Introduction
In the previous lesson, we introduced matrices. There is a class of matrices that have inverses, and such matrices are knowns as invertible. This means that they can be “undone” or reversed by multiplying them by another matrix. An invertible matrix is sometimes referred to as a nonsingular or non-degenerate matrix.
Objectives: At the end of this lesson, you should be able to:
Definition of invertible matrices
Such a matrix $B$, when it exists, is unique. Indeed, suppose that $B$ and $C$ satisfies $AB=BA=I_n$ and $AC=CA=I_n$. We want to show that $B=C$. Since $AB=I_n$, multiplying both sides on the left by $C$ yields $CAB=CI_n$. But $CAB=(CA)B=I_nB=B$ and $CI_=C$. It follows that $B=C$. This proves the uniqueness of the inverse.
In this case, we say that $B$ is the inverse of $A$ and is denoted by $A^{-1}$.
1&0\\
2&1
\end{bmatrix}$ and $B=\begin{bmatrix}
1&0\\
-2&1
\end{bmatrix}$. Verify that $B$ is the inverse of $A$.
We have
\begin{eqnarray*}
AB&=&\begin{bmatrix}
1&0\\
2&1
\end{bmatrix}\begin{bmatrix}
1&0\\
-2&1
\end{bmatrix}=\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
\end{eqnarray*}
and
\begin{eqnarray*}
BA&=&\begin{bmatrix}
1&0\\
-2&1
\end{bmatrix}\begin{bmatrix}
1&0\\
2&1
\end{bmatrix}=\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
\end{eqnarray*}
This proves that $B$ is the inverse of $A$. That is, $A^{-1}=B$.
-1 & 0 & 1 \\
0 & 1 & 0 \\
2 & 3 & -1
\end{array}\right] $ and $B\left[\begin{array}{ccc}
1 & -3 & 1\\
0 & 1 & 0\\
2 & -3 & 1
\end{array}\right]$. Verify that $B$ is the inverse of $A$.
It suffices to verify that $AB=BA=I$. We have:
\begin{eqnarray*}
AB&=&\begin{bmatrix}
-1 & 0 & 1 \\
0 & 1 & 0 \\
2 & 3 & -1
\end{bmatrix}\begin{bmatrix}
1 & -3 & 1\\
0 & 1 & 0\\
2 & -3 & 1
\end{bmatrix}=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\end{eqnarray*}
and
\begin{eqnarray*}
BA&=&\begin{bmatrix}
1 & -3 & 1\\
0 & 1 & 0\\
2 & -3 & 1
\end{bmatrix}\begin{bmatrix}
-1 & 0 & 1 \\
0 & 1 & 0 \\
2 & 3 & -1
\end{bmatrix}=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\end{eqnarray*}
Therefore, $A^{-1}= \left[\begin{array}{ccc}
1 & -3 & 1\\
0 & 1 & 0\\
2 & -3 & 1
\end{array}\right]$.
\begin{eqnarray*}
A^{-1}AX=A^{-1}\textbf{b} &\Longrightarrow &I_nX=A^{-1}\textbf{b}\qquad \mbox{since $A^{-1}A=I_n$}\\
&\Longrightarrow &X=A^{-1}\textbf{b}\qquad \mbox{since $I_X=X$}
\end{eqnarray*}
This proves that he matrix equation $A.X=\textbf{b}$ has a unique solution given by $A^{-1}\textbf{b}$
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}$ and let $B=\begin{bmatrix}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{bmatrix} $.
- Show that $A^{-1}=B$.
- Consider the linear system
\begin{eqnarray*}
\begin{array}{ccccccccc}
5x&- &y&+&3z&=&2\\
x& & &+&z&=&-1\\
& & y&+& z&=& 3
\end{array}
\end{eqnarray*}
Solve this linear system using the information from part (1)
- We have
\begin{eqnarray*}
AB&=&\begin{bmatrix}
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}\begin{bmatrix}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{bmatrix}=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\end{eqnarray*}
and\begin{eqnarray*}
BA&=&\begin{bmatrix}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{bmatrix}\begin{bmatrix}
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\end{eqnarray*}Therefore, $A^{-1}= B=\left[\begin{array}{ccc}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{array}\right]$. - The linear system
\begin{eqnarray*}
\begin{array}{ccccccccc}
5x&- &y&+&3z&=&2\\
x& & &+&z&=&-1\\
& & y&+& z&=& 3
\end{array}
\end{eqnarray*}
can be written in matrix form as\begin{eqnarray*}
\begin{bmatrix}
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
2\\
-1\\
3
\end{bmatrix}
\end{eqnarray*}Since the matrix $\begin{bmatrix}
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}$ is invertible with inverse $\left[\begin{array}{ccc}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{array}\right]$, it follows that
\begin{eqnarray*}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
5 & -1 & 3 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}^{-1}\begin{bmatrix}
2\\
-1\\
3
\end{bmatrix}=\begin{bmatrix}
1 & -4 & 1 \\
1 & -5 & 2 \\
-1 & 5 & -1
\end{bmatrix}\begin{bmatrix}
2\\
-1\\
3
\end{bmatrix}=\begin{bmatrix}
9\\
13\\
-10
\end{bmatrix}
\end{eqnarray*}
- The matrix $A^{-1}$ is invertible and its inverse is $A$. That is, $\left(A^{-1}\right)^{-1}=A$
- The matrix $AB$ is invertible and $(AB)^{-1}=B^{-1}A^{-1}$
- The matrix $A^T$ is invertible and $\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$
- If $c\neq 0$ then $cA$ is invertible and $(cA)^{-1}=\dfrac{1}{c}A^{-1}$
- The matrix $A^n$ and $\left(A^n\right)^{-1}=\left(A^{-1}\right)^n$
- The identity $I_n$ is invertible and $I_n^{-1}=I_n$.
- Since $A$ is invertible, then $A^{-1}$ exists and $AA^{-1}=A^{-1}A=I_n$. This can be rewritten as $A^{-1}A=AA^{-1}=I_n$. This proves that $A^{-1}$ is invertible and $\left(A^{-1}\right)^{-1}=A$.
- Since $A$ and $B$ are invertible, then $B^{-1}$ and $A^{-1}$ exist and $AA^{-1}=A^{-1}A=I_n$ and $BB^{-1}=B^{-1}B=I_n$. We have
\begin{eqnarray*}
(AB)\left(B^{-1}A^{-1}\right)&=&A\left(BB^{-1}\right) A^{-1}\\
&=&AI_nA^{-1}\qquad\qquad \,\mbox{since $BB^{-1}=I_n$}\\
&=&(AI_n)A^{-1}\\
&=&AA^{-1}\qquad\qquad \quad\,\mbox{since $AI_n=A$}\\
&=&I_n\qquad\qquad \qquad\quad\mbox{since $AA^{-1}=I_n$}
\end{eqnarray*}
and
\begin{eqnarray*}
\left(B^{-1}A^{-1}\right)(AB)&=&B^{-1}\left( A^{-1}A\right)B\\
&=&B^{-1}I_nB\qquad\qquad\, \mbox{since $AA^{-1}=I_n$}\\
&=&B^{-1}(I_nB)\\
&=&BB^{-1}\qquad\qquad\quad\, \mbox{since $I_nB=B$}\\
&=&I_n\qquad\qquad\qquad\quad \mbox{since $BB^{-1}=I_n$}.
\end{eqnarray*}
This proves that $(AB)^{-1}$ exists and $(AB)^{-1}=B^{-1}A^{-1}$. - We have
\begin{eqnarray*}
A^T\left(A^{-1}\right)^T&=&\left(AA^{-1}\right)^T=I_n^T=I_n
\end{eqnarray*}
and
\begin{eqnarray*}
\left(A^{-1}\right)^TA^T&=&\left(A^{-1}A.\right)^T=I_n^T=I_n.
\end{eqnarray*}
This proves that $A^T$ is invertible and $\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$. - We have
$$(cA)\left(\frac{1}{c}A^{-1}\right)=c\frac{1}{c}(AA^{-1})=I_n$$
and
$$\left(\frac{1}{c}A^{-1}\right)(cA)=\frac{c}{c}(A^{-1}A)=I_n.$$
Hence $cA$ is invertible and $(cA)^{-1}=\frac{1}{c}A^{-1}$. - We have
\begin{eqnarray*}
A^n\left(A^{-1}\right)^n&=&\left(\underbrace{AA\cdots A}_{\text{$n$ times}}\right)\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n$ times}}\right)\\
&=&\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)\left(AA^{-1}\right)\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)I_n\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{AA\cdots A}_{\text{$n-2$ times}}\right)\left(AA^{-1}\right)\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-2$ times}}\right)\\
&=&\left(\underbrace{AA\cdots A}_{\text{$n-2$ times}}\right)\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-2$ times}}\right)\\
&\vdots &\\
&=&AA^{-1}=I_n
\end{eqnarray*}
and\begin{eqnarray*}
\left(A^{-1}\right)^nA^n&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n$ times}}\right)\left(\underbrace{AA\cdots A}_{\text{$n$ times}}\right)\\
&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)\left(A^{-1}A\right)\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)I_n\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-1$ times}}\right)\left(\underbrace{AA\cdots A}_{\text{$n-1$ times}}\right)\\
&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-2$ times}}\right) \left(A^{-1}A\right) \left(\underbrace{AA\cdots A}_{\text{$n-2$ times}}\right)\\
&=&\left(\underbrace{A^{-1}A^{-1}\cdots A^{-1}}_{\text{$n-2$ times}}\right)\left(\underbrace{AA\cdots A}_{\text{$n-2$ times}}\right)\\
&\vdots &\\
&=&A^{-1}A=I_n
\end{eqnarray*}
This proves that $A^n$ is invertible and $\left(A^n\right)^{-1}=\left(A^{-1}\right)^n$. - Since $I_nI_n=I_n$ then $I_n$ is invertible and $I_n^{-1}=I_n$.
\begin{examp}
Let $A$, $B$, and $C$ be invertible matrices of the same size $n\times n$. Simplify the following expression:
$$A^{-1}\left(B^TA^{-1}\right)^{-1} B^TB+B^{-1}C^{-1}(CB-A)A^{-1}CB^2.$$
\end{examp}
\begin{answer}
\begin{eqnarray*}
&&A^{-1}\left(B^TA^{-1}\right)^{-1} B^TB+B^{-1}C^{-1}(CB-A)A^{-1}CB^2\\
&=&A^{-1}\left(\left(A^{-1}\right)^{-1}\left(B^T\right)^{-1}\right) B^TB+\left(B^{-1}C^{-1}CB-B^{-1}C^{-1}A\right) A^{-1}CB^2\\
&=&A^{-1}\left(A\left(B^T\right)^{-1}\right) B^TB+B^{-1}C^{-1}CBA^{-1}CB^2-B^{-1}C^{-1}AA^{-1}CB^2\\
&=&\left(A^{-1}A\right)\left(\left(B^T\right)^{-1} B^T\right)B+ B^{-1}\left( C^{-1}C\right)BA^{-1}CB^2-B^{-1}C^{-1}\left( AA^{-1}\right)CB^2\\
&=&I_nI_nB+ B^{-1}I_nBA^{-1}CB^2-B^{-1}C^{-1}I_nCB^2\\
&=&B+ B^{-1}BA^{-1}CB^2-B^{-1}C^{-1}CB^2\\
&=&B+ (B^{-1}B)A^{-1}CB^2-B^{-1}(C^{-1}C)B^2\\
&=&B+ I_nA^{-1}CB^2-B^{-1}I_nB^2\\
&=&B+A^{-1}CB^2-B^{-1}B^2\\
&=&B+A^{-1}CB^2-(B^{-1}B)B\\
&=&B+A^{-1}CB^2-I_nB\\
&=&B+A^{-1}CB^2-B\\
&=&A^{-1}CB^2\\
\end{eqnarray*}
\end{answer}
There are several methods for finding the inverse of a matrix, including:
- Gaussian elimination: Gaussian elimination is a systematic method of solving systems of linear equations and can be used to find the inverse of a matrix. By performing elementary row operations on the augmented matrix, we can transform the matrix into its row echelon form and then into its reduced row echelon form, which will yield the inverse.
- Adjoint method: The adjoint method involves finding the adjoint of a matrix and dividing it by the determinant of the matrix to obtain the inverse. The adjoint of a matrix is obtained by taking the transpose of the matrix of cofactors.
- Elementary matrices: Elementary matrices are matrices obtained by performing elementary row operations on the identity matrix. By performing the same elementary row operations on the original matrix, we can obtain an elementary matrix that is the inverse of the original matrix.
- Inverse formula: For a $2\times 2$ matrix, there is a formula that can be used to find the inverse directly.
Invertible $2\times 2$ matrices
\begin{thm}
Let $A=\begin{bmatrix}
a&b\\c&d
\end{bmatrix}$ be a $2\times 2$ matrix. Then $A$ is invertible if and only if $ad-bc\neq 0$. In this case
$$A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\
-c&a
\end{bmatrix}$$
\end{thm}
\begin{proof} Let $A=\begin{bmatrix}
a&b\\c&d
\end{bmatrix}$ be a $2\times 2$ matrix such that $ad-bc\neq 0$. We have:
\begin{eqnarray*}
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}\left(\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\
-c&a
\end{bmatrix}\right)&=&\dfrac{1}{ad-bc}\begin{bmatrix}
a&b\\c&d
\end{bmatrix}\begin{bmatrix}
d&-b\\
-c&a
\end{bmatrix}\\
&=&\dfrac{1}{ad-bc}\begin{bmatrix}
ad-bc&0\\0&ad-bc
\end{bmatrix}\\
\\
&=&\begin{bmatrix}
\frac{ad-bc}{ad-bc}&0\\0&\frac{ad-bc}{ad-bc}
\end{bmatrix}=\begin{bmatrix}
1&0\\0&1
\end{bmatrix}
\end{eqnarray*}
and
\begin{eqnarray*}
\left(\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\
-c&a
\end{bmatrix}\right)\begin{bmatrix}
a&b\\c&d
\end{bmatrix}&=&\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\
-c&a
\end{bmatrix}\begin{bmatrix}
a&b\\c&d
\end{bmatrix}\\
&=&\dfrac{1}{ad-bc}\begin{bmatrix}
ad-bc&0\\0&ad-bc
\end{bmatrix}\\
\\
&=&\begin{bmatrix}
\frac{ad-bc}{ad-bc}&0\\0&\frac{ad-bc}{ad-bc}
\end{bmatrix}=\begin{bmatrix}
1&0\\0&1
\end{bmatrix}.
\end{eqnarray*}
This proves that $A$ is invertible and $A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\
-c&a
\end{bmatrix}$.
1&-3\\3&0
\end{bmatrix}$, $B=\begin{bmatrix}
1&-3\\1&-3
\end{bmatrix}$, and $C=\begin{bmatrix}
4&1\\-6&t
\end{bmatrix}$.
- Show that $A$ is invertible and find its inverse.
- Show that $B$ is not invertible.
- Find the values of $t$ for which the matrix $C$ is invertible.
- Since $1(0)-3(-3)=9\neq 0$, then $A$ is invertible and $A^{-1}=\dfrac{1}{9}\begin{bmatrix}
0&3\\-3&1
\end{bmatrix}$ - Since $1(-3)-1(-3)=-3+3=0$, then $B$ is not invertible.
- $C$ is invertible when $4(t)-(-6)(1)=4t+6\neq 0$. That is, $t\neq -\dfrac{3}{2}$. Hence $C$ is invertible if and ony if $t\neq -\dfrac{3}{2}$.
