Matrix Inverses

In an earlier handout, we hinted about the possibility of solving a matrix equation $AX=B$ by multiplying both sides of the equation by an inverse $A^{-1}$ to the matrix $A$. Now we are going to define the inverse matrix and see how to compute it. First, we need to define another concept:

The $n \times n$ identity matrix $I$ (sometimes denoted $I_n$, if the dimension is not clear from context) is the $n \times n$ matrix that has 1's down its main diagonal and 0's everyplace else:

$$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) \qqua... ...0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \qquad\qquad\dots$$

The identity matrices are the 1's (the ``units'' in technical mathematical language) of matrix multiplication. That is, if $I$ is the $n \times n$ identity matrix, and $A$ and $B$ are any matrices of the right dimensions so the following products are defined, then

$$IA = A \qquad\qquad\hbox{and}\qquad\qquad BI=B.$$

Now an $n \times n$ matrix $A$ has an inverse $A^{-1}$ if there is an $n \times n$ matrix $A^{-1}$ such that

$$AA^{-1} = A^{-1}A =I.$$

This means, in particular, that multiplying by $A^{-1}$ is like dividing by $A$:

$$A^{-1}(AX) = (A^{-1}A)(X) = IX = X.$$

Now that we have a definition of a matrix inverse, how do we find one? First, we need a small but important fact:

Fact: If $A$ and $B$ are $n \times n$ matrices and $AB= I$, then also $BA = I$. (Remember that in general, $AB \neq BA$, so the truth of this fact is not obvious.)

This fact means that if we can find an $n \times n$ matrix $X$ with the property that $AX = I$, then we will know that $XA = I$ also, so $X = A^{-1}$. And (guess what?) we know how to solve the matrix equation $AX = I$: Write down the augmented matrix $A\vdots I$, row-reduce it, then look at the equivalent matrix equation obtained from the new, row-reduced augmented matrix.

For example, let's try to find an inverse to the matrix

$$A = \left(\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 1 \end{array}\right).$$

We try to solve the matrix equation $AX = I$ by writing down the augmented matrix

$$A\vdots I = \left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 ... ... 1 & \vdots & 0 & 1 & 0 \\ 2 & 1 & 1 & \vdots & 0 & 0 & 1 \end{array}\right)$$

and row-reducing it. (Some of these steps combine two operations in one. For example, the first step consists of first adding $-1$ times row 1 to row 2 and then adding $-2$ times row 1 to row 3. The second step is simply to multply row 2 by $-1$.)

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 1... ... 1 & \vdots & 0 & 1 & 0 \\ 2 & 1 & 1 & \vdots & 0 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 0... ... \vdots & -1 & 1 & 0 \\ 0 & -3 & -1 & \vdots & -2 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 0... ... \vdots & -1 & 1 & 0 \\ 0 & -3 & -1 & \vdots & -2 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 1 & \vdots & -1 & 2 & 0 \\ ... ...& \vdots & 1 & -1 & 0 \\ 0 & 0 & -1 & \vdots & 1 & -3 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 1 & \vdots & -1 & 2 & 0 \\ ... ...& \vdots & 1 & -1 & 0 \\ 0 & 0 & 1 & \vdots & -1 & 3 & -1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 0 & \vdots & 0 & -1 & 1 \\ ... ... \vdots & 1 & -1 & 0 \\ 0 & 0 & 1 & \vdots & -1 & 3 & -1 \end{array}\right).$$

This is the augmented matrix of the matrix equation

$$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \... ...ray}{ccccccc} 0 & -1 & 1 \\ 1 & -1 & 0 \\ -1 & 3 & -1 \end{array}\right) ,$$

or

$$X = \left(\begin{array}{ccccccc} 0 & -1 & 1 \\ 1 & -1 & 0 \\ -1 & 3 & -1 \end{array}\right) .$$

Therefore we have our solution:

$$A^{-1} = \left(\begin{array}{ccccccc} 0 & -1 & 1 \\ 1 & -1 & 0 \\ -1 & 3 & -1 \end{array}\right) .$$

This is one way things can work out when we try to find the inverse of a matrix A. Here's another: let us try to find an inverse to the matrix

$$A = \left(\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 2 \end{array}\right).$$

We try to solve the matrix equation $AX = I$ by writing down the augmented matrix

$$A\vdots I = \left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 ... ... 1 & \vdots & 0 & 1 & 0 \\ 2 & 3 & 2 & \vdots & 0 & 0 & 1 \end{array}\right)$$

and row-reducing it:

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 1... ... 1 & \vdots & 0 & 1 & 0 \\ 2 & 3 & 2 & \vdots & 0 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 0... ...& \vdots & -1 & 1 & 0 \\ 0 & -1 & 0 & \vdots & -2 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 2 & 1 & \vdots & 1 & 0 & 0 \\ 0... ...& \vdots & -1 & 1 & 0 \\ 0 & -1 & 0 & \vdots & -2 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 1 & \vdots & -1 & 2 & 0 \\ ... ...& \vdots & 1 & -1 & 0 \\ 0 & 0 & 0 & \vdots & -1 & -1 & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 1 & \vdots & -1 & 2 & 0 \\ ... ... & \vdots & 1 & -1 & 0 \\ 0 & 0 & 0 & \vdots & 1 & 1 & -1 \end{array}\right)$$

$$\left(\begin{array}{ccccccc} 1 & 0 & 1 & \vdots & 0 & 3 & -1 \\ ... ...& \vdots & 0 & -2 & 1 \\ 0 & 0 & 0 & \vdots & 1 & 1 & -1 \end{array}\right).$$

This is the augmented matrix of the matrix equation

$$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \... ...rray}{ccccccc} 0 & 3 & -1 \\ 0 & -2 & 1 \\ 1 & 1 & -1 \end{array}\right) .$$

Now this equation has no solutions. How do we know this? The entries in the third row of the product

$$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) X$$

will be the products of the third row of the first factor with the columns of $X$. Since the third row of the first factor consists entirely of zeroes, these products will all be zero as well, so whatever $X$ is, the third row of

$$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) X$$

will be $(0,0,0)$; it is impossible for it to be $(1,1,-1)$. Therefore, the matrix $A$ does not have an inverse.

These are the two possibilities. We can collect this information into a procedure:

To find the inverse $A^{-1}$ of a square matrix $A$, write down the matrix $A\vdots I$ and then row-reduce it. Either it will row-reduce to a matrix of the form $I \vdots B$, in which case $B = A^{-1}$, or it will row-reduce to a matrix of the form $C \vdots B$ where $C$ has a row consisting entirely of zeroes, in which case $A$ has no inverse.