Matrix Inverses are unique

Left Inverse = Right Inverse

Let $A$ be an $n \times n$ matrix and $AB = \mathbb{I}$, so that $B = A^{-1}$.
It follows that $BA = \mathbb{I}$...
$$\begin{aligned} AB & = \mathbb{I} \\ B \left( AB \right) & = B \mathbb{I} \\ \left( B A \right) B & = B \\ \therefore \left( B A \right) & = \mathbb{I} \end{aligned}$$

Inverses Are Unique

Suppose another inverse of $A$ existed, $C$, so that $AC = CA = \mathbb{I}$.
It follows that $B = C = A^{-1}$...
$$\begin{aligned} A B & = \mathbb{I} \\ C \left( A B \right) & = C \mathbb{I} \\ \left( C A \right) B & = C \\ \left( \mathbb{I} \right) B & = C \\ \therefore B & = C \end{aligned}$$