Activity 3.1.2

This activity demonstrates a procedure for finding the inverse of a matrix $A$.

1. Suppose that $A = \begin{bmatrix} 3 & -2 \\ 1 & -1 \\ \end{bmatrix}$. To find an inverse $B$, we write its columns as $B = \begin{bmatrix}\bvec_1 & \bvec_2 \end{bmatrix}$ and require that $$\begin{aligned} AB & = I \\ \begin{bmatrix} A\bvec_1 & A\bvec_2 \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}. \end{aligned}$$ In other words, we can find the columns of $B$ by solving the equations $$A\bvec_1 = \twovec10,~~~ A\bvec_2 = \twovec01.$$ Solve these equations to find $\bvec_1$ and $\bvec_2$. Then write the matrix $B$ and verify that $AB=I$. This is enough for us to conclude that $B$ is the inverse of $A$.

2. Find the product $BA$ and explain why we now know that $B$ is invertible and $B^{-1}=A$.

3. What happens when you try to find the inverse of $C = \begin{bmatrix} -2 & 1 \\ 4 & -2 \\ \end{bmatrix}$?

4. We now develop a condition that must be satisfied by an invertible matrix. Suppose that $A$ is an invertible $n\times n$ matrix with inverse $B$ and suppose that $\bvec$ is any $n$-dimensional vector. Since $AB=I$, we have $$A(B\bvec) = (AB)\bvec = I\bvec = \bvec.$$ This says that the equation $A\xvec = \bvec$ is consistent and that $\xvec=B\bvec$ is a solution.

   Since we know that $A\xvec = \bvec$ is consistent for any vector $\bvec$, what does this say about the span of the columns of $A$?

5. Since $A$ is a square matrix, what does this say about the pivot positions of $A$? What is the reduced row echelon form of $A$?

6. In this activity, we have studied the matrices $$A = \begin{bmatrix} 3 & -2 \\ 1 & -1 \\ \end{bmatrix},~~~ C = \begin{bmatrix} -2 & 1 \\ 4 & -2 \\ \end{bmatrix}.$$ Find the reduced row echelon form of each and explain how those forms enable us to conclude that one matrix is invertible and the other is not.