##### Exercise8

In this exercise, you will construct the inverse of a matrix, a subject that we will investigate more fully in the next chapter. Suppose that $$A$$ is the $$2\times2$$ matrix:

\begin{equation*} A = \left[\begin{array}{rr} 3 \amp -2 \\ -2 \amp 1 \\ \end{array}\right] \text{.} \end{equation*}
1. Find the vectors $$\bvec_1$$ and $$\bvec_2$$ such that the matrix $$B=\left[\begin{array}{rr} \bvec_1 \amp \bvec_2 \end{array}\right]$$ satisfies

\begin{equation*} AB = I = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] \text{.} \end{equation*}

2. In general, it is not true that $$AB = BA\text{.}$$ Check that it is true, however, for the specific $$A$$ and $$B$$ that appear in this problem.

3. Suppose that $$\xvec = \twovec{x_1}{x_2}\text{.}$$ What do you find when you evaluate $$I\xvec\text{?}$$

4. Suppose that we want to solve the equation $$A\xvec = \bvec\text{.}$$ We know how to do this using Gaussian elimination; let's use our matrix $$B$$ to find a different way:

\begin{equation*} \begin{aligned} A\xvec \amp {}={} \bvec \\ B(A\xvec) \amp {}={} B\bvec \\ (BA)\xvec \amp {}={} B\bvec \\ I\xvec \amp {}={} B\bvec \\ \xvec \amp {}={} B\bvec \\ \end{aligned} \text{.} \end{equation*}

In other words, the solution to the equation $$A\xvec=\bvec$$ is $$\xvec = B\bvec\text{.}$$

Consider the equation $$A\xvec = \twovec{5}{-2}\text{.}$$ Find the solution in two different ways, first using Gaussian elimination and then as $$\xvec = B\bvec\text{,}$$ and verify that you have found the same result.

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