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 a matrix \(B=\left[\begin{array}{rr} \bvec_1 \amp \bvec_2 \end{array}\right]\) such that

    \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.