Exercise 9
We defined an \(n\times n\) matrix to be invertible if there is a matrix \(B\) such that \(BA=I_n\text{.}\) In this exercise, we will explain why \(B\) is also invertible and that \(AB = I\text{.}\) This means that, if \(B=A^{1}\text{,}\) then \(A = B^{1}\text{.}\)
Given the fact that \(BA = I_n\text{,}\) explain why the matrix \(B\) must also be a square \(n\times n\) matrix.
Suppose that \(\bvec\) is a vector in \(\real^n\text{.}\) Since we have \(BA = I\text{,}\) it follows that \(B(A\bvec) = \bvec\text{.}\) Use this to explain why the columns of \(B\) span \(\real^n\text{.}\) What does this say about the pivot positions of \(B\text{?}\)
Explain why the equation \(B\xvec = \zerovec\) has only the trivial solution.

Beginning with the equation, \(BA=I\text{,}\) multiply both sides by \(B\) to obtain \(BAB = B\text{.}\) We will rearrange this equation:
\begin{equation*} \begin{aligned} BAB \amp {}={} B \\ BAB  B\amp {}={} 0 \\ B(ABI) \amp {}={} 0\text{.} \\ \end{aligned} \end{equation*}Since the homogeneous equation \(B\xvec =\zerovec\) has only the trivial solution, explain why \(ABI = 0\) and therefore, \(AB = I\text{.}\)