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 \(A = B^{-1}\text{.}\)

  1. Given the fact that \(BA = I_n\text{,}\) explain why the matrix \(B\) must also be a square \(n\times n\) matrix.

  2. 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 pivots of \(B\text{?}\)

  3. Explain why the equation \(B\xvec = \zerovec\) has only the trivial solution.

  4. 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(AB-I) \amp {}={} 0\text{.} \\ \end{aligned} \end{equation*}

    Since the homogeneous equation \(B\xvec =\zerovec\) has only the trivial solution, explain why \(AB-I = 0\) and therefore, \(AB = I\text{.}\)