# Maths - Matrix algebra - Inverse 3x3

## Notation

On this page we number the matrix elements with two index numbers the first represents the column, the second the row. The index numbers start at 0 as follows:

 m00 m10 m20 m01 m11 m21 m02 m12 m22

## Description

The invese of a 3x3 matrix is:

[M]-1 = 1/det[M] *
 m11*m22 - m12*m21 m02*m21 - m01*m22 m01*m12 - m02*m11 m12*m20 - m10*m22 m00*m22 - m02*m20 m02*m10 - m00*m12 m10*m21 - m11*m20 m01*m20 - m00*m21 m00*m11 - m01*m10

where:

det = m00*m11*m22 + m01*m12*m20 + m02*m10*m21 - m00*m12*m21 - m01*m10*m22 - m02*m11*m20

## Speed Optimisation

The above program is valid for a general 3×3 matrix which will work in all circumstances but when the matrix is being used to represent a rotation (as described on this page) then the matrix carries a lot of redundant information. So if we want to speed up the code on this page then, for this case only, we can take advantage of this redundant information. To invert a pure rotation then we just take the transpose of the 3x3 part of the matrix.

## Calculator

The following calculator allows you to calculate the inverse for a 3x3 matrix. Enter the values into the matrix and then press "calc inverse " to display the result:

## Code

```   public void inverse() {
double det = m00*m11*m22 + m01*m12*m20 + m02*m10*m21 - m00*m12*m21 - m01*m10*m22 - m02*m11*m20;
m00 = (m11*m22 - m12*m21)/det;
m01 = (m02*m21 - m01*m22)/det;
m02 = (m01*m12 - m02*m11)/det;
m10 = (m12*m20 - m10*m22)/det;
m11 = (m00*m22 - m02*m20)/det;
m12 = (m02*m10 - m00*m12)/det;
m20 = (m10*m21 - m11*m20)/det;
m21 = (m01*m20 - m00*m21)/det;
m22 = (m00*m11 - m01*m10)/det;
}```