Section 9.1 Invertible matrices
Definition 9.1.1.
An \(n\times n\) matrix \(A\) is called invertible if there is a matrix \(B\) such that \(AB = I_n\text{,}\) where \(I_n\) is the \(n\times n\) identity matrix. The matrix \(B\) is called the inverse of \(A\) and denoted \(A^{-1}\text{.}\)
Notice that we only define invertibility for matrices that have the same number of rows and columns in which case we say that the matrix is square.
Example 9.1.2.
Suppose that \(A\) is the matrix that rotates two-dimensional vectors counterclockwise by \(90^\circ\) and that \(B\) rotates vectors clockwise by \(90^\circ\text{.}\) We have
\begin{equation*}
A=\left[\begin{array}{rr}
0 \amp -1 \\
1 \amp 0 \\
\end{array}\right],~~~
B=\left[\begin{array}{rr}
0 \amp 1 \\
-1 \amp 0 \\
\end{array}\right].
\end{equation*}
We can check that
\begin{equation*}
AB = \begin{bmatrix}
0 \amp -1 \\
1 \amp 0 \\
\end{bmatrix}
\begin{bmatrix}
0 \amp 1 \\
-1 \amp 0 \\
\end{bmatrix}
= \begin{bmatrix}
1 \amp 0 \\
0 \amp 1 \\
\end{bmatrix}
= I
\end{equation*}
which shows that \(A\) is invertible and that \(A^{-1}=B\text{.}\)
Notice that if we multiply the matrices in the opposite order, we find that \(BA=I\text{,}\) which says that \(B\) is also invertible and that \(B^{-1} = A\text{.}\) In other words, \(A\) and \(B\) are inverses of each other.
Activity 9.1.1.
This activity demonstrates a procedure for finding the inverse of a matrix \(A\text{.}\)
Suppose that \(A = \begin{bmatrix}
3 \amp -2 \\
1 \amp -1 \\
\end{bmatrix}
\text{.}\) To find an inverse \(B\text{,}\) we write its columns as \(B = \begin{bmatrix}\bvec_1 \amp \bvec_2
\end{bmatrix}\) and require that
\begin{equation*}
\begin{aligned}
AB \amp = I \\
\begin{bmatrix}
A\bvec_1 \amp A\bvec_2
\end{bmatrix}
\amp =
\begin{bmatrix}
1 \amp 0 \\
0 \amp 1 \\
\end{bmatrix}.
\end{aligned}
\end{equation*}
In other words, we can find the columns of \(B\) by solving the equations
\begin{equation*}
A\bvec_1 = \twovec10,~~~
A\bvec_2 = \twovec01.
\end{equation*}
Solve these equations to find
\(\bvec_1\) and
\(\bvec_2\text{.}\) Then write the matrix
\(B\) and verify that
\(AB=I\text{.}\) This is enough for us to conclude that
\(B\) is the inverse of
\(A\text{.}\)
Find the product
\(BA\) and explain why we now know that
\(B\) is invertible and
\(B^{-1}=A\text{.}\)
What happens when you try to find the inverse of \(C = \begin{bmatrix}
-2 \amp 1 \\
4 \amp -2 \\
\end{bmatrix}\text{?}\)
In this activity, we have studied the matrices
\begin{equation*}
A = \begin{bmatrix}
3 \amp -2 \\
1 \amp -1 \\
\end{bmatrix},~~~
C = \begin{bmatrix}
-2 \amp 1 \\
4 \amp -2 \\
\end{bmatrix}.
\end{equation*}
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.
Solution.
Solving the two equations for \(\bvec_1\) and \(\bvec_2\) gives \(B = \begin{bmatrix}
1 \amp -2 \\
1 \amp -3 \\
\end{bmatrix}\text{.}\) We can verify that, as we expect, \(AB=I\text{.}\)
We find that \(BA=I\text{,}\) which is the condition that tells us that \(B\) is invertible.
Seeking the first column of \(C^{-1}\text{,}\) we see that the equation \(C\xvec = \twovec10\) is not consistent. This means that \(C\) is not invertible.
We see that
\begin{equation*}
A\sim \begin{bmatrix}
1 \amp 0 \\
0 \amp 1 \\
\end{bmatrix},~~~
C\sim
\begin{bmatrix}
1 \amp -\frac12 \\
0 \amp 0 \\
\end{bmatrix},
\end{equation*}
which shows that \(A\) is invertible and \(C\) is not.
