Exercise 12

We defined an $n\times n$ matrix to be invertible if there is a matrix $B$ such that $AB=I_n$. In this exercise, we will explain why it is also true that $BA = I$, which is the statement of this proposition. This means that, if $B=A^{-1}$, then $A = B^{-1}$.

1. Suppose that $\xvec$ is an $n$-dimensional vector. Since $AB=I$, explain why $AB\xvec = \xvec$ and use this to explain why the only vector for which $B\xvec = \zerovec$ is $\xvec = \zerovec$.

2. Explain why this implies that $B$ must be invertible. We will call the inverse $C$ so that $BC = I$.

3. Beginning with $AB = I$, explain why $B(AB)C = BIC$ and why this tells us that $BA = I$.