Find all the right inverses of a matrix

Question

How do I find the right inverse of a non square matrix? The matrix i have is $$M = \begin{bmatrix} 1 & 1 & 0 \\ 2 & 3 & 1\\ \end{bmatrix}$$

Im really not sure how to even start this?

Answer 1

Right inverse means a matrix $A_{3 \times 2}$ such that $MA=I_{2 \times 2}$. So you are looking for a matrix $A=\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}$ such that $$MA = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 3 & 1\\ \end{pmatrix}\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$ This gives the following system: \begin{align*} x+y & = 1\\ 2x+3y+z & = 0\\ p+q & = 0\\ 2p+3q+r & = 1. \end{align*} Solving this gives $$A=\begin{pmatrix}3+z & r-1\\-2-z & 1-r\\z & r\end{pmatrix},$$ where $r,z \in \mathbb{R}$.

Answer 2

More generally, assume that $A\in M_{n,m}(\mathbb{C})$, where $n<m$, has full row rank $n$. Then the pseudo-inverse is $A^+=A^*(AA^*)^{-1}$ and is a right-inverse of $A$. Moreover, the general right-inverse of $A$ has the form $A^+ +(I_m-A^+A)U$ where $U\in M_{m,n}$ is an arbitrary matrix.

Here $A^+=1/3\begin{pmatrix}4&-1\\-1&1\\-5&2\end{pmatrix}$ and $(I_m-A^+A)U$ has the form $\begin{pmatrix}u&v\\-u&-v\\u&v\end{pmatrix}$.

The Anurag's result is recovered with $z=-5/3+u,r=2/3+v$.