In this activity, we will begin with the matrix

\begin{equation*} A = \left[\begin{array}{rr} 1 \amp 2 \\ 1 \amp 3 \\ \end{array}\right] \end{equation*}

and construct its inverse \(A^{-1}\text{.}\) For the time being, let's denote the inverse by \(B\) so that \(B=A^{-1}\text{.}\)

  1. We know that \(AB = I\text{.}\) If we write \(B = \left[\begin{array}{rr}\bvec_1\amp \bvec_2\end{array}\right]\text{,}\) then we have

    \begin{equation*} AB = \left[\begin{array}{rr} A\bvec_1 \amp A\bvec_2 \end{array}\right] = \left[\begin{array}{rr} \evec_1 \amp \evec_2 \end{array}\right] = I \text{.} \end{equation*}

    This means that we need to solve the equations

    \begin{equation*} \begin{aligned} A\bvec_1 \amp {}={} \evec_1 \\ A\bvec_2 \amp {}={} \evec_2 \\ \end{aligned} \text{.} \end{equation*}

    Using the Sage cell below, solve these equations for the columns of \(B\text{.}\)

  2. What is the matrix \(B\text{?}\) Check that \(AB = I\) and \(BA = I\text{.}\)

  3. To find the columns of \(B\text{,}\) we solved two equations, \(A\bvec_1=\evec_1\) and \(A\bvec_2=\evec_2\text{.}\) We can do by this by augmenting \(A\) two separate times, forming matrices

    \begin{equation*} \begin{aligned} \left[\begin{array}{r|r} A \amp \evec_1 \end{array}\right] \amp \\ \left[\begin{array}{r|r} A \amp \evec_2 \end{array}\right] \amp \\ \end{aligned} \end{equation*}

    and finding their reduced row echelon forms.

    Rather than solving these two equations separately, we can solve them together by forming the augmented matrix \(\left[\begin{array}{r|rr} A \amp \evec_1 \amp \evec_2 \end{array}\right]\) and finding the row reduced echelon form. In other words, we augment \(A\) by the matrix \(I\) to form \(\left[\begin{array}{r|r} A \amp I \end{array} \right] \text{.}\)

    Form this augmented matrix and find its reduced row echelon form to find \(A^{-1}\text{.}\)

    Assuming \(A\) is invertible, we have shown that

    \begin{equation*} \left[\begin{array}{r|r} A \amp I \end{array}\right] \sim \left[\begin{array}{r|r} I \amp A^{-1} \end{array}\right] \text{.} \end{equation*}
  4. If you have defined a matrix \(A\) in Sage, you can find it's inverse as A.inverse(). Use Sage to find the inverse of the matrix

    \begin{equation*} A = \left[\begin{array}{rrr} 1 \amp -2 \amp -1 \\ -1 \amp 5 \amp 6 \\ 5 \amp -4 \amp 6 \\ \end{array}\right] \text{.} \end{equation*}

  5. What happens when we try to find the inverse of the matrix

    \begin{equation*} \left[\begin{array}{rr} -4 \amp 2 \\ -2 \amp 1 \\ \end{array}\right] \text{?} \end{equation*}
  6. Suppose that \(n\times n\) matrices \(C\) and \(D\) are both invertible. What do you find when you simplify the product \((D^{-1}C^{-1})(CD)\text{?}\) Explain why the product \(CD\) is invertible and \((CD)^{-1} = D^{-1}C^{-1}\text{.}\)