Section 18.1 Inverse Matrices
Definition 18.1.1. Inverse Matrix.
Given the square matrix \(A \text{,}\) if there exists a square matrix \(B \) such that
\begin{equation*}
AB=BA=I
\end{equation*}
then we call the matrix \(B \) the inverse of \(A \) and write \(B=A^{-1} \text{.}\)Note:
If matrix \(B \) is the inverse of matrix \(A \) then matrix \(A \) is the inverse of matrix \(B, \text{,}\) i.e.
\begin{equation*}
(A^{-1})^{-1}=A.
\end{equation*}
If matrix \(A \) has an inverse then we say that \(A \) is invertible or non-singular.
The inverse of a matrix (if it exists) is unique.
For matrix \(A \text{,}\) if there exists a matrix \(B \) such that \(AB=I \) then it follows that \(BA=I \) as well.
Example 18.1.2.
Let \(A=\begin{pmatrix} 1 \amp 1 \\ 1 \amp 2 \end{pmatrix} \quad \mbox{and} \quad B=\begin{pmatrix} 2 \amp -1 \\ -1 \amp 1 \end{pmatrix}.\) Calculate \(AB\; \text{and}\; BA. \)
Answer.
Solution.
\(\displaystyle AB=\begin{pmatrix} 1 \amp 1 \\ 1 \amp 2 \end{pmatrix} \begin{pmatrix} 2 \amp -1 \\ -1 \amp 1 \end{pmatrix} =\begin{pmatrix} 1 \amp 0 \\ 0 \amp 1 \end{pmatrix}=I \)
\(\displaystyle BA=\begin{pmatrix} 2 \amp -1 \\ -1 \amp 1 \end{pmatrix} \begin{pmatrix} 1 \amp 1 \\ 1 \amp 2 \end{pmatrix} = \begin{pmatrix} 1 \amp 0 \\ 0 \amp 1 \end{pmatrix}=I \)
Thus
\(A^{-1}=B\; \text{and}\; B^{-1}=A. \)
Example 18.1.3.
Show that \(A=\begin{pmatrix} 1 \amp -1 \\ 1 \amp -1 \end{pmatrix} \) is not invertible.
Solution.Assume that
\(A^{-1} \) exists. Then, since
\(A^{2}=0, \) we have that
\begin{equation*}
A=IA=(A^{-1}A)A=A^{-1}A^{2}=A^{-1}0=0,
\end{equation*}
which is a contradiction. Thus we conclude that
\(A^{-1} \) does not exist.
Given a square matrix \(A \) to find its inverse we need to find a matrix \(A^{-1} \) such that \(AA^{-1}=I \text{.}\) Let’s begin by considering the \(2\times 2 \) case. Let
\begin{equation*}
A=\begin{pmatrix} 1 \amp -1 \\ 1 \amp -1 \end{pmatrix},
\end{equation*}
where \(a\text{,}\) \(b\text{,}\) \(c\) and \(d\) are given. We want to find the entries in
\begin{equation*}
A^{-1}=\begin{pmatrix} x_{1} \amp y_{1} \\ x_{2} \amp y_{2} \end{pmatrix}.
\end{equation*}
Since \(AA^{-1}=I \) we have that
\begin{equation*}
\begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix} \begin{pmatrix} x_{1} \amp y_{1} \\ x_{2} \amp y_{2} \end{pmatrix}=\begin{pmatrix} 1 \amp 0 \\ 0 \amp 1 \end{pmatrix},
\end{equation*}
or equivalently,
\begin{equation*}
\begin{cases}
ax_{1} + bx_{2} = 1\\
cx_{1} + dx_{2} = 0
\end{cases}
\;\;\; \text{ and } \;\;\;
\begin{cases}
ay_{1} + by_{2} = 0\\
cy_{1} + dy_{2} = 1
\end{cases}
\end{equation*}
Both systems of equations have the same coefficient matrix, i.e.
\begin{equation*}
\begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix}.
\end{equation*}
The augmented matrices for these systems are
\begin{equation*}
\begin{pmatrix} a \amp b \amp 1 \\ c \amp d \amp 0 \end{pmatrix}
\;\;\; \text{ and } \;\;\;
\begin{pmatrix} a \amp b \amp 0 \\ c \amp d \amp 1 \end{pmatrix}
\end{equation*}
and since these have the same coefficient matrix we can combine the augmented matrices to get
\begin{equation*}
\left(\begin{array}{c c | c c} a \amp b \amp 1 \amp 0 \\c \amp d \amp 0 \amp 1 \end{array}\right)
\end{equation*}
By reducing this matrix to reduced row echelon form we can solve both sets of equations at the same time. If \(A \) has an inverse then the reduced row-echelon form will be
\begin{equation*}
\left(\begin{array}{c c | c c} 1 \amp 0 \amp \alpha \amp \beta \\ 0 \amp 1 \amp \chi \amp \delta \end{array}\right)
\end{equation*}
and hence \(x_{1}=\alpha,\; x_{2}=\chi,\; y_{2}=\beta \) and \(\; y_{2}=\delta \text{.}\) Thus, the augmented section of this matrix will contain \(A^{-1} \text{.}\)
Example 18.1.4.
Find the inverse of matrix \(A=\begin{pmatrix} 1 \amp 1 \\ 1 \amp 2 \end{pmatrix}. \)
Answer.\(A^{-1}=\begin{pmatrix} 2 \amp -1 \\ -1 \amp 1 \end{pmatrix}. \)
Begin by augmenting matrix
\(A \) with the identity matrix
\(I, \)
\begin{equation*}
\left(\begin{array}{c c | c c} 1 \amp 1 \amp 1 \amp 0 \\1 \amp 2 \amp 0 \amp 1 \end{array}\right)
\end{equation*}
Now use the elementary row operations to reduce this to reduced row echelon form
\begin{align*}
\left(\begin{array}{c c | c c} 1 \amp 1 \amp 1 \amp 0 \\1 \amp 2 \amp 0 \amp 1 \end{array}\right) \amp \sim\left(\begin{array}{c c | c c} 1 \amp 1 \amp 1 \amp 0 \\0 \amp 1 \amp -1 \amp 1 \end{array}\right) \hspace{8mm} R'_{2}= R_{2}-R_{1} \\
\amp \sim \left(\begin{array}{c c | c c} 1 \amp 0 \amp 2 \amp -1 \\0 \amp 1 \amp -1 \amp 1 \end{array}\right) \hspace{4mm} R'_{1}= R_{1}-R_{2}
\end{align*}
We can read off the inverse as
\begin{equation*}
A^{-1}=\begin{pmatrix} 2 \amp -1 \\ -1 \amp 1 \end{pmatrix}.
\end{equation*}
Example 18.1.5.
Find the inverse of matrix \(A=\begin{pmatrix} 1 \amp -1 \\ 1 \amp -1 \end{pmatrix}. \)
Answer.
Solution.Using the same procedure as in
Example 18.1.4 begin by augmenting matrix
\(A \) with the identity matrix
\(I \text{,}\)
\begin{equation*}
\left(\begin{array}{c c | c c} 1 \amp -1 \amp 1 \amp 0 \\1 \amp -1 \amp 0 \amp 1 \end{array}\right)
\end{equation*}
Now use the elementary row operations to reduce this to reduced row echelon form.
\begin{equation*}
\left(\begin{array}{c c | c c} 1 \amp -1 \amp 1 \amp 0 \\1 \amp -1 \amp 0 \amp 1 \end{array}\right)\sim \left(\begin{array}{c c | c c} 1 \amp -1 \amp 1 \amp 0 \\0 \amp 0 \amp -1 \amp 1 \end{array}\right) \hspace{5mm} R'_{2}= R_{2}-R_{1}
\end{equation*}
Since there is a row of
\(0 \)'s in the coefficient part of the reduced row echelon form while the remainder of the row is non-zero, we can see that there is no solution to the equations for finding the entries in the inverse matrix for
\(A \text{.}\) Thus, we conclude that matrix
\(A \) is not invertible, i.e. has no inverse.
The reasoning applied above to find a procedure for finding the inverse of a \(2\times 2 \) matrix applies equally well to any sized square matrix. Thus we have a general procedure for finding the inverse of a square matrix.
Theorem 18.1.6.
Given the square matrix \(A \text{,}\) to find its inverse \(A^{-1} \text{:}\)
Form the matrix \(\begin{pmatrix}
A \bigm | I \\
\end{pmatrix} \) by augmenting \(A \) with the identity matrix \(I, \)
Row reduce \(\begin{pmatrix}
A \bigm | I \\
\end{pmatrix} \) to reduced row echelon form,
If the reduced row echelon matrix is of the form \(\begin{pmatrix}
I \bigm | A^{-1} \\
\end{pmatrix} \) read off \(A^{-1} \text{.}\) Otherwise the inverse does not exist.
Example 18.1.7.
Find the inverse, if it exists, of \(A=\begin{pmatrix} 2 \amp 1 \amp 6 \\ -4 \amp 5 \amp -3 \\ 2 \amp -1 \amp 3 \end{pmatrix}. \)
Answer.\(A^{-1}=\begin{pmatrix} -2 \amp \frac{3}{2} \amp \frac{11}{2} \\ -1 \amp 1 \amp 3 \\ 1 \amp -\frac{2}{3} \amp -\frac{7}{3} \end{pmatrix}. \)
Form the augmented matrix and row reduce to reduced row echelon form:
\begin{align*}
\amp \left(\begin{array}{c c c | c c c} 2 \amp -1 \amp 6 \amp 1 \amp 0 \amp 0 \\-4 \amp 5 \amp -3 \amp 0 \amp 1 \amp 0 \\2 \amp -1 \amp 3 \amp 0 \amp 0 \amp 1 \end{array}\right)\\
\amp \sim \left(\begin{array}{c c c | c c c} 1 \amp \frac{1}{2} \amp 3 \amp \frac{1}{2} \amp 0 \amp 0 \\0 \amp 7 \amp 9 \amp 2 \amp 1 \amp 0 \\0 \amp -2 \amp -3 \amp -1 \amp 0 \amp 1 \end{array}\right)
\begin{matrix}
R'_{1} = \amp \frac{R_{1}}{2}\;\;\;\qquad\\
R'_{2} = \amp R_{2}+4R'_{1}\\
R'_{3} = \amp R_{3}-R_{1}
\end{matrix} \\
\amp \sim \left(\begin{array}{c c c | c c c} 1 \amp 0 \amp \frac{33}{14} \amp \frac{5}{14} \amp -\frac{1}{14} \amp 0 \\0 \amp 1 \amp \frac{9}{7} \amp \frac{2}{7} \amp \frac{1}{7} \amp 0 \\0 \amp 0 \amp -\frac{3}{7} \amp -\frac{3}{7} \amp 0 \amp 1 \end{array}\right)
\begin{matrix}
R'_{1} = \amp R_{1}-\frac{R'_{2}}{2} \\
R'_{2} = \amp \frac{R_{2}}{7}\;\;\;\qquad\\
R'_{3} = \amp R_{3}+2R'_{2}
\end{matrix} \\
\amp \sim \left(\begin{array}{c c c | c c c} 1 \amp 0 \amp 0 \amp -2 \amp \frac{3}{2} \amp \frac{11}{2} \\0 \amp 1 \amp 0 \amp -1 \amp 1 \amp 3 \\0 \amp 0 \amp 1 \amp 1 \amp -\frac{2}{3} \amp -\frac{7}{3} \end{array}\right)
\begin{matrix}
R'_{1} = \amp R_{1}-\frac{33R'_{3}}{14} \\
R'_{2} = \amp R_{2}-\frac{9R'_{3}}{7} \\
R'_{3} = \amp -\frac{7R'_{3}}{3} \;\;\; \qquad
\end{matrix}
\end{align*}
Thus
\begin{equation*}
A^{-1}=\begin{pmatrix} -2 \amp \frac{3}{2} \amp \frac{11}{2} \\ -1 \amp 1 \amp 3 \\ 1 \amp -\frac{2}{3} \amp -\frac{7}{3} \end{pmatrix}.
\end{equation*}
Of course we can always check our answer by confirming that
\(AA^{-1}=I \text{.}\)
Example 18.1.8.
Find the inverse, if it exists, of \(A=\begin{pmatrix} 2 \amp 1 \amp 6 \\ -4 \amp 5 \amp -3 \\ 2 \amp 8 \amp 15 \end{pmatrix}. \)
Answer.
Solution.Form the augmented matrix and row reduce to reduced row echelon form:
\begin{align*}
\amp \left(\begin{array}{c c c | c c c} 2 \amp 1 \amp 6 \amp 1 \amp 0 \amp 0 \\-4 \amp 5 \amp -3 \amp 0 \amp 1 \amp 0 \\2 \amp 8 \amp 15 \amp 0 \amp 0 \amp 1\end{array}\right)\\
\amp \sim\left(\begin{array}{c c c | c c c} 1 \amp \frac{1}{2} \amp 3 \amp \frac{1}{2} \amp 0 \amp 0 \\0 \amp 7 \amp 9 \amp 2 \amp 1 \amp 0 \\0 \amp 7 \amp 9 \amp -1 \amp 0 \amp 1\end{array}\right)
\begin{matrix}
R'_{1} = \amp \frac{R_{1}}{2}\;\;\;\qquad\\
R'_{2} = \amp R_{2}+4R'_{1}\\
R'_{3} = \amp R_{3}-R_{1}
\end{matrix} \\
\amp \sim\left(\begin{array}{c c c | c c c} 1 \amp 0 \amp \frac{33}{14} \amp \frac{5}{14} \amp -\frac{1}{14} \amp 0 \\0 \amp 1 \amp \frac{9}{7} \amp \frac{2}{7} \amp \frac{1}{7} \amp 0 \\0 \amp 0 \amp 0 \amp -3 \amp -1 \amp 1\end{array}\right)
\begin{matrix}
R'_{1} = \amp R_{1}-\frac{R'_{2}}{2} \\
R'_{2} = \amp \frac{R_{2}}{7}\;\;\;\qquad\\
R'_{3} = \amp R_{3}-R'_{2}
\end{matrix}
\end{align*}
Since the coefficient part of the reduced row echelon matrix is not the identity matrix
\(A \) does not have an inverse.
For later reference, some properties of the inverse of a matrix are listed below.
Theorem 18.1.9. Properties of the Matrix Inverse.
Let \(A \) and \(B \) be square invertible matrices of order \(n \) and let \(k \) be a real number. Then
\(\displaystyle (A^{-1})^{-1}=A \)
\(\displaystyle (A^{T})^{-1}=(A^{-1})^{T} \)
\(\displaystyle (kA)^{-1}=\frac{1}{k}A^{-1} \)
\(\displaystyle (AB)^{-1}=B^{-1}A^{-1} \)
\(\displaystyle (A^{r})^{-1}=(A^{-1})^{r}\;\;\;\; \text{where}\;\; r\in \mathbb{N} \)
Example 18.1.10.
Confirm that \((AB)^{-1}=B^{-1}A^{-1} \) holds for the matrices.
\begin{equation*}
A=\begin{pmatrix} 3 \amp 1 \\ -1 \amp 2 \end{pmatrix} \;\;\; \text{and}\;\;\; B=\begin{pmatrix} 1 \amp 5 \\ 0 \amp -2 \end{pmatrix}
\end{equation*}
Solution.Firstly,
\begin{equation*}
AB=\begin{pmatrix} 3 \amp 1 \\ -1 \amp 2 \end{pmatrix} \begin{pmatrix} 1 \amp 5 \\ 0 \amp -2 \end{pmatrix} = \begin{pmatrix} 3 \amp 13 \\ -1 \amp -9 \end{pmatrix},
\end{equation*}
and so
\begin{equation*}
(AB)^{-1}=\begin{pmatrix} 3 \amp 13 \\ -1 \amp 9 \end{pmatrix}^{-1} = \frac{1}{14}\begin{pmatrix} -9 \amp -13 \\ 1 \amp 3 \end{pmatrix}.
\end{equation*}
Next
\begin{equation*}
A^{-1}=\begin{pmatrix} 3 \amp 1 \\ -1 \amp 2 \end{pmatrix}^{-1} = \frac{1}{7}\begin{pmatrix} 2 \amp -1 \\ 1 \amp 3 \end{pmatrix},
\end{equation*}
\begin{equation*}
B^{-1}=\begin{pmatrix} 1 \amp 5 \\ 0 \amp -2 \end{pmatrix}^{-1} = -\frac{1}{2}\begin{pmatrix} -2 \amp -5 \\ 0 \amp 1 \end{pmatrix},
\end{equation*}
and so
\begin{equation*}
B^{-1}A^{-1}= -\frac{1}{2}\begin{pmatrix} -2 \amp -5 \\ 0 \amp 1 \end{pmatrix} \frac{1}{7}\begin{pmatrix} 2 \amp -1 \\ 1 \amp 3 \end{pmatrix} = - \frac{1}{14}\begin{pmatrix} -9 \amp -13 \\ 1 \amp 3 \end{pmatrix}.
\end{equation*}
The idea of a matrix inverse can be related to the problem of solving systems of linear equations in the case where the number of equations in the system is the same as the number of variables. As we have seen previously, we can write the system of \(n \) linear equations in \(n \) unknowns
\begin{align*}
a_{11} x_{1}+a_{12}x_{2} + \dots +a_{1n} x_{n}= \amp b_{1}\\
a_{21} x_{1}+a_{22}x_{2} + \dots +a_{2n} x_{n}= \amp b_{2}\\
\vdots \amp\\
a_{n1} x_{1}+a_{n2}x_{2} + \dots +a_{nn} x_{n}= \amp b_{n}
\end{align*}
using matrix notation as
\begin{equation}
A \bm{x} = \bm{b}\label{Ax_equal_b}\tag{18.1.1}
\end{equation}
where \(A \) is the \(n\times n \) coefficient matrix, \(\bm{x} \) is the \(n\times 1 \) column vector of variables and \(\bm{b} \) is the column vector of the constants. If the coefficient matrix \(A \) has an inverse, then from (18.1.1)
\begin{align*}
A^{-1}(A \bm{x})= \amp A^{-1} \bm{b},\\
(A^{-1}A) \bm{x}= \amp A^{-1} \bm{b},\\
\bm{x}= \amp A^{-1}\bm{b}.
\end{align*}
Example 18.1.11.
Solve the system of equations
\begin{align*}
2x + y + 6z = \amp 9,\\
-4x + 5y - 3z = \amp -7,\\
2x - y + 3z = \amp 5.
\end{align*}
Answer.\(x=-1, \; y=-1,\; \text{and} \; z=2. \)
In matrix notation this system can be written as
\begin{equation*}
\begin{pmatrix}
2 \amp 1 \amp 6 \\
-4 \amp 5 \amp -3 \\
2 \amp -1 \amp 3
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=
\begin{pmatrix}
9 \\
-7 \\
5
\end{pmatrix}
\end{equation*}
We found the inverse of the coefficient matrix in
Example 18.1.7 and, using that result, we have
\begin{equation*}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=
\begin{pmatrix}
-2 \amp \frac{3}{2} \amp \frac{11}{2} \\
-1 \amp 1 \amp 3 \\
1 \amp -\frac{2}{3} \amp -\frac{7}{3}
\end{pmatrix}
\begin{pmatrix}
9 \\
-7 \\
5
\end{pmatrix}=
\begin{pmatrix}
-1 \\
-1 \\
2
\end{pmatrix}
\end{equation*}
Thus the solution is
\(x=-1, \; y=-1,\; \text{and} \; z=2. \)
Theorem 18.1.12.
Consider the non-homogenous system of \(n \) linear equations in \(n \) variables
\begin{equation*}
A \bm{x} = \bm{b} ,\; \bm{b} \neq 0.
\end{equation*}
The following statements are equivalent:
The system has a unique solution,
\(A \bm{x} =0 \) has only the trivial solution \(\bm{x} =0, \)
The columns of \(A \) are linearly independent,
\(A \) is invertible.
Exercises Example Tasks
1.
Find the inverse, if it exists, of
\begin{align*}
A = \amp \begin{pmatrix}
2 \amp 1 \amp 3 \\
-1 \amp 2 \amp 4 \\
8 \amp -1 \amp 1
\end{pmatrix}\\
B = \amp \begin{pmatrix}
1 \amp 1 \amp 2 \\
-1 \amp 2 \amp -1 \\
1 \amp -1 \amp 1
\end{pmatrix}
\end{align*}
2.
Find the matrix for a rotation in the plane about the origin through \(\frac{\pi}{4}^{c} \) . Find the inverse of this matrix and interpret it geometrically.