3 The inverse of a 3 $\times$ 3 matrix - Gauss elimination method

It is true, in general, that if the determinant of a matrix is zero then that matrix has no inverse. If the determinant is non-zero then the matrix has a (unique) inverse. In this Section and the next we look at two ways of finding the inverse of a $3\times 3$ matrix; larger matrices can be inverted by the same methods - the process is more tedious and takes longer. The $2\times 2$ case could be handled similarly but as we have seen we have a simple formula to use.

The method we now describe for finding the inverse of a matrix has many similarities to a technique used to obtain solutions of simultaneous equations. This method involves operating on the rows of a matrix in order to reduce it to a unit matrix.

The row operations we shall use are

1. interchanging two rows
2. multiplying a row by a constant factor
3. adding a multiple of one row to another.

Note that in (ii) and (iii) the multiple could be negative or fractional, or both.

The Gauss elimination method is outlined in the following Key Point:

Suppose we wish to find the inverse of the matrix

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right]$

We first place $A$ and $I$ adjacent to each other.

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Phase 1

We now proceed by changing the columns of $A$ left to right to reduce $A$ to the form $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \ast \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ where $\ast$ can be any number. This form is called upper triangular .

First we subtract row 1 from row 2 and twice row 1 from row 3. ‘Row’ refers to both matrices.

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill R2-R1\hfill \\ \hfill R3-2R1\hfill \end{array}\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -2& \hfill 0& \hfill 1\end{array}\right]$

Now we subtract row 2 from row 3

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -2& \hfill 0& \hfill 1\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill \hfill \\ \hfill R3-R2\hfill \end{array}\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

Phase 2

This consists of continuing the row operations to reduce the elements above the leading diagonal to zero.

We proceed right to left . We subtract 3 times row 3 from row 1 (the elements in row 2 column 3 is already zero.)

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 1& \hfill 1\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill \hfill \\ \hfill R1-3R3\hfill \end{array}\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 4& \hfill 3& \hfill -3\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

Finally we subtract 3 times row 2 from row 1.

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 4& \hfill 3& \hfill -3\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill 1\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill \hfill \\ \hfill R1-3R2\hfill \end{array}\Rightarrow \left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 7& \hfill 0& \hfill -3\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

Then we have $A^{-1}=\left[\begin{array}{ccc}\hfill 7& \hfill 0& \hfill -3\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

(This can be verified by showing that $A{A}^{-1}=I$ or $A^{-1}A=I$ .)

Task!

Consider $A=\left[\begin{array}{ccc}\hfill 0& \hfill 1& \hfill 1\\ \hfill 2& \hfill 3& \hfill -1\\ \hfill -1& \hfill 2& \hfill 1\end{array}\right],I=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ .

Use the Gauss elimination method to obtain $A^{-1}$ .

First interchange rows 1 and 2, then carry out the operation (row 3)+ $\frac{1}{2}$ (row 1):

Answer

Now carry out the operation (row 3)- $\frac{7}{2}$ (row 2) followed by (row 1)- $\frac{1}{3}$ (row 3) and (row 2)+ $\frac{1}{3}$ (row 3):

Answer

Next, subtract 3 times row 2 from row 1, then, divide row 1 by 2 and row 3 by $\left(-3\right)$ .

Finally identify $A^{-1}$ :

Answer