Exercise 8

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: $$A = \left[\begin{array}{rr} 3 & -2 \\ -2 & 1 \\ \end{array}\right]$$.

1. Find the vectors $\bvec_1$ and $\bvec_2$ such that the matrix $B=\left[\begin{array}{rr} \bvec_1 & \bvec_2 \end{array}\right]$ satisfies $$AB = I = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right]$$.

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

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

4. Suppose that we want to solve the equation $A\xvec = \bvec$. We know how to do this using Gaussian elimination; let's use our matrix $B$ to find a different way: $$\begin{aligned} A\xvec & {}={} \bvec \\ B(A\xvec) & {}={} B\bvec \\ (BA)\xvec & {}={} B\bvec \\ I\xvec & {}={} B\bvec \\ \xvec & {}={} B\bvec \\ \end{aligned}$$. In other words, the solution to the equation $A\xvec=\bvec$ is $\xvec = B\bvec$.

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