In the previous section we introduced the idea of inverse matrices and elementary matrices.  In this section we need to devise a method for actually finding the inverse of a matrix and as we’ll see this method will, in some way, involve elementary matrices, or at least the row operations that they represent.

The first thing that we’ll need to do is take care of a couple of theorems.

Before we get into the proof let’s say a couple of words about just what this theorem tells us and how we go about proving something like this.  First, when we have a set of statements and when we say that they are equivalent then what we’re really saying is that either they are all true or they are all false.  In other words, if you know one of these statements is true about a matrix A then they are all true for that matrix.  Likewise, if one of these statements is false for a matrix A then they are all false for that matrix.

To prove a set of equivalent statements we need to prove a string of implications.  This string has to be able to get from any one statement to any other through a finite number of steps.  In this case we’ll prove the following chain .  By doing this if we know one of them to be true/false then we can follow this chain to get to any of they others.

The actual proof will involve four parts, one for each implication.  To prove a given implication we’ll assume the statement on the left is true and show that this must in some way also force the statement on the right to also be true.  So, let’s get going.

Proof :

: So we’ll assume that A is invertible and we need to show that this assumption also implies that  will have only the trivial solution.  That’s actually pretty easy to do.  Since A is invertible we know that  exists.  So, start by assuming that  is any solution to the system, plug this into the system and then multiply (on the left) both sides by  to get,

So,  has only the trivial solution and we’ve managed to prove this implication.

: Here we’re assuming that  will have only the trivial solution and we’ll need to show that A is row equivalent to .  Recall that two matrices are row equivalent if we can get from one to the other by applying a finite set of elementary row operations.

Let’s start off by writing down the augmented matrix for this system.

Now, if we were going to solve this we would use elementary row operations to reduce this to reduced row-echelon form,  Now we know that the solution to this system must be,

by assumption.  Therefore, we also know what the reduced row-echelon form of the augmented matrix must be since that must give the above solution.  The reduced-row echelon form of this augmented matrix must be,

Now, the entries in the last column do not affect the values in the entries in the first n columns and so if we take the same set of elementary row operations and apply them to A we will get  and so A is row equivalent to  since we can get to  by applying a finite set of row operations to A.  Therefore this implication has been proven.

 : In this case we’re going to assume that A is row  equivalent to  and we’ll need to show that A can be written as a product of elementary matrices.

So, since A is row equivalent to  we know there is a finite set of elementary row operations that we can apply to A that will give us .  Let’s suppose that these row operations are represented by the elementary matrices .  Then by Theorem 4 of the previous section we know that applying each row operation to A is the same thing as multiplying the left side of A by each of the corresponding elementary matrices in the same order.  So, we then know that we will have the following.

Now, by Theorem 5 from the previous section we know that each of these elementary matrices is invertible and their inverses are also elementary matrices.  So multiply the above equation (on the left) by  (in that order) to get,

So, we see that A is a product of elementary matrices and this implication is proven.

 :  Here we’ll be assuming that A is a product of elementary matrices and we need to show that A is invertible.  This is probably the easiest implication to prove.

First, A is a product of elementary matrices.  Now, by Theorem 5 from the previous section we know each of these elementary matrices is invertible and by Theorem 2(a) also from the previous section we know that a product of invertible matrices is also invertible.  Therefore, A is invertible since it can be written as a product of invertible matrices and we’ve proven this implication.

This theorem can actually be extended to include a couple more equivalent statements, but to do that we need another theorem.

Proof :

(a) This proof will need part (b) of Theorem 1.  If we can show that  has only the trivial solution then by Theorem 1 we will know that A is invertible.  So, let  be any solution to .  Plug this into the equation and then multiply both sides (on the left by B.

So, this shows that any solution to  must be the trivial solution and so by Theorem 1 if one statement is true they all are and so A is invertible.  We know from the previous section that inverses are unique and because  we must then also have .

(b) In this case let’s let  be any solution to .  Then multiplying both sides (on the left) of this by A we can use a similar argument to that used in (a) to show that  must be the trivial solution and so B is an invertible matrix and that in fact .  Now, this isn’t quite what we were asked to prove, but it does in fact give us the proof.  Because B is invertible and its inverse is A (by the above work) we know that,

but this is exactly what it means for A to be invertible and that .  So, we are done.

So, what’s the big deal with this theorem?  We’ll recall in the last section that in order to show that a matrix, B, was the inverse of A we needed to show that .  In other words, we needed to show that both of these products were the identity matrix.  Theorem 2 tells us that all we really need to do is show one of them and we get the other one for free.

This theorem gives us is the ability to add two equivalent statements to Theorem 1.  Here is the improved Theorem 1.