##### Activity3.1.3

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 could 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. But instead of solving these two equations separately, we could also 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 $$(CD)^{-1} = D^{-1}C^{-1}\text{.}$$

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