# Inverse of matrices

A square matrix $\mathbf{A}$ of order $n$ is a regular (invertible) matrix if exists a matrix $\mathbf{B}$  such that

$$\mathbf{A}\mathbf{B} = \mathbf{B} \mathbf{A} = \mathbf{I},$$

where $\mathbf{I}$ is an identity matrix. If a matrix $\mathbf{A}$ is not regular, then is singular.

A matrix $\mathbf{B}$ is unique, what we can show from the definition above.

Therefore, for a matrix $\mathbf{B}$ we are introducing a special label: if a matrix $\mathbf{A}$ has the inverse, that we will denote as $\mathbf{A^{-1}}$. Now we have, by definition:

$$\mathbf{A} \mathbf{A^{-1}} =\mathbf{A^{-1}} \mathbf{A} = \mathbf{I}.$$

An identity matrix is the inverse of itself, that is, $\mathbf{I} \cdot \mathbf{I} = \mathbf {I}$ and zero matrix does not have an inverse matrix.

For ever two invertible matrices $\mathbf{A}$ and $\mathbf{B}$ of order $n$ and the identity matrix $\mathbf{I}$ of the same order is valid:

1.) $(\mathbf {A^{-1}})^{-1} = \mathbf{A}$,

2.) $(\mathbf{A} \cdot \mathbf{B})^{-1} = \mathbf{B^{-1}} \cdot \mathbf{A^{-1}}$,

3.) $\mathbf{I^{-1}} = \mathbf{I}$.

Our mission is to explore how to determine the inverse of matrices and which matrices even have the inverse matrix.

Example 1. Find the inverse matrix of the following matrix, if exists: $$\mathbf{A}=\left [\begin{array} {cc} 3 & -1 \\ 1& 0 \\ \end{array} \right].$$

Solution.

We need to find a matrix $\mathbf{B} \in \mathbb{R} ^{2}$ such that $\mathbf{A} \mathbf{B}= \mathbf{I}$. Let

$\mathbf{B}=\left [\begin{array} {cc} a & b \\ c & d \\ \end{array} \right]$. Therefore, we need to find numbers $a, b, c$ and $d$ in such that $$\left [\begin{array} {cc} 3 & -1 \\ 1& 0 \\ \end{array} \right] \cdot \left [\begin{array} {cc} a & b \\ c & d \\ \end{array} \right] = \left [\begin{array} {cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right]$$

is valid.

Now we have

$$\left [\begin{array} {cc} 3a-c & 3b-d \\ a & b \\ \end{array} \right] = \left [\begin{array} {cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right].$$

It follows

$$3a-c=1,$$

$$3 b – d = 0,$$

$$a=0,$$

$$b=1.$$

Finally, $a=0$, $b=1$, $c= -1$, $d=3$, that is

$$\mathbf{A}^{-1} = \mathbf{B} = \left [\begin{array} {cc} a & b \\ c & d \\ \end{array} \right] = \left [\begin{array} {cc} 0 & 1 \\ -1 & 3 \\ \end{array} \right] .$$

We would obtain the same result in which we observed the condition $\mathbf{B} \mathbf{A} = \mathbf{I}$.

The formula for the inverse matrix of order $2$

For every matrix $\mathbf{A} \in \mathbb{R}^{2}$

$$\mathbf{A}^{-1} = \frac{1}{ad – bc} \left[ \begin {array} {cc} d & -b \\ -c& a \\ \end{array} \right]$$

is valid.

If $ad-bc=0$, then matrix do not have an inverse matrix.

An inverse matrix is a neutral element for multiplication of matrices.