 So we know how to add and subtract two matrices. We know how to multiply two matrices. What about dividing two matrices? Well, when we divide real numbers, it's easier for us to think about that in terms of multiplying by the inverse of another real number. And so that leads us to finding the inverse of a matrix. And we'll present an algorithm for finding an inverse. But remember that algorithms can be replaced by software that is free or very, very, very cheap. So while we'll present the algorithm here, you should understand why the algorithm works. And you can either do this by thinking about what the problem of finding an inverse really is. It's really setting up a system of equations for a particular matrix product. And then that translates into what we're about to do. So here's the algorithm. Suppose I want to find the inverse of some matrix A. So what I'm going to do is I'm going to augment this matrix A with the identity matrix of the appropriate size. And I'm going to apply row reductions to produce the identity matrix as the first part of the augmented matrix and then something else as the other part. And if I can do this, and there's no guarantee I can always do this, but if it's possible to do this, then this matrix over here is going to be the inverse of matrix A. So for example, let's say I want to find the inverse of this matrix. And again, if you just learned the algorithm, then your entire learning can be replaced by a piece of software. So this is Wolfram Alpha. I'm not affiliated in any way, shape, or form with Wolfram Alpha, but anybody who does mathematics should be aware of this particular site. This is free. I can go to this if you want to download and copy onto your smartphone. It'll cost you about three bucks. But if I can ask a mathematical question, it can tell me the answer. So I'll shift things around a little bit so I can say, so I want to find the, oh my God, I have to type this up. So I'll find the inverse of 2, negative 1, 0, 7, 2, 3, 2, 1, 1. And I say, okay, Wolfram Alpha, tell me what the inverse of this matrix is. And we play the jeopardy themes on that. Oh, that was quick. Okay, so I found the inverse of the matrix and there it is. And well, hopefully, we'll actually get this answer at the end of the day. But here's the thing to realize that if you only learn how to find the inverse by following an algorithm, what you've done can be replaced by somebody using Wolfram Alpha who didn't take the course, who didn't pay however much you paid for the course. Your education is more valuable than something you can find free on the internet. So again, while you should learn the following algorithm, realize that this is the beginning of the educational process. It is not the goal of the educational process. So let's go ahead and find the inverse. So I want to augment that by a 3 by 3 identity matrix. I'll drop it down there. And now I'm going to perform row reductions. And what I'm going to try to do is try to get the identity matrix in the first part. And then I'll have the inverse in the second part. So let's see how that goes. So let's see. I'll keep this first row the same. My second row, the coefficient is 7. So I'll multiply the first row by 7 and the second row by negative 2. And then I'll add the two rows. So 7 times the first row, negative 2 times the second. Throw that into the second row. I add those and the coefficients will drop out. The leading term will drop out. And then I'll get a whole bunch of other terms. To get the first and third row, I'll multiply the first row by 1 and multiply the third row by negative 1. And I'll add those two rows. And that will eliminate the leading coefficient. And so now I have my new third row. Now I'll continue. Again, I like the first row, second row. And now I want to get this third row. So I'll take that second row. I'll multiply it by 2. I'll take the third row, multiply it by negative 11. And that will give me, that will line up my coefficients. And I'll get my new third row. And we have a matrix that's in, we can think about this as quasi-row echelon form. First non-zero entry is above a bunch of zeros. Now, again, with Gauss-Shorten reduction, we do want to get ones here. We ultimately want to get ones here. But if we don't worry about getting the ones there until the end of the process, we can avoid fractions. So let's go through that. We have our quasi-row echelon matrix. I'm going to keep that last row. And now I want to eliminate the stuff above the first non-zero entry in that last row. So I need to eliminate this minus 6 here. So I'm going to take my two rows here. And I'm going to multiply the third row by minus 6. And I'm going to add it to that second row. So there's my third row. There's my third row times minus 6. There's my second row. I'm going to add them. And that gets me a new second row. And my first row already has a zero above that leading one. So I'll just keep that third row without changing anything. All right, so continuing my process. So now here is my new matrix. So I want to reduce the second column. So again, my leading entry of that second row, I want things above and below that to be zero. So I'll keep that second row. Third row is fine. And I need to eliminate the negative 1 here. So I can take the first row, multiply it by negative 11. And I'll add it to the second row. And that'll eliminate that coefficient. And so my new first row looks something like that. Now remember, I'm trying to get the identity matrix over in these first few columns. So that means that the entries along the diagonals have to be 1. And then I have to have zeros everywhere else. Well, I did the zeros everywhere else. So now let's get our entries along that diagonal to be 1. So I'll copy my matrix over. And let's see, to make this entry 1, I need to multiply by negative 1 over 22. And again, remember, each row of a matrix corresponds to an equation. So what this really requires me to do is I multiply all of the coefficients of that equation by negative 1 over 22. So that's going to change that first row to that. Second row, I want this term to be a 1. So I'm going to multiply by negative 11. And third row, I want this term to be a 1. So I'll multiply everything by negative 1. And now the first part of my augmented matrix is the identity, which means that the second part is going to be the multiplicative inverse, the inverse of the matrix. And so we say that the inverse of this matrix is equal to this. And so we found the answer. And again, we've applied an algorithm. It's taken us about five minutes to do what Wolfram Alpha was able to find in about two seconds. So again, the algorithm is useful to know because, well, what if there's a disaster and the entire internet vanishes and then you have to find the multiplicative inverse of a matrix? Well, realistically, if any sort of disaster of that magnitude happened, your inability to find the multiplicative inverse of a matrix will be the least of your worries. But here's another bunch of other important questions. How do we know this is even true? It's on the internet, so it must be true. But it'd be kind of nice to have some independent way of verifying that this is actually the multiplicative inverse of this matrix. But the other thing, again, is that the process for finding the multiplicative inverse, this particular row reduction technique, does emerge as a way of solving a particular type of problem. And if you set up the problem that you're really trying to solve to find the multiplicative inverse, then why this process works becomes self-evident. And it is the understanding of the origin of the algorithm. It is the creating of the algorithm that is really the important thing. Following the algorithm, again, a cheap, free computer application could do that and your education should be worth more than a free or cheap computer application.