![[sol-5.pdf#page=18]] --- $\textbf{Exercise 24.}$ Suppose $A$ and $B$ are square matrices of the same size and $AB = I$. Prove that $BA = I$. $\textbf{Solution 24.}$ Suppose $A$ and $B$ are $n$-by-$n$ matrices and $AB = I$. There exist $S, T \in \mathcal{L}(\mathbb{F}^n)$ such that $\begin{equation*} \mathcal{M}(S) = A \text{ and } \mathcal{M}(T) = B; \end{equation*}$ here we are using the standard basis of $\mathbb{F}^n$ (the existence of $S, T \in \mathcal{L}(\mathbb{F}^n)$ satisfying the equations above follows from 3.71). Because $AB = I$, we have $\mathcal{M}(S)\mathcal{M}(T) = I$, which implies that $\mathcal{M}(ST) = \mathcal{M}(I)$, which implies that $ST = I$, which implies that $TS = I$ (by 3.68). Thus $\begin{align*} BA &= \mathcal{M}(T)\mathcal{M}(S) \\ &= \mathcal{M}(TS) \\ &= \mathcal{M}(I) \\ &= I. \end{align*}$