Elementary Linear Algebra For University of Lethbridge Math 1410

Of course, there is a lot of speculation here. We don’t know in general that such a matrix like \(A^{-1}\) exists. (And if it does, whether that matrix is unique, despite the use of the definite article in stating that \(X\) is “the” inverse of \(A\text{.}\)) However, we do know how to solve the matrix equation \(AX = B\text{,}\) so we can use that technique to solve the equation \(AX = I\) for \(X\text{.}\) This seems like it will get us close to what we want. Let’s practice this once and then study our results.

Looking at our previous example, we are tempted to jump in and call the matrix \(X\) that we found “\(A^{-1}\text{.}\)” However, there are two obstacles in the way of us doing this.

First, we know that in general \(AB \neq BA\text{.}\) So while we found that \(AX = I\text{,}\) we can’t automatically assume that \(XA=I\text{.}\)

Secondly, we have seen examples of matrices where \(AB = AC\text{,}\) but \(B\neq C\text{.}\) So just because \(AX = I\text{,}\) it is possible that another matrix \(Y\) exists where \(AY = I\text{.}\) If this is the case, using the notation \(A^{-1}\) would be misleading, since it could refer to more than one matrix.

Perhaps this first obstacle isn’t much of an obstacle after all. Of course, we only have one example where it worked, so this doesn’t mean that it always works. We have good news, though: it always does work. The only “bad” news is that this is a bit harder to prove. For now, we will state it as theorem, but the proof will have to wait until later: see the proof of Theorem 4.5.1.

Combining the two ideas above, we see that \(X = XAY = Y\text{;}\) that is, \(X=Y\text{.}\) We conclude that there is only one matrix \(X\) where \(XA = I = AX\text{.}\) (Even if we think we have two, we can do the above exercise and see that we really just have one.)

We have just proved the following theorem.

Thus, we were justified in Definition 4.4.1 in calling \(A^{-1}\) “the” inverse of \(A\) (rather than merely “an” inverse). Theorem 4.4.4 is incredibly important in practice. It tells us that if we are able to establish that either \(AX=I_n\) or \(XA = I_n\) for some matrix \(X\text{,}\) then we can immediately conclude two things: first, that \(A\) is invertible, and second, that \(A=A^{-1}\text{.}\) We put this observation to use in the next example.

At this point, it is natural to wonder which \(n\times n\) matrices will be invertible. Will any non-zero matrix do? (No.) Are such matrices a rare occurrence? (No.) As we proceed through this chapter and the next, we will see that there are many different conditions one can place on an \(n\times n\) matrix that are equivalent to the statement “The matrix \(A\) is invertible.” Before we begin our attempt to answer this question in general, let’s look at a particular example.

We have just seen that not all matrices are invertible. The attentive reader might have been able to spot the source of the trouble in the previous example: notice that the second row of \(A\) is a multiple of the first, so that the row operation \(R_2-2R_1\to R_2\) created a row of zeros. Can you think what sort of condition would signal trouble for a general \(n\times n\) matrix? Here, we need to think back to our discussions of the various theoretical concepts we’ve encountered, such as rank, span, linear independence, and so on. Let us think of the rows of \(A\) as row vectors.

The elementary row operations that we perform on a matrix either rearrange these vectors, or create new vectors that are linear combinations of the old ones. The only way we end up with a row of zeros in the reduced row echelon form of \(A\) is if one of the rows of \(A\) can be written as a linear combination of the others; that is, if the rows of \(A\) are linearly dependent. We also know that if there is a row of zeros in the reduced row echelon form of \(A\text{,}\) then not every row contains a leading 1. Recalling that the rank of \(A\) is equal to the number of leading 1s in the reduced row echelon form of \(A\text{,}\) we have the following:

The claim that “the following statements are equivalent” in Theorem 4.4.7 means that as soon as we know that one of the statements on the list is true, we can immediately conclude that the others are true as well. This is also the case if we know one of the statements is false. For example, if we know that \(\operatorname{rank}(A)\lt n\text{,}\) then we can immediately conclude that \(A\) will not be invertible.

Let’s sum up what we’ve learned so far. We’ve discovered that if a matrix has an inverse, it has only one. Therefore, we gave that special matrix a name, “the inverse.” Finally, we describe the most general way to find the inverse of a matrix, and a way to tell if it does not have one.

Let’s try again.

In general, given a matrix \(A\text{,}\) to find \(A^{-1}\) we need to form the augmented matrix \(\bbm A\amp I \ebm\) and put it into reduced row echelon form and interpret the result. In the case of a \(2\times 2\) matrix, though, there is a shortcut. We give the shortcut in terms of a theorem.

Although this idea is simple, we should practice it.

Let’s step back and think about this for a moment. The only thing we know about the equation \(A\vx = \vb\) is that \(A\) is invertible. We also know that solutions to \(A\vx = \vb\) come in three forms: exactly one solution, infinitely many solutions, and no solution. We just showed that if \(A\) is invertible, then \(A\vx = \vb\) has at least one solution. We showed that by setting \(\vx\) equal to \(A^{-1}\vb\text{,}\) we have a solution. Is it possible that more solutions exist?

This shows that all solutions to \(A\vx = \vb\) are exactly \(\vx = A^{-1}\vb\) when \(A\) is invertible. We have just proved the following theorem.

A corollary to this theorem is: If \(A\) is not invertible, then \(A\vx = \vb\) does not have exactly one solution. It may have infinitely many solutions and it may have no solution, and we would need to examine the reduced row echelon form of the augmented matrix \(\bbm A \amp \vb \ebm\) to see which case applies.

We demonstrate our theorem with an example.

Knowing a matrix is invertible is incredibly useful. Among many other reasons, if you know \(A\) is invertible, then you know for sure that \(A\vx = \vb\) has a solution (as we just stated in Theorem 4.4.12). In the next section we’ll demonstrate many different properties of invertible matrices, including stating several different ways in which we know that a matrix is invertible.