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? 

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, $AB\ne I_{3\times 3}$.
