Our definition of an invertible matrix requires that \(A\) be a square \(n\times n\) matrix. Let's examine what happens when \(A\) is not square. For instance, suppose that

\begin{equation*} A = \left[\begin{array}{rr} -1 \amp -1 \\ -2 \amp -1 \\ 3 \amp 0 \\ \end{array}\right], \hspace{24pt} B = \left[\begin{array}{rrr} -2 \amp 2 \amp 1 \\ 1 \amp -2 \amp -1 \\ \end{array}\right] \text{.} \end{equation*}
  1. Verify that \(BA = I_2\text{.}\) In this case, we say that \(B\) is a left inverse of \(A\text{.}\)

  2. If \(A\) has a left inverse \(B\text{,}\) we can still use it to find solutions to linear equations. If we know there is a solution to the equation \(A\xvec = \bvec\text{,}\) we can multiply both sides of the equation by \(B\) to find \(\xvec = B\bvec\text{.}\)

    Suppose you know there is a solution to the equation \(A\xvec = \threevec{-1}{-3}{6}\text{.}\) Use the left inverse \(B\) to find \(\xvec\) and verify that it is a solution.

  3. Now consider the matrix

    \begin{equation*} C = \left[\begin{array}{rrr} 1 \amp -1 \amp 0 \\ -2 \amp 1 \amp 0 \\ \end{array}\right] \end{equation*}

    and verify that \(C\) is also a left inverse of \(A\text{.}\) This shows that the matrix \(A\) may have more than one left inverse.

  4. When \(A\) is a square matrix, we said that \(BA=I\) implies that \(AB=I\text{.}\) In this problem, we have a non-square matrix \(A\) with \(BA = I\text{.}\) What happens when we try to compute \(AB\text{?}\)