# Finding the Inverse of a Matrix

Posted by Jeff Schuler on Mar 5, 2011 in Uncategorized |

### Matrix manipulations and properties

Finding the inverse of a matrix is much more complex than finding the inverse of a number. All real numbers have an inverse (i.e. $6^{-1}= \frac{1}{6}$). However, not all matrices have an inverse. There are several characteristics that allow us to visibly determine whether a matrix has an inverse but we will only focus on one. A matrix must be square (i.e. 2×2, 3×3, etc.) to have an inverse. Performing the following manipulations will be a waste of time if a matrix is not square. It is also important to know the inverse matrix property. Using my example above, $6^{-1} * \frac{1}{6} = 1$ and similarly with matrices, $A * A^{-1} = In$ where In is the identity matrix (diagonal from top left to bottom right contains all 1’s, and everything else is 0) . We take advantage of this property when solving systems of matrices.

In words, the general algorithm for determining the existence of an inverse matrix is to manipulate the matrix into row reduced echelon form (rref). If the rref matrix is an identity matrix, then the inverse matrix exists. Hang on now, earlier I mentioned that there were other, visible characteristics that allow us to determine the existence of an inverse matrix, but now I’m asking you to perform a tedious process (without a calculator) with the same goal? Wouldn’t it be easier to first determine if finding the rref of the matrix is worthwhile? You’re right, except we are going to make a simple manipulation, and at the same time that we finish our rref process and determine that an inverse matrix exists, we will have found the inverse matrix! How do we do that? We will create an augmented matrix between our matrix in question, $A$, and the appropriate identity matrix where the size of matrix $A$ is equal to the size of matrix $In$. We will perform the same rref process to the augmented matrix $| A$ $In |$. If the portion of our augmented matrix previously belonging to matrix $A$ reduces to an identity matrix (indicating the existence of $A^{-1}$ ), then the portion previously belonging to the identity matrix, will equal $A^{-1}$.

### Some matrix math

Now, for the math…

Suppose we are asked to find the inverse of the following matrix:

$\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}$

First, we must set up the augmented matrix discussed above. Notice that I have simply placed the identity matrix (of the same size as $A$ ) on the right of matrix $A$.

$\begin{bmatrix}1&3&3&1&0&0\\1&4&3&0&1&0\\1&3&4&0&0&1\end{bmatrix}$

### Finding the rref of an augmented matrix

Next, we will attempt to find the rref of the augmented matrix. If the portion of the augmented matrix previously belonging to $A$ yields an identity matrix, $A$ is invertible.

rref $\begin{bmatrix}1&3&3&1&0&0\\1&4&3&0&1&0\\1&3&4&0&0&1\end{bmatrix}$ = $\begin{bmatrix}1&0&0&7&-3&-3\\0&1&0&-1&1&0\\0&0&1&-1&0&1\end{bmatrix}$

Ok great! The left half of our augmented matrix reduced to an identity matrix. That means two things to us: the matrix has an inverse and we’ve already found the inverse. If you recall from above, $A^{-1}$ is the right half of the augmented matrix (after finding it’s rref, of course). So we can conclude:

$\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}^{-1} = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$

If our rref of the augmented matrix had yielded anything other than an identity matrix, we would conclude that $A^{-1}$ does not exist. This method will simply allow us to determine the existence of and entries to $A^{-1}$ for any size matrix.