Activity 3.1.3
In this activity, we will begin with the matrix
and construct its inverse \(A^{1}\text{.}\) For the time being, let's denote the inverse by \(B\) so that \(B=A^{1}\text{.}\)

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

To find the columns of \(B\text{,}\) we solved two equations, \(A\bvec_1=\evec_1\) and \(A\bvec_2=\evec_2\text{.}\) We could do this by augmenting \(A\) two separate times, forming matrices
\begin{equation*} \begin{aligned} \left[\begin{array}{rr} A \amp \evec_1 \end{array}\right] \amp \\ \left[\begin{array}{rr} A \amp \evec_2 \end{array}\right] \amp \\ \end{aligned} \end{equation*}and finding their reduced row echelon forms. But instead of solving these two equations separately, we could also solve them together by forming the augmented matrix \(\left[\begin{array}{rrr} A \amp \evec_1 \amp \evec_2 \end{array}\right]\) and finding the row reduced echelon form. In other words, we augment \(A\) by the matrix \(I\) to form \(\left[\begin{array}{rr} A \amp I \end{array} \right] \text{.}\)
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}{rr} A \amp I \end{array}\right] \sim \left[\begin{array}{rr} I \amp A^{1} \end{array}\right]\text{.} \end{equation*} 
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*} 
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*} 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 \((CD)^{1} = D^{1}C^{1}\text{.}\)