I saw this question somewhere and made me think do 3x4 matrices have an inverse, as I previously that that only square matrices have an inverse. If non-square matrices have an inverse, especially if 3x4 has one please let me know, the reason why.
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Only square matrices can have an inverse. To see why, let $A$ be a $3 \times 4$ matrix. An inverse of $A$, by definition, is a matrix $B$ which satisfies $AB=BA=I$. We have already run into trouble here. For $AB$ and $BA$ to both be defined, $B$ must be a $4 \times 3$ matrix. But then $AB$ is a $3 \times 3$ matrix and $BA$ is a $4 \times 4$ matrix, so we can't have $AB=BA$.
Now, one might ask "what if we ask that both $AB$ and $BA$ equal are an identity matrix, but not necessarily the same size identity?" E.g. what if we only ask that $A B=I_3$ and $BA=I_4$. While $AB=I_3$ is possible, $BA=I_4$ is not. $A$ is a matrix which maps $\mathbb{R}^4$ into $\mathbb{R}^3$, so the pigeon hole principal says that $A$ cannot be one to one. That is there is a nonzero vector $v \in \mathbb{R}^4$ so that $Av=0$. It follows that $BAv=0$. However if $BA=I_4$ then we would have to have $BAv=v$, which is a contradiction.
The answer is no. You can have an inverse on one side, but not on both. The main reason is rank (which is the dimension of the image). If $A$ and $B$ are two matrices that can be multiplied together, $\operatorname{rank}(AB)\leq \min(\operatorname{rank}(A),\operatorname{rank}(B))$. So if $A$ is an $n\times k$ matrix and $B$ is a $k\times n$ matrix, then $\operatorname{rank}(AB)$ and $\operatorname{rank}(BA)$ are at most $\min(n,k)$, so if $k<n$, then $AB$ cannot be equal to the $n\times n$ identity matrix, which has rank $n$.