# Section 5.8

The number $1$ is the multiplicative identity since

$$
1\cdot a=a\cdot1=a
$$

For matrices we have the same identity.

$I_{2}$ represents the $2\times2$ identity matrix and is defined
as
$$
I_{2}=\left[\begin{matrix}1 & 0\\
0 & 1
\end{matrix}\right]
$$

For example:

Let $A=\left[\begin{matrix}-5 & 2\\
3 & 4
\end{matrix}\right]$. Compute $A\cdot I_{2}$ and $I_{2}\cdot A$.

\begin{align*}
\left[\begin{matrix}-5 & 2\\
3 & 4
\end{matrix}\right]\cdot\left[\begin{matrix}1 & 0\\
0 & 1
\end{matrix}\right] & =\left[\begin{matrix}(-5)(1)+(2)(0) & (-5)(0)+(2)(1)\\
(3)(1)+(4)(0) & (3)(0)+(4)(1)
\end{matrix}\right]\\
 & =\left[\begin{matrix}-5 & 2\\
3 & 4
\end{matrix}\right]
\end{align*}

and

\begin{align*}
\left[\begin{matrix}1 & 0\\
0 & 1
\end{matrix}\right]\cdot\left[\begin{matrix}-5 & 2\\
3 & 4
\end{matrix}\right] & =\left[\begin{matrix}(1)(-5)+(0)(3) & (1)(2)+(0)(4)\\
(0)(-5)+(1)(3) & (0)(2)+(1)(4)
\end{matrix}\right]\\
 & =\left[\begin{matrix}-5 & 2\\
3 & 4
\end{matrix}\right]
\end{align*}

:::{prf:definition} $n\times n$ Identity Matrix
:label: identityMatrix

The $n\times n$ identity matrix is $I_{n}$ and is defined as

$$
I_{n}=\left[\begin{matrix}1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & a_{ij} & \vdots\\
0 & 0 & \cdots & 1
\end{matrix}\right].
$$

The element $a_{ij}=1$ when $i=j$ (the diagonal elements), and $a_{ij}=0$ otherwise.
:::

We say $\frac{1}{a}$ is the multiplicative inverse of $a$ provided
$a\ne0$. That is,

$$
\frac{1}{a}\cdot a=a\cdot\frac{1}{a}=1.
$$

:::{prf:definition} Multiplicative Inverse of Matrices
:label: multInverseMatrix

We say $A^{-1}$ is the multiplicative inverse matrix of matrix $A$
if and only if

$$
AA^{-1}=A^{-1}A=I_{n}
$$

:::

::::{prf:example} 
:label: findInverseExam1

Find $B^{-1}$ if 

$$
B=\left[\begin{matrix}-4 & 2 & 0\\
1 & -1 & 2\\
0 & 1 & 4
\end{matrix}\right]
$$

:::{dropdown} Solution:

First, we will construct the augmented matrix

$$
\left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\
1 & -1 & 2 & 0 & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]
$$

then perform reduce row echelon form on this new matrix

\begin{align*}
\left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\
1 & -1 & 2 & 0 & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right] & \to_{-\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
1 & -1 & 2 & 0 & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & -\frac{1}{2} & 2 & \frac{1}{4} & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{-2\times R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{(-1)\times R_{2}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -6 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 8 & \frac{1}{2} & 2 & 1
\end{matrix}\right]\\
 & \to_{\frac{1}{8}\times R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\\
 & \to_{4\times R_{3}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\\
 & \to_{\frac{1}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{4} & -\frac{1}{2} & \frac{1}{4}\\
0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]
\end{align*}

Therefore, $B^{-1}=\left[\begin{matrix}-\frac{3}{4} & -\frac{1}{2} & \frac{1}{4}\\
-\frac{1}{4} & -1 & \frac{1}{2}\\
\frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]$.

:::
::::

::::{prf:example}
:label: multInverseExam2

Find $A^{-1}$, if possible, given that 

$$
A=\left[\begin{matrix}4 & -2 & 5\\
0 & 1 & 0\\
-8 & 4 & -10
\end{matrix}\right].
$$

:::{dropdown} Solution:

First, we construct the augmented matrix and the perform reduced row echelon form

\begin{align*}
\left[\begin{matrix}4 & -2 & 5 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
-8 & 4 & -10 & 0 & 0 & 1
\end{matrix}\right] & \to_{\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & \frac{5}{4} & \frac{1}{4} & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
-8 & 4 & -10 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{8\times R_{1}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & \frac{5}{4} & \frac{1}{4} & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 2 & 0 & 1
\end{matrix}\right]
\end{align*}

Since $a_{31}=a_{32}=a_{33}=0$ we will not be able to reach the desired transformation. Therefore, there does not exists an $A^{-1}$.

:::
::::

:::{prf:theorem} Solving Matrix Equation
:label: inverseApplication

Suppose $A$ is an $n\times n$ matrix with inverse $A^{-1}$, $X$
is an $n\times1$ matrix of variables, and $B$ is an $n\times1$
matrix. The matrix equation

$$
AX=B
$$

has the solution

$$
X=A^{-1}B.
$$

:::

::::{prf:example}
:label: systemMatrixExam1

Solve the system

$$
\begin{cases}
2x-3y & =4\\
x+5y & =2
\end{cases}
$$

:::{dropdown} Solution:

First, we will write the system in matrix equation form

$$
\left[\begin{matrix}2 & -3\\
1 & 5
\end{matrix}\right]\cdot\left[\begin{matrix}x\\
y
\end{matrix}\right]=\left[\begin{matrix}4\\
2
\end{matrix}\right]
$$

where $A=\left[\begin{matrix}2 & -3\\
1 & 5 \end{matrix}\right]$, $X=\left[\begin{matrix}x\\ y \end{matrix}\right]$, and $B=\left[\begin{matrix}4\\ 2 \end{matrix}\right]$.

Next, we want to find $A^{-1}$

\begin{align*}
\left[\begin{matrix}2 & -3 & 1 & 0\\
1 & 5 & 0 & 1
\end{matrix}\right] & \to_{\frac{1}{2}\times R_{1}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\
1 & 5 & 0 & 1
\end{matrix}\right]\\
 & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\
0 & \frac{13}{2} & -\frac{1}{2} & 1
\end{matrix}\right]\\
 & \to_{\frac{2}{13}\times R_{2}}\left[\begin{matrix}1 & -\frac{3}{2} & \frac{1}{2} & 0\\
0 & 1 & -\frac{1}{13} & \frac{2}{13}
\end{matrix}\right]\\
 & \to_{\frac{3}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & \frac{5}{13} & \frac{3}{13}\\
0 & 1 & -\frac{1}{13} & \frac{2}{13}
\end{matrix}\right]\implies A^{-1}=\left[\begin{matrix}\frac{5}{13} & \frac{3}{13}\\
-\frac{1}{13} & \frac{2}{13}
\end{matrix}\right].
\end{align*}

Using the solution of the matrix equation theorem we have

\begin{align*}
X & =A^{-1}B\\
 & =\left[\begin{matrix}\frac{5}{13} & \frac{3}{13}\\
-\frac{1}{13} & \frac{2}{13}
\end{matrix}\right]\cdot\left[\begin{matrix}4\\
2
\end{matrix}\right]\\
 & =\left[\begin{matrix}\frac{5}{13}\cdot 4+\frac{3}{13}\cdot 2\\
-\frac{1}{13}\cdot 4+\frac{2}{13}\cdot 2
\end{matrix}\right]\\
 & =\left[\begin{matrix}2\\
0
\end{matrix}\right]=\left[\begin{matrix}x\\
y
\end{matrix}\right]=X
\end{align*}

which means the solution set of the system is $\{(2,0)\}$.

:::
::::

::::{prf:example}
:label: systemMatrixExam2

Solve the system

$$
\begin{cases}
-4x+2y & =12\\
x-y+2z & =7\\
y+4z & =20
\end{cases}
$$

:::{dropdown} Solution:

First, we write the system in matrix equation form

$$
\left[\begin{matrix}-4 & 2 & 0\\
1 & -1 & 2\\
0 & 1 & 4
\end{matrix}\right]\cdot\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]=\left[\begin{matrix}12\\
7\\
20
\end{matrix}\right]
$$

where $A=\left[\begin{matrix}-4 & 2 & 0\\ 1 & -1 & 2\\ 0 & 1 & 4 \end{matrix}\right]$, $X=\left[\begin{matrix}x\\ y\\ z \end{matrix}\right]$, and $B=\left[\begin{matrix}12\\ 7\\ 20 \end{matrix}\right]$.

Next, we will find $A^{-1}$

\begin{align*}
\left[\begin{matrix}-4 & 2 & 0 & 1 & 0 & 0\\
1 & -1 & 2 & 0 & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right] & \to_{-\frac{1}{4}\times R_{1}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
1 & -1 & 2 & 0 & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{(-1)\times R_{1}+R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & -\frac{1}{2} & 2 & \frac{1}{4} & 1 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{-2\times R_{2}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 1 & 4 & 0 & 0 & 1
\end{matrix}\right]\\
 & \to_{(-1)\times R_{2}+R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 8 & \frac{1}{2} & 2 & 1
\end{matrix}\right]\\
 & \to_{\frac{1}{8}\times R_{3}}\left[\begin{matrix}1 & -\frac{1}{2} & 0 & -\frac{1}{4} & 0 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\\
 & \to_{\frac{1}{2}\times R_{2}+R_{1}}\left[\begin{matrix}1 & 0 & -2 & -\frac{1}{2} & -1 & 0\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\\
 & \to_{2\times R_{3}+R_{1}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\
0 & 1 & -4 & -\frac{1}{2} & -2 & 0\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\\
 & \to_{4\times R_{3}+R_{2}}\left[\begin{matrix}1 & 0 & 0 & -\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\
0 & 1 & 0 & -\frac{1}{4} & -1 & \frac{1}{2}\\
0 & 0 & 1 & \frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\implies A^{-1}=\left[\begin{matrix}-\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\
-\frac{1}{4} & -1 & \frac{1}{2}\\
\frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]
\end{align*}

Using the solution of the matrix theorem we have

\begin{align*}
X & =A^{-1}B\\
 & =\left[\begin{matrix}-\frac{3}{8} & -\frac{1}{2} & \frac{1}{4}\\
-\frac{1}{4} & -1 & \frac{1}{2}\\
\frac{1}{16} & \frac{1}{4} & \frac{1}{8}
\end{matrix}\right]\cdot\left[\begin{matrix}12\\
7\\
20
\end{matrix}\right]\\
 & =\left[\begin{matrix}-\frac{3}{8}\cdot 12+(-\frac{1}{2})\cdot 7+\frac{1}{4}\cdot 20\\
-\frac{1}{4}\cdot 12+(-1)\cdot 7+\frac{1}{2}\cdot 20\\
\frac{1}{16}\cdot 12+\frac{1}{4}\cdot 7+\frac{1}{8}\cdot 20
\end{matrix}\right]\\
 & =\left[\begin{matrix}-3\\
0\\
5
\end{matrix}\right]=\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]=X
\end{align*}
Therefore, the solution to the system is $\{(-3,0,5)\}$.

:::
::::