# Find 3×3 Matrix Inverse,Matrix Elementary Row Operations

On the matrix inverse introduction page, it is shown how to find the inverse of a
2 × 2  size matrix.

While that case doesn’t take too much effort, finding the inverse of a  3 × 3  is something that requires a bit more work.

We can find the inverse of a  3 × 3  matrix by doing some matrix elementary row operations.

There are three types of matrix elementary row operations that can be performed.

### Matrix Elementary Row Operations:

Swap/interchange matrix rows with each other.
Multiply a row by a number, including fractions.
Multiply a row by a number, including fractions, then add it to or subtract it from another row.

So how do we use these matrix elementary row operations to find the inverse of a  3 × 3  matrix exactly?

Well, the first step is take a given  3 × 3  matrix and write it down alongside a same size identity matrix.

So if we have a matrix   \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}.

This would look like:   \begin{bmatrix} a & b & c & | & 1 & 0 & 0 \\ d & e & f & | & 0 & 1 & 0 \\ g & h & i & | & 0 & 0 & 1 \end{bmatrix}.

The name for this matrix is an ‘augmented matrix’.

## Using the Augmented Matrix

The aim now is to turn the original matrix on the left, in to the identity matrix by performing some matrix elementary row operations.

But each operation we do on the left matrix, will also be done at the same stage to the identity matrix on the right.

When we finally have the identity matrix on the left, the resulting matrix on the right from the row operations performed, will be the inverse matrix we want.
Another name for this method is Gauss Jordan Elimination.

Example

(1.1)

Find the inverse of the matrix   A  =  \begin{bmatrix} 3 & 2 & 2 \\ 0 & 4 & 1 \\ 1 & 2 & 0 \end{bmatrix}.

Solution

The approach we use with the row operations, is to focus on one column at a time, obtaining the leading 1’s on the main diagonal first.

Then obtaining the 0’s in the other rows of the column.

The row operations are in red, with R1, R2 and R3 representing rows 1, 2 and 3.

1)
\begin{bmatrix} 3 & 2 & 2 & | & 1 & 0 & 0 \\ 0 & 4 & 1 & | & 0 & 1 & 0 \\ 1 & 2 & 0 & | & 0 & 0 & 1 \end{bmatrix}     R1 − 2R3         =         \begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 4 & 1 & | & 0 & 1 & 0 \\ 1 & 2 & 0 & | & 0 & 0 & 1 \end{bmatrix}

2)
\begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 4 & 1 & | & 0 & 1 & 0 \\ 1 & 2 & 0 & | & 0 & 0 & 1 \end{bmatrix}     R3 − R1         =         \begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 4 & 1 & | & 0 & 1 & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}

3)
\begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 4 & 1 & | & 0 & 1 & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}     R2 × \bf{\frac{1}{4}}         =         \begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}

4)
\begin{bmatrix} 1 & {\text{-}}2 & 2 & | & 1 & 0 & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}     R1 + 2R2         =         \begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}

5)
\begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 4 & {\text{-}}2 & | & {\text{-}}1 & 0 & 3 \end{bmatrix}     R3 − 4R2         =         \begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & {\text{-}}3 & | & {\text{-}}1 & {\text{-}}1 & 3 \end{bmatrix}

6)
\begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & {\text{-}}3 & | & {\text{-}}1 & {\text{-}}1 & 3 \end{bmatrix}     R3 × \bf{\frac{1}{3}}         =         \begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & 1 & | & {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}

7)
\begin{bmatrix} 1 & 0 & {\small{\frac{5}{2}}} & | & 1 & {\small{\frac{1}{2}}} & {\text{-}}2 \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & 1 & | & {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}     R1 − \bf{\frac{5}{2}}R3         =         \begin{bmatrix} 1 & 0 & 0 & | & {\small{\frac{1}{6}}} & {\text{-}}{\small{\frac{1}{3}}} & {\small{\frac{1}{2}}} \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & 1 & | & {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}

8)
\begin{bmatrix} 1 & 0 & 0 & | & {\small{\frac{1}{6}}} & {\text{-}}{\small{\frac{1}{3}}} & {\small{\frac{1}{2}}} \\ 0 & 1 & {\small{\frac{1}{4}}} & | & 0 & {\small{\frac{1}{4}}} & 0 \\ 0 & 0 & 1 & | & {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}     R2 − \bf{\frac{1}{4}}R3         = &nbsp       \begin{bmatrix} 1 & 0 & 0 & | & {\small{\frac{1}{6}}} & {\text{-}}{\small{\frac{1}{3}}} & {\small{\frac{1}{2}}} \\ 0 & 1 & 0 & | & {\text{-}}{\small{\frac{1}{12}}} & {\small{\frac{1}{6}}} & {\small{\frac{1}{4}}} \\ 0 & 0 & 1 & | & {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}

Now that we have the identity matrix on the left following the matrix elementary row operations that were performed, we have found the inverse of matrix A.

A-1  =  \begin{bmatrix} {\small{\frac{1}{6}}} & {\text{-}}{\small{\frac{1}{3}}} & {\small{\frac{1}{2}}} \\ {\text{-}}{\small{\frac{1}{12}}} & {\small{\frac{1}{6}}} & {\small{\frac{1}{4}}} \\ {\small{\frac{1}{3}}} & {\small{\frac{1}{3}}} & {\text{-}}1 \end{bmatrix}

The matrix elementary row operations we used aren’t the only ones that could have been used to find the inverse matrix.

However as long as you perform the matrix elementary row operations on both matrices, and aim towards obtaining the identity matrix in place of the original matrix, you should eventually find the inverse of a  3 x 3  matrix.

But like with finding the determinant of a matrix, when you want to obtain the inverse of larger matrices, it’s best to make use of a computer or a matrix inverse calculator.

There is one available that can be used when required at the  emathhelp  website.

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