ne I_{3\times 3}$.
A: You need to solve:
$$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
All matrices
$$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} $$
satisfying the above equation will be left inverses. You can use the same argument to check whether any right inverses exist. However, this will not be the case since your matrix is not quadratic (if both exist they agree and the matrix must be quadratic).
