Subsection Prepare
Consider the matrix
\begin{equation*}
A=\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\text{.}
\end{equation*}
\begin{align*}
A^{-1}\amp = \frac{1}{ad-bc}\left[\begin{array}{rr|rr} d\amp -b\\-c\amp a \end{array}\right]\\
\amp = \frac{1}{(1)(5)-(2)(3)}\left[\begin{array}{rr|rr} 5\amp -2\\-3\amp 1 \end{array}\right] \\
\amp = \left[\begin{array}{rr|rr} -5\amp 2\\3\amp -1 \end{array}\right] \text{.}
\end{align*}
Let’s now find \(A^{-1}\) in another way, using a method that will also work on larger matrices. Let
\begin{equation*}
A^{-1}=\begin{bmatrix} x_1 \amp y_1 \\ x_2 \amp y_2 \end{bmatrix}\text{.}
\end{equation*}
Calculating \(A^{-1}\) means we need to solve for the values of \(x_1, y_1, x_2, y_2\text{.}\) We know that \(A A^{-1}=I\text{,}\) which means that
\begin{equation*}
\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\begin{bmatrix} x_1 \amp y_1 \\ x_2 \amp y_2 \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\text{,}
\end{equation*}
and using matrix multiplication on the two matrices above means that
\begin{equation*}
\begin{bmatrix} 1x_1+2x_2 \amp 1y_1+2y_2 \\ 3x_1+5x_2 \amp 3y_1+5y_2 \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\text{.}
\end{equation*}
Two matrices are equal only when each of their corresponding entries is equal, so this means that
\begin{align*}
1x_1+2x_2 \amp =1 \amp \amp \amp 1y_1+2y_2 \amp =0 \\
3x_1+5x_2 \amp =0 \amp \amp \amp 3y_1+5y_2 \amp =1 \text{.}
\end{align*}
This looks like two linear systems of equations, one with variables \(x_1, x_2\) and another one with variables \(y_1, y_2\text{.}\) We know how to solve a system of linear equations: use an augmented matrix and row reduce. Let’s solve the first system first.
\begin{align*}
\amp \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 1 \\ 3 \amp 5 \amp 0 \end{array}\right] \\
-3R_1+R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 1 \\ 0 \amp -1 \amp -3 \end{array}\right] \\
-R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 1 \\ 0 \amp 1 \amp 3 \end{array}\right] \\
-2R_2+R_1\amp \rightarrow R_1 \amp \amp \left[\begin{array}{rr|r} 1 \amp 0 \amp -5 \\ 0 \amp 1 \amp 3 \end{array}\right]
\end{align*}
This means that \(x_1=-5\) and \(x_2=3\text{.}\)
Now let’s solve the second system.
\begin{align*}
\amp \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 0 \\ 3 \amp 5 \amp 1 \end{array}\right] \\
-3R_1+R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 0 \\ 0 \amp -1 \amp 1 \end{array}\right] \\
-R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|r} 1 \amp 2 \amp 0 \\ 0 \amp 1 \amp -1 \end{array}\right] \\
-2R_2+R_1\amp \rightarrow R_1 \amp \amp \left[\begin{array}{rr|r} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp -1 \end{array}\right] \text{.}
\end{align*}
This means that \(y_1=2\) and \(y_2=-1\text{.}\)
Putting those values we calculated together results in
\begin{equation*}
A^{-1}=\begin{bmatrix} x_1 \amp y_1 \\ x_2 \amp y_2 \end{bmatrix} = \left[\begin{array}{rr|rr} -5 \amp 2 \\ 3 \amp -1 \end{array}\right]\text{,}
\end{equation*}
and this is indeed the same matrix we had calculated using the formula for \(2\times 2\) inverses.
Activity 2.4.1. What do you notice?
We solved for the inverse of \(A=\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\) by setting up
\begin{equation*}
\begin{bmatrix} 1 \amp 2 \\ 3 \amp 5 \end{bmatrix}\begin{bmatrix} x_1 \amp y_1 \\ x_2 \amp y_2 \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\text{,}
\end{equation*}
and multiplying out and setting corresponding entries equal led to 2 systems of equations to solve:
\begin{align*}
1x_1+2x_2 \amp =1 \amp \amp \amp 1y_1+2y_2 \amp =0 \\
3x_1+5x_2 \amp =0 \amp \amp \amp 3y_1+5y_2 \amp =1 \text{.}
\end{align*}
What do you notice about these 2 systems and the process of solving them?
The two systems have the same coefficients.
-
The two systems have the different coefficients.
-
The two systems have the same constants.
-
The two systems have the different constants.
-
We used the same sequence of row operations on the augmented matrices to solve the system.
-
We used a different sequence of row operations on the augmented matrices to solve the system.
-
The two systems have the same solutions.
-
The two systems have different solutions.
-
The way it worked out in this example is not a coincidence, and the same things are true for larger matrices:
-
Calculating
\(AA^{-1}=I\text{,}\) where
\(A\) is a known matrix and
\(A^{-1}\) is a matrix of unknowns, leads to multiple linear systems that all have coefficient matrix
\(A\text{.}\)
-
In each system, the column matrix corresponding to the constants will be a column of the identity matrix.
-
Since the coefficient matrices are all the same, we will need the exact same row operations to solve each of the systems.
Since we did the exact same row operations and put the answers into the columns of \(A^{-1}\text{,}\) we could have been more efficient and solved both systems at the same time.
\begin{align*}
\amp \amp \amp \left[\begin{array}{rr|rr} 1 \amp 2 \amp 1 \amp 0 \\ 3 \amp 5 \amp 0 \amp 1 \end{array}\right] \\
-3R_1+R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|rr} 1 \amp 2 \amp 1 \amp 0 \\ 0 \amp -1 \amp -3 \amp 1\end{array}\right] \\
-R_2\amp \rightarrow R_2 \amp \amp \left[\begin{array}{rr|rr} 1 \amp 2 \amp 1 \amp 0 \\ 0 \amp 1 \amp 3 \amp -1\end{array}\right] \\
-2R_2+R_1\amp \rightarrow R_1 \amp \amp \left[\begin{array}{rr|rr} 1 \amp 0 \amp -5 \amp 2\\ 0 \amp 1 \amp 3 \amp -1\end{array}\right]
\end{align*}
Note that we started with \(A\) on the left and the identity matrix on the right. After row reducing, the left side turned into the identity matrix, and the right side turned into \(A^{-1}\text{.}\)
In math notation,
\begin{equation*}
\bigg[\begin{array}{c|c} A \amp I \end{array}\bigg] \rrefarrow \bigg[\begin{array}{c|c} I \amp A^{-1} \end{array}\bigg]\text{.}
\end{equation*}
Let’s see if that happens again with another example, using Sage to help with the calculations.
Example 2.4.2. Calculating the inverse of a larger matrix.
We calculate
\(A^{-1}\) for
\(A = \left[\begin{array}{rrr} 1\amp 1\amp -1\\1\amp -1\amp 1\\1\amp 2\amp 3 \end{array} \right]\text{.}\)
We’re trying to find an unknown matrix \(A^{-1}\) such that \(AA^{-1}=I\text{.}\) So we set up three augmented systems of equations to solve at the same time, resulting in
\begin{equation*}
\left[\begin{array}{rrr|rrr} 1\amp 1\amp -1\amp 1\amp 0\amp 0\\1\amp -1\amp 1\amp 0\amp 1\amp 0\\1\amp 2\amp 3\amp 0\amp 0\amp 1 \end{array} \right]\text{.}
\end{equation*}
Let’s let Sage row reduce this for us.
Then let’s check that the matrix in the last 3 columns really is the inverse of
\(A\text{,}\) by extracting the last three columns of the reduced echelon form, and then multiplying that matrix by
\(A\) to make sure we get the identity matrix.
Be sure you’ve already hit “Evaluate(Sage)” above.
What happens if we row reduce
\(\big[\begin{array}{c|c} A \amp I \end{array}\big]\) and don’t get the identity matrix appearing on the left?
Example 2.4.3. When an inverse doesn’t exist.
We attempt to calculate
\(A^{-1}\) for
\(A = \left[\begin{array}{ccc} 1\amp 1\amp -1\\1\amp -1\amp 1\\3\amp -1\amp 1 \end{array} \right]\text{.}\)
We’re trying to find an unknown matrix \(A^{-1}\) such that \(AA^{-1}=I\text{.}\) So we set up three augmented systems of equations to solve at the same time, resulting in
\begin{equation*}
\left[\begin{array}{ccc|ccc} 1\amp 1\amp -1\amp 1\amp 0\amp 0\\1\amp -1\amp 1\amp 0\amp 1\amp 0\\3\amp -1\amp 1\amp 0\amp 0\amp 1 \end{array} \right]\text{.}
\end{equation*}
Let’s let Sage row reduce this for us.
We don’t see
\(I\) in the first 3 columns. Let’s check that the matrix in the last 3 columns is
not the inverse of
\(A\text{,}\) by extracting the last three columns of the reduced echelon form, and then multiplying that matrix by
\(A\) to see if we get the identity matrix.
Be sure you’ve already hit “Evaluate(Sage)” above.
We did not get the identity matrix.
In fact, there is no possible inverse of \(A\text{.}\) We can see this is true by observing that
\begin{equation*}
\left[\begin{array}{rrr} 1\amp 1\amp -1\\1\amp -1\amp 1\\3\amp -1\amp 1 \end{array} \right]\left[\begin{array}{r} 0 \\-1 \\ 1 \end{array}\right]=\left[\begin{array}{r} 0 \\0 \\ 0 \end{array}\right]
\end{equation*}
If \(A^{-1}\) exists, then we could multiply both sides on the left by \(A^{-1}\) and get
\begin{equation*}
A^{-1}A\left[\begin{array}{r} 0 \\-1 \\ 1 \end{array}\right]=A^{-1}\left[\begin{array}{r} 0 \\0 \\ 0 \end{array}\right]
\end{equation*}
and since \(A^{-1}\mathbf{0}=\mathbf{0}\text{,}\) this would mean that
\begin{equation*}
\left[\begin{array}{r} 0 \\-1 \\ 1 \end{array}\right]=\left[\begin{array}{r} 0 \\0 \\ 0 \end{array}\right]\text{,}
\end{equation*}
which is not true. Thus, there is no possible matrix which is the inverse of \(A\text{.}\)
Let’s summarize what we’ve learned about calculating larger inverses.
Calculating inverses of any size.
Let \(A\) be an \(n \times n\) matrix. To calculate \(A^{-1}\text{,}\) augment the matrix \(A\) by the \(n\times n\) identity matrix \(I\)
\begin{equation*}
\big[\begin{array}{c|c} A \amp I \end{array} \big]
\end{equation*}
and put the resulting \(n\times 2n\) matrix into reduced row echelon form.
-
If the reduced row echelon form is of the form
\begin{equation*}
\big[\begin{array}{c|c} I \amp X \end{array} \big]\text{,}
\end{equation*}
then \(A^{-1} = X\text{.}\)
-
If the first
\(n\) columns of the reduced row echelon form are not
\(I\text{,}\) then
\(A\) is not invertible.
Activity 2.4.4. Interpreting the rref of \(\big[\begin{array}{c|c} A \amp I \end{array} \big]\).
Interpret the given information to answer the following questions about inverses.
(a)
Suppose that
\begin{equation*}
A=\left[\begin{array}{rrr} 1\amp 2\amp -1\\2\amp 0\amp -1\\-2\amp -3\amp 2 \end{array} \right]
\end{equation*}
and
\begin{equation*}
\left[\begin{array}{rrr|rrr} 1\amp 2\amp -1\amp 1\amp 0\amp 0\\2\amp 0\amp -1\amp 0\amp 1\amp 0\\-2\amp -3\amp 2\amp 0\amp 0\amp 1 \end{array} \right]
\rrefarrow \left[\begin{array}{rrr|rrr} 1\amp 0\amp 0\amp 3\amp 1\amp 2\\0\amp 1\amp 0\amp 2\amp 0\amp 1\\0\amp 0\amp 1\amp 6\amp 1\amp 4 \end{array} \right]\text{.}
\end{equation*}
Which of the following is true?
The matrix
\(A\) doesn’t have an inverse.
After row reducing \(\big[A | I \big]\) we see the identity matrix appear on the left.
The inverse of
\(A\) is
\(I\text{.}\)
-
The inverse of a matrix \(A\) is the matrix we multiply \(A\) by in order to get \(I\text{.}\)
But \(AI=IA=A\text{,}\) not \(I\text{.}\)
The inverse of
\(A\) is
\(\left[\begin{array}{rrr} 3\amp 1\amp 2\\2\amp 0\amp 1\\6\amp 1\amp 4 \end{array} \right]\)
Correct!
(b)
Suppose that
\begin{equation*}
A=\left[\begin{array}{rrrr} 1\amp 2\amp -1\amp 5\\2\amp 0\amp -1\amp 2\\-2\amp -3\amp 2 \amp 1 \\1\amp -1\amp 0 \amp 8\end{array} \right]
\end{equation*}
and
\begin{equation*}
\left[\begin{array}{rrrr|rrrr} 1\amp 2\amp -1\amp 5 \amp 1\amp 0\amp 0 \amp 0\\2\amp 0\amp -1 \amp 2 \amp 0\amp 1\amp 0 \amp 0\\-2\amp -3\amp 2 \amp 1 \amp 0\amp 0\amp 1 \amp 0 \\ 1\amp -1\amp 0 \amp 8 \amp 0 \amp 0 \amp 0 \amp 1\end{array} \right]
\rrefarrow \left[\begin{array}{rrrrrrrr} 1\amp 0\amp 0\amp 19\amp 0 \amp -2\amp -1\amp 3\\0\amp 1\amp 0 \amp 11 \amp 0\amp -2\amp -1 \amp 2\\0\amp 0\amp 1 \amp 36\amp 0 \amp -5\amp -2\amp 6 \\ 0\amp 0\amp 0 \amp 0\amp 1 \amp 1\amp 1\amp -1 \end{array} \right]\text{.}
\end{equation*}
Which of the following is true?
The matrix
\(A\) doesn’t have an inverse.
After row reducing \(\big[A | I \big]\) we don’t see the \(4 \times 4\) identity matrix appear on the left.
The inverse of
\(A\) is
\(\left[\begin{array}{rrrr} 0\amp -2\amp -1\amp 3\\0\amp -2\amp -1 \amp 2 \\ 0\amp -5\amp -2\amp 6\\1\amp 1\amp 1 \amp -1\end{array} \right]\text{.}\)
We don’t see the \(4 \times 4\) identity matrix appear on the left.
The inverse of
\(A\) is
\(\left[\begin{array}{rrr} 19\amp 0\amp -2 \\ 11\amp 0\amp -2\\36\amp 0\amp -5 \end{array} \right]\text{.}\)
We don’t see the \(4 \times 4\) identity matrix appear on the left.
(c)
Suppose that \(A\) is a matrix such that
\begin{equation*}
A \rrefarrow \left[\begin{array}{rrrr} 1\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 1 \amp 0 \\0\amp 0\amp 0 \amp 1\end{array} \right]\text{.}
\end{equation*}
Which of the following is true?
The matrix
\(A\) doesn’t have an inverse.
After row reducing \(\big[A | I \big]\) the identity would appear on the left, using the same row operations used to row reduce \(A\) by itself.
The matrix
\(A\) has an inverse.
After row reducing \(\big[A | I \big]\) the identity would appear on the left, using the same row operations used to row reduce \(A\) by itself.
Without knowing what
\(A\) is, we can’t tell whether or not
\(A\) has an inverse.
We don’t know what the inverse is without knowing what \(A\) is, but the same sequence of row operations that turns \(A\) into \(I\) would turn \(\big[A | I \big]\) into \(\big[I | X \big]\text{,}\) and \(X\) would be the inverse of \(A\text{.}\)
Reading Questions Reading Questions
1. Parsons Problem, Calculate inverse.
Put the correct steps in the correct order to calculate the inverse of a square matrix
\(A\) of any size.
Make sure \(ad-bc \neq 0\text{.}\)
#distractor
---
Augment the matrix \(A\) by the same size identity matrix \(I\) on right.
#tag:augment; depends:;
---
Perform Gauss-Jordan elimination on \(\big[\begin{array}{c|c} A \amp I \end{array} \big]\)
#tag:GJ; depends:augment;
---
Row reduce \(A\text{.}\)
#distractor
---
The inverse of \(A\) will appear on the right.
#distractor
---
If the identity matrix appears on the left, the inverse of \(A\) is on the right.
#tag:; depends:augment,GJ;
---
If the identity matrix does not appear on the left, the inverse of \(A\) does not exist.
#tag:; depends:augment,GJ;
2. How many inverses?
For a square matrix
\(A\text{,}\) we know it’s possible that
\(A\) doesn’t have an inverse, or that we can calculate
\(A^{-1}\) using the method in this section. Is it possible that
\(A\) could have
more than one inverse? That is, that there could be two different matrices
\(X\) and
\(Y\) such that
\(AX=XA=I\) and
\(AY=YA=I\text{?}\) Explain why or why not.
3. Reflection.
(a)
Reflect on your confidence level.
(b)
Ask a question about the material. What additional information do you think someone would need to become more confident in their understanding?
