# Inverse Matrix (2x2)

How to find and use the inverse matrix of a matrix (2x2): definition, 2 formulas, 3 examples, and their solutions.

## Definition

### Definition

The inverse matrix, A-1, is a matrix
that satisfies
AA-1 = A-1A = I.

I: Identity matrix

## Formula: A-1

### Formula

For a 2x2 square matrix A = [a b / c d],
A-1 = (1/[det (A)]) [d -b / -c a].

det (A): Determinant of A, ad - bc

Write 1/[det (A)].
Switch a and d.
And change the signs of b and c.

## Example 1

### Solution

First, find the determinant.

det (A) = 4⋅1 - 1⋅3

4⋅1 = 4
-1⋅3 = -3

4 - 3 = 1

det (A) = 1

If det (A) ≠ 0,
then the inverse matrix A-1 exists.

Then find A-1.

A-1
is equal to,
1 over, det (A), 1

times,
switch a and d, 1 and 4

and change the signs of b and c, -1 and -3.

So
A-1 = (1/1)[1 -1 / -3 4].

(1/1)[1 -1 / -3 4] = [1 -1 / -3 4]

So A-1 = [1 -1 / -3 4].

## Example 2

### Solution

First, find the determinant.

det (A) = 6⋅4 - 8⋅3

6⋅4 = 24
-8⋅3 = -24

24 - 24 = 0

det (A) = 0

If det (A) = 0,
then the inverse matrix A-1
does not exist.

So
A-1 does not exist

## Formula: AX = B

### Formula

AX = B is an example of a matrix equation.
The goal is to find the matrix X.

To solve AX = B,
change this to X = A-1B.
(= multiply A-1 to the front of both sides.)

## Example 3

### Solution

First, find the determinant of the matrix
in front of the X.

det (A) = 5⋅2 - 3⋅3

5⋅2 = 10
-3⋅3 = -9

10 - 9 = 1

det (A) = 1 ≠ 0
Then the inverse matrix A-1 = [5 3 / 3 2]-1 exists.

So X = [5 2 / 3 2]-1 [8 5 / 5 3].

Find [5 2 / 3 2]-1.

Write
1 over, det (A), 1

times,
switch a and d, 2 and 5

and change the signs of b and c, -3 and -3.

And write the back matrix [8 5 / 5 3].

So
[5 3 / 3 2]-1[8 5 / 5 3]
= (1/1)[2 -3 / -3 5][8 5 / 5 3].

(1/1)[2 -3 / -3 5] = [2 -3 / -3 5]

Solve [2 -3 / -3 5][8 5 / 5 3].

Multiply Matrices

Row 1, column 1:
2⋅8 + (-3)⋅5

Row 1, column 2:
2⋅5 + (-3)⋅3

Row 2, column 1:
-3⋅8 + 5⋅5

Row 2, column 2:
-3⋅5 + 5⋅3

So
[2 -3 / -3 5][8 5 / 5 3]
= [2⋅8 + (-3)⋅5 2⋅5 + (-3)⋅3 / -3⋅8 + 5⋅5 -3⋅5 + 5⋅3].

2⋅8 + (-3)⋅5
= 16 - 15

2⋅5 + (-3)⋅3
= 10 - 9

-3⋅8 + 5⋅5
= -24 + 25

-3⋅5 + 5⋅3
= -15 + 15

16 - 15 = 1

10 - 9 = 1

-24 + 25 = 1

-15 + 15 = 0

So X = [1 1 / 1 0].