 Welcome to episode 27. This is a review for exam number four. And in this episode we've been talking about matrices, applications of matrices for solving systems of equations and other things. Let's go to our first graphic here and look at some of the ideas you need to be familiar with before you take this test. First of all, you should be prepared to solve problems similar to those that we've been discussing in class and those that have been assigned for homework. On this exam, you can't use a calculator. So I'll try to keep the numerical computations fairly simple when we reduce matrices so that a calculator won't be that significant. As usual, you can't use notes, you can't use your text. The scratch paper that you're given with your test will have to be turned in even the paper that you don't use. And finally, we suggest that you show all your work to get partial credit in case you miss a problem. We try to be fairly generous with the partial credit but if there's no work shown and you just have a wrong answer then we don't have much option except to just count off full credit. So keep that in mind when you're taking the test. Okay, well let's go to episode 22 and look at some of the ideas we want to review from that. First of all, you want to be able to use a Gaussian elimination with back substitution to solve systems of linear equations. Now before we look at the second item there I think we should work an example of a problem with Gaussian elimination with back substitution to remind you how this goes. So I have an example here that I've made up that I'll solve using this method. The problem goes like this. x minus y plus 3z is equal to 4 and x plus 2y minus 2z is equal to 10 and then 3x minus y plus 5z is 14. Now I want to solve this system of equations using matrix methods. In particular I want to use the Gaussian method with back substitution, Gaussian elimination. And you know that means we're going to convert this to a matrix and then there are only three things I can do to the matrix that will allow me to keep the solution while simplifying the system. And that is I can interchange two rows. I can multiply any row by a non-zero constant. It's got to be non-zero of course. And then I can add a multiple of one row to another row. So those are called the three elementary row operations. So here's how we do this. First of all I abbreviate this with a matrix where I only show the coefficients in the constants. That would be 1 minus 1, 3 and 4. 1 minus 1, 3 and 4. And then 1, 2, negative 2, 10. And then 3, 1, 3, negative 1, 5, 14. Okay, so what I'm going to do is to alter this system of equations so that I get ones along the main diagonals and I get zeros below that. Well I already have a one here so I think I'm ready to get a zero below that one and a zero at the very bottom. So I'm working on the first column only here. So what I'm going to do is take row 2 and add on negative 1 times row 1. And also I'll take row 3 and add on negative 3 times row 1. That's a negative 3 there. But I'm not going to change row 1 so I could rewrite that. 1, negative 1, 3 and 4. But if I take negative 1 times row 1 and add it to row 2 this will be a zero. And then negative 1 times this plus 2 is a 3. And then a negative 5 and then a 6. And on the bottom row, let's see, I'll be taking negative 3 times row 1 and adding it to row 3. This will be a zero. And then I get a plus 2. Negative 3 times negative 1 is 3. Add it to negative 1 is 2. And then negative 4 and then 2. Okay so I've taken care of the first column. Now I've moved to the second column. And if you remember in this case what I want to do is get a 1 in this position and then zeroes below it. I don't care about getting zeroes above it. That'll come later in a different, when we refer to a different episode. Well to get a 1 here it doesn't look like there's any easy, well I think there is an easy way to get a 1. Let's take a negative 1 times row 3 and add it to row 2. That'll give me a 1. So let's do that. Let's take row 2 and add on negative 1 times row 3. So my first row is 1, negative 1, 3 and 4. And then, let's see, the bottom row isn't going to change so I'll rewrite that. But if I take negative 1 times row 3 and add it to row 2, this will be a zero here. And I'll get a 1 there. That's what I was after. And then I'll get a negative 1 and then a 4. Now the reason I'd like to get a 1 here is because I can use that as a lever to get a zero below it. So let's go ahead and do that right now. I'd like to get a zero right here so I'm going to take row 3 and add on negative 2 times row 2. So row 1 doesn't change. And row 2 doesn't change. But row 3 changes dramatically. This becomes a zero and this becomes a zero. And then I get a negative 2 here because that's a negative 4. Let me emphasize that. And then negative 6. Okay, now at this point I'd like to get a 1 right there but you know I think I can step out right now and just solve for z. So I'm going to write down my three equations. They say x minus y plus 3z is 4. And the second equation says that y minus z is equal to 4. And the last equation because I never got a 1 in that position, I have negative 2z is equal to negative 6. Well we can see right here that that means z is equal to 3. Okay then I use back substitution to solve for y and then I use back substitution to solve for x. So I'm going to back substitute right here to get y and then I'll back substitute both of those to calculate x. And so it goes like this. y minus z is equal to 4 and I know that z is equal to 3. So y minus 3 is equal to 4 which means y is equal to 7. And then if I substitute into the first equation which is x minus y plus 3z is equal to 4. Substituting in I have x minus 7 plus let's see now 3z would be 9 is equal to 4. In other words x plus 2 is 4 and so x is equal to 2. So my solution is 2, 7, 3. So I'll write it as an order triple 2, 7, 3. And you notice when I list it I listed an alphabetical order. Now if I were to substitute those back in just to check them. If I put a 2 here and a 7 here and a 3 here let's just see if those work out. This says the 2 minus 7 is negative 5. Negative 5 plus 9 is 4. In the second equation 2 plus 14 is 16. 16 take away 6 is 10. That one checks. And 6 take away 7 is negative 1. Negative 1 plus 15 is 14. So it checks in all three equations. So this procedure is called Gaussian elimination with back substitution. And it's one of two significant methods that we back substitution here. I wrote the wrong word. It's one of two methods that we've discussed in this chapter for solving systems of equations. Let's take one more equation and this time I'm going to pick one that has no solution and I want to use Gaussian elimination to solve it. Now as you might guess the place where people generally make their mistakes in this procedure is just making simple arithmetic mistakes. I make simple arithmetic mistakes sometimes too. So the trouble is if you make a simple mistake like that it can really throw off the problem and it can lead you to rather complex looking solutions like messy fractions and so forth. So not only is it the wrong answer but you have to sometimes work even harder to get your answer than the answer that was actually intended. So you have to be careful about that. As long as you're proceeding in generally the right direction you'll get some partial credit even if you make an error along the way. Okay, in this next problem I'd like to make up a situation where there's no solution and let me show you how we can do that. Suppose I have the equation x plus y plus z equals one. I'll just make that one up at random. And actually the next equation then I've actually chosen to be x minus y plus 2z is let's say three. Now to get a third equation what I want to do is to combine these by subtraction. So what if I subtract these? x minus x is zero, y take away negative y, that'll be a 2y, and z take away 2z is minus z. And then what number should go over here? I guess negative 2? Well instead of negative 2 I'm going to put something else in there. Let's say we pick ten. You say but if you subtract these that shouldn't be ten, that should be negative two. Well because I've changed that number there will be no solution because this is not the number that I should have gotten. Let's see what happens when I solve this and I think we'll see that there's no solution in this problem. The augmented matrix will be one, one, one one, and one, negative one, two, three, and zero, two, negative ten. Now of course just by looking at this we wouldn't know that the answer is going to be no solution, but by the way we've made this up I think we'll find out that there is no solution because we've made an adjustment from what the answer we would have expected. Well I have a one in the first position again so that certainly helps me so I'd like to get a zero here. So I'm going to take row two and add on negative one times row one. So I get one, one, one, and I get zero, negative two, one, and two. And the bottom row is okay I have a zero right here so this is zero, two, negative one, and ten. Okay well I'd like to get a zero right there, okay. But you know excuse me I'd like to get a one right here, but what I'm noticing is if I look further ahead when I go to get a zero here I'll be adding row two to row three that'll be a zero. This will be a zero and this one won't. So I'm going to jump ahead to the situation where to the case where I want to get a zero in the row three column two position. So this is going to be, let's see I'll take row three and I'll add on row two. So the top row is one, one, one, one. The second row is zero, negative two, one, two. I've avoided getting the one there because it's going to introduce fractions and I think the significant part of this problem is going to be what happens on the bottom row. So when I add rows two and three I get a zero, a zero, a zero, and oh I get a twelve when I add those. Well the problem with this last equation is it says zero equals twelve. My equations are x plus y plus z is equal to one. There's nothing wrong with that in fact that was my original equation. The second equation is negative two y plus z is equal to two. And the last equation says zero equals twelve. Well that's a contradiction of course. And by using elementary row operations I know that I haven't destroyed the solution. I haven't changed the solution. So if there's no solution at the end there was no solution at the very beginning. So this problem has no solution. Now you could either stop right here and say no solution because of that contradiction. Or in some textbooks you'll see they just write empty set for no solution. So in this case we can stop early. We don't have to proceed to get a one here or anything else. As soon as you can get a contradiction within a row then you're home free. Okay so much for that problem. Let's go back to our list of objectives in episode twenty-two. And there's one more thing on there that I haven't mentioned that we should. And that is the second item. Be able to solve applications of systems of linear equations such as for example partial fractions. Now you only saw me work one partial fraction problem in class. There are many more in the textbook in the section on partial fractions. But I think maybe we should perhaps take a partial fraction problem as another example. So I've picked out one here from the text and it goes like this. I'll just write it here on the board. Find the partial fraction decomposition of let's see, 7x minus 3 over x times x plus 3 times x minus 1. Okay now first of all let me remind you what I mean by a partial fraction decomposition. You see if I have a fraction such as let's say 5 over 6 this denominator will factor. That's really 5 over 2 times 3. And this is a proper fraction because the numerator is smaller than the denominator. Now whenever you have a proper fraction and the numerator will factor, you can break this up into two what are called partial fractions. And I'm going to put these different factors in the denominators a half, halves, and thirds. And these two expressions are called the partial fractions and I'm getting the partial fraction decomposition of 5 over 6. Now I'm looking for a numerator either positive or negative that makes this a proper fraction and a numerator either positive or negative that makes this one a proper fraction so that this sum ends up being 5, 6. And I think what I need is 1 half and a 1 third. Let's see 1 half is 3 over 6, 1 third is 2 over 6, and when you add those up sure enough you get 5 over 6. Now as a matter of fact this can always be done. This isn't a fact that's generally that you generally see say in secondary school because it has no particular importance there but the partial fraction decomposition does have some importance in some later courses. And it says if you have a proper fraction and if the denominator factors then you can write this in the form of well because these are two different factors 2 and 3 in the halves and thirds the numerators might be negative or positive but they have to be proper fractions otherwise and the sum adds up to be 5 over 6. Now with that idea let's try to find the partial fraction decomposition of 7x minus 3 over x times x plus 3 times x minus 1. Now you know I should mention at the outset that in the textbook these denominators are not always factored and you may have to factor this on your own like take out the common factor of x and then you have a quadratic which you factor in this case to be x plus 3 and x minus 1. Okay so the proper fraction or the partial fraction decomposition will have an x and an x plus 3 minus 1. And this original expression is a proper fraction in the sense that I have a smaller degree on top than I have one bottom. On top I have a linear polynomial first degree and on the bottom I have a cubic polynomial if I multiply it out so it's a proper fraction in the sense of its degree. And so these will be proper fractions and if the denominators have degree 1 the numerators must be degree 0 which means they're constant. So I'll say a b and c. Those are all constants. I'd like to find out what constants I could put there that would make that sum be my original expression. Well to solve this I'm going to multiply both sides by x times x plus 3 times x minus 1. And over here I'll multiply by x times x plus 3 times x minus 1. And on the left hand side I'm left with only 7x minus 3. On the right hand side when I multiply this quantity by those three factors I get a times x plus 3 times x minus 1 plus b times x times x minus 1 plus c times x times x plus 3. Now I'm going to make a comparison of the left side and the right side and see what the constant term is on the right and compare it with negative 3. What the linear or first power term is on the right and compare it with 7. And what the quadratic term is the square term on the right and compare it with 0 because I have 0 x squares on the other side. But I'll need to multiply this out so let's do that. 7x minus 3 is going to be a times x squared plus 2x minus 3 plus if I multiply this out it's bx squared minus bx. And then cx squared plus 3 cx. Okay now I still need to multiply through by a so I have 7x minus 3 equals ax squared plus 2ax minus 3a plus bx squared minus bx plus cx squared plus 3cx. Now I want to group all the x squares together, all the x's together, and all the constants together and compare these two. So how many x squares will there be? Well let's see I have ax squared I have bx squared and cx squared that's a plus b plus c. How many x's do we have? First powers. Well I have 2ax minus bx plus 3cx. So that says 2a minus b plus 3c. And now what's left over? There's only one constant term. This is what I'd call my constant term because a is representing some constant. We just don't know what it is yet. So that's a negative 3a. And if I compare a plus b plus c must be 0 because over here on the other side I have 0x squared. So a plus b plus c is 0. And in the second term 2a minus b plus 3c must be 7. 2a minus b plus 3c is 7. And finally negative 3a is equal to negative 3. Well that one I can solve quickly and see that a is equal to 1. So if a is equal to 1 I have 9 plus b plus c is 0. That says that b plus c is negative 1. And in the second equation 2 minus b plus 3c is 7. So that says that negative b plus 3c is equal to 5. So I have two equations. This one and this one. Actually I had three equations originally but one of them was very easy to solve. So now I want to solve these two. And I want to continue demonstrating Gaussian elimination with back substitution. So this is going to be fairly short. But let's solve those two equations and I'll erase this and solve it up above. Now you know when you're solving a 2 by 2 system of equations any method you know of is probably going to be fairly short. And it may seem like this is overkill to use Gaussian elimination with back substitution to solve this. But since that's what you're being tested on I think that's what we ought to demonstrate. So we're solving b plus c is equal to negative 1. And negative b plus 3c is equal to 5. So what I'll do is convert that to matrix form and that's 1, 1, negative 1. And then negative 1, 3 and 5. That's my so called augmented matrix. I have a 1 in the first position. I want to get a 0 right below it and I think we'll be home free. So I'm going to take row 2 and add on row 1. 1, 1, negative 1. Adding I get 0, 4 and 4. And while we're at it let's go ahead and divide through. So I'm going to multiply by 1 fourth. 1 fourth times row 2 and that's 1, 1, negative 1 0, 1, 1. Okay and I now have this reduced to a form where I can back substitute. Okay so our system of equations is going to be b plus c is equal to negative 1 and c is equal to 1. Now you notice this first equation hasn't changed from what it was back up here. But if I back substitute for c that says that b plus 1 is equal to negative 1. So b is equal to negative 2. Now if we put this together with the fact that we already found that a was equal to, let's see what was a, I forgot here, oh a was equal to 1. Then what that tells me then is that the partial fraction decomposition of this original rational expression was going to be, let's say we said a over x plus b over x plus 3 plus c over x minus 1. Now a is equal to 1, b is equal to negative 2 and c is equal to 1. Which means if you add these three fractions together now, you'll have the original fraction. That is if you add 1 over x minus 2 over x plus 3, so I bring the negative out in front, plus 1 over x minus 1, that's the partial fraction decomposition. Now there are some courses that lie ahead particularly if you're going to study mathematics where you need to know about the partial fraction decomposition. So this is I think only our second example in the telecourse but you'll find other examples in the textbook. Okay let's go to episode 23. In episode 23 we talked about using an alternative means of solving systems of equations called the Gauss-Jordan elimination method. This is where you get zeros both above and below the diagonal of ones. I think we should take an example of that and just see what are some of the nuances that come up when we solve that using the Gauss-Jordan elimination method. So this time we're using Gauss-Jordan. Okay I have a system of equations here that I've picked out. Let's solve x plus y plus z is equal to 2 and 2x minus 3y plus 2z is 4 and rather than adding or subtracting those equations I have what you would refer to as an independent equation that is something completely different and that is that 4x plus y minus 3z is equal to 1. And this time I want to use the Gauss-Jordan method. Now on the exam if I don't specify which of these two matrix methods you should use you can use either one that you feel most comfortable with. If you want to back substitute you could use the first method. If you want to use Gauss-Jordan there's no Gauss there's no back substitution involved but there is a little bit more work in reducing the matrix. If I specify the method then of course you should use that method completely. Okay so the augmented matrix will be 1, 1, 1 and 2 and then 2, negative 3 2 and 4 and then 4, 1 negative 3, 1. Now we have a 1 in the row 1, column 1 position and of course if there hadn't been a 1, if I could find a 1 somewhere else I would interchange 2 rows to get a 1 up there. Otherwise I'd find some way to add or subtract 2 rows or even multiply through by a non-zero constant to get a 1 but that last option I usually postpone until I have to use it because if it introduces fractions all the arithmetic gets more complicated. But we have a 1 so let's just keep going. I need to get the rest of the column. So I'll take row 2 and add on negative 2 times row 1 and then I'll take row 3 and add on negative 4 times row 1. But row 1 doesn't change. However on the second row I'll get a zero and a negative 5 and a zero again and a zero in the last position. On the bottom row I'll get a zero and then I'll get negative 3 and then I'll get negative 7 and then I'll get negative 7 again. Now you notice in this second row what this says is negative 5 y is equal to zero. So that tells me y is going to equal zero but rather than stepping out and writing that down I want to demonstrate the procedure so I'm going to proceed as if I didn't realize I solved that equation just by looking at it. So let's just keep going. Now my goal after finishing the first column is to go to the second column and get a 1 in this position and then afterwards get zeros above and below it. So let's get a 1 right there first and I can do that by multiplying by a negative 1 fifth on row 2. Row 1 hasn't changed here and row 3 hasn't changed here. But row 2 does zero, one, zero, zero. And you see the reason I want to get a 1 is I use it as a lever to produce the zeros below and above so that I can continue. Well to get a zero up above I think I'll take row 1 and add on negative 1 times row 2. That's going to be fairly easy because three of the four numbers there are zeros. And on the bottom row I'll take row 3 and I'll add on times row 2 and to produce a zero right here. So let's do that. Row 2 doesn't change. But row 1 becomes one, zero, one, two. And row 3 becomes zero, zero, negative 7, negative 7. Okay well I've taken care of the first two columns. I come to the last column and the first thing I do is get the 1 along the main diagonal. So I want to get a 1 right there. So I'll multiply it by negative 1 seventh. Because I see that I'm not going to produce any fractions there it's going to just cancel out very nicely. So on top I have one, zero, one, two. Then I have zero, one, zero, zero. And then I have zero, zero, one, one. Okay I've almost finished. I still need to get a zero at the top of that column because I want only ones along the main diagonal. So to do, to accomplish that I'll take row 1 and add on negative 1 times row 3. That's a negative 1 right there. So let's see the bottom row hasn't changed. And the second row hasn't changed. But the top row becomes one, zero, zero, and one. Well you see what's nice about the Gauss-Jordan method is although it takes me a few more steps to get the zeros above and below I can just read the answers off. This says that x is equal to 1. It says that y is equal to zero. And it says that z is equal to one. So my solution is one, zero, one. Most textbooks are going to write it this way because it takes up less space. Let's just check these numbers and see if they really do work. Now I'm pointing this out because when you're doing this on an exam and you come up with your answer, on time go back and check your work and you'll know if you got it right or wrong by substituting it back in. So I have 1 plus 1 is 2. I have 2 plus 2 is 4. And I have 4 minus 3 is 1. So they all check. So that makes me think I've gotten the right answer here. Okay, one more example of Gauss-Jordan and this time I want to work an example in which I'll get infinitely many solutions to show and demonstrate how we parametrize the solutions. So let's use the same method, different problem, and it should have a different outcome. What if we have a system of equations that looks like this? 2x plus y minus z equals, I'll just pick a number. Let's say equals 3. And the second equation is plus 2y plus 3z equals let's say negative 5. Now for my third equation, since I want this to have infinitely many solutions, I want my third equation not to be independent, that is just to make it up separately from the others, but I want it to be a combination of the previous equations. So what let's do is just subtract these 2x minus x is x, y minus 2y is negative y. And negative z, take away 3z is negative 4z. And over here, 3 take away negative 5 is 8. Now you remember when I made up the problem with no solution a few minutes ago? What I did is I changed that number so it didn't give me what I thought it was supposed to give, that gave me no solution. This time I'm going to write down the answer as it actually turns out. Now if I solve this, I think I'll get infinitely many solutions because you see this equation is not giving me any new information, it's just giving me a sort of a compilation of what I already knew. So the augmented matrix is 2, 1, negative 1, 3. And then 1, 2, 3, negative 5. And then 1, negative 1, negative 4, 8. Now the trouble is when you look at a system of equations like this, or when you look at the augmented matrix, you don't recognize how it came about and that it's going to have infinitely many solutions. So you just have to proceed as usual. So let's continue as if we didn't know what's going to happen. I want to get a 1 in the first position. So why don't we interchange 2 rows? It doesn't matter which 2. I'm going to interchange the first and the third. So I'm going to interchange the first and the third in rewrite number 2. Now you might say, well why aren't you interchanging rows 1 and 2? Well you could do that too. And everything will work out fine. Now the first row becomes 1, negative 1, negative 4, and 8. And the last equation is 2, 1, negative 1, 3. Okay so if I continue as usual, I want to get 0's in these positions where the 1 and the 2 are. So row 1 won't change. But row 2 will. I'm going to take row 2 and add on negative 1 times row 1. And for row 3 I'll replace it with row 3 plus negative 2 times row 1. So this becomes 0 and then 3 and then 7 and then negative 13. And on the second, on the third row I'll get 0 and then 3 and then 7 and then negative 13. Now do you notice anything unusual here? The last two rows are the same. Yeah it looks like the last two rows are identical. So you see if I just take row 2 and subtract it from row 3 I'll have all 0's across the bottom. So let's do that. I'm going to take row 3 and add on negative 1 times row 2. So row 1 is 1, negative 1, negative 4 and 8. Row 2 hasn't changed. It's 0, 3, 7 and negative 13. And now if when I take negative 1 times row 2 and add it to row 3 I get 0, 0, 0 and 0. Now when I get all 0's across here this is not a contradiction. This says 0 equals 0. It just means there's no new information there. I knew that 0 is equal to 0 before I started the problem. So all the information comes from the first two equations. And you can see now why this is happening because of the way we designed this. So let's see, one more step if I get a 1 right here this won't be pretty but it looks like I'll just have to divide through by 3 or multiply by a 1 third. So let's do that. 1 third times row 2. Now let's see row 1 is still as it has been. Row 3 is the same but on row 2 in order to get that 1 I have to go to fractions and I get 0, 1, 7 thirds and negative 13 thirds. And then because I'm using Gauss Jordan elimination I want to get a 0 up above. So let's do that. I think it's going to be worth the arithmetic to see this answer. I'm going to take row 1 and add on row 2. Row 2 is this and row 3 looks like that. So on row 1 it's going to become 1, 0. Now let's see, when I add negative 4 and 7 thirds, negative 12 thirds plus 7 thirds is negative 5 thirds. And then this is 24 thirds and I'm adding on negative 13 thirds gives me 11 thirds. Okay so what can I get from that? Well let me erase this part. Maybe this will be enough. And let's write down what this augmented matrix says. It says that x minus 5 thirds z is equal to 11 thirds. And the second equation says that y plus 7 thirds z is equal to negative 13 thirds. Now solving for x it looks like x is equal to 11 thirds plus 5 thirds z. And y is equal to negative 13 thirds minus 7 thirds z. So I have x, I have y, the only thing left I need is z and z is equal to z. I mean that's certainly true. So now I have x, y and z defined in terms of z. So on the right hand side z is referred to as a parameter because if you know the value of that parameter you can calculate the value of our three unknowns. Now in some textbooks they'll write this answer as 11 thirds plus 5 thirds z, negative 13 thirds minus 7 thirds z and z. And in other books they'll change the parameter to a different letter, usually t, and they'll write this answer as 11 thirds plus 5 thirds t, negative 13 thirds minus 7 thirds t, and then t. Either form of the answer is perfectly fine with me, but I do want you to isolate all three variables x, y and z and then write this in the form of an ordered triple as your answer. And of course what this means is as you pick different values of z or t if you plug in 0 or 1 or 2 you generate infinitely many solutions for the original system of equations. But rather than doing that I think we ought to move on. So let's go to episode 24 and look at some things you need to know in that next episode. First of all you should be able to perform matrix arithmetic. Now what I mean by that is you should be able to add matrices of the appropriate dimensions. You should be able to subtract them. You should be able to multiply by scalars. Keep in mind that you can only add and subtract matrices if they have exactly the same dimensions. If they don't have the same dimensions then we just say that the answer is undefined. Now you should be able to compute the inverse of an invertible matrix by beginning with the matrix of the form A, I, and reducing it to I, A inverse. I think maybe we better take an example of that because there's a lot involved in those words. Suppose I have this matrix. Matrix A, let's make it a 2 by 2. Matrix A is 3, 4, 2, and 3. 3, 4, 2, 3. And I'd like to find A inverse if it exists. Now you remember not every matrix has an inverse, not even every square matrix has an inverse. If it has an inverse then matrix A is said to be non-singular or matrix A is said to be invertible. Now to find A inverse what I do is I set up not a 2 by 2 matrix but a 2 by 4 matrix. 2 rows, 4 columns. I put matrix A on the left and I put the identity matrix on the right. Now matrix A is 3, 4, 2, 3. And the identity matrix is 1, 0, 0, 1. So then what I do is proceed to convert this into the identity matrix. And when I get the identity matrix here, if I get the identity matrix here, then A inverse will come up on the other side. If there is no A inverse then what will happen is I won't be able to produce the identity matrix on the left-hand side. So that's my goal. And first of all I want to get a 1 right here. Of course one way to do it would be to multiply through by 1 third but then I'd get fractions. Now why don't we take row 1 and add on the negative of row 2? So row 1, add on negative 1 times row 2 and the 3 and the 2 I think would give me a difference of 1. So the bottom row doesn't change but the top row does. So this will be a 1. 4, adding on negative 3 is a 1. 1, adding on a 0 is 1. And negative 1 goes here. Okay, so I have a 1 there. I want to get a 0 in this position. And to do that I think I'll take row 2 and add on negative 2 times row 1. So row 1 hasn't changed and row 2 becomes a 0 and 1 and negative 2 and 3. Well that's good. I have a 1 here so now I can use that to produce a 0 up above it. So let's take row 1 and add on negative 1 times row 2. So this looks a lot like Gauss in Elimination and Gauss-Jordan Elimination but we're computing the inverse of a matrix. Let's see now, I'm subtracting row 2 from row 1 so it'll be a 1 and a 0 and a 3 and a negative 4. A negative 4. Now you notice this is the identity on the left hand side which means this must be the inverse of a on the right hand side. So when we said calculate a inverse, if it exists we now have an answer for a inverse and it is this 2 by 2 matrix. It's 3, negative 4, negative 2 and 3. If I check that to see if it really is the identity matrix well I could multiply those two together and see if it gives me the identity when I'm multiplying. So let's do that right here and that is a inverse, excuse me, a inverse times a is 3, negative 4, negative 2 and 3 multiplied by 3, 4, 2 and 3 and we get 9, take away 8 is 1 and we get 12, take away 12 is 0 and we get negative 6 plus 6 is 0 and we get negative 8 plus 9 is 1. Yes that is the identity matrix and if you multiply in the reverse order I won't work that out but you'll end up getting the identity matrix again that way. So we've calculated the inverse of a matrix using matrix methods. Okay let's look again at episode 24 and see what we're after here. We've just reduced a matrix to get its inverse so the next thing is be able to solve a matrix equation of the form A x equals B using the inverse matrix. Well I think we can use this as an example right here. So coming back to this board we have a matrix A and we've already computed its inverse so what if I wanted to solve this system of equations? 3x plus 4y equals 10 and 2x plus 3y equals 19. Now I don't know what the solution for this is going to be because I just made up two numbers to go on the end here but I want to convert this into a matrix equation of the form A times x equals B. Now this would be the coefficient matrix and this would be the variable matrix and this is the constant matrix. So the coefficient matrix is 3, 4, 2, 3 and the unknown matrix that I've called x here is actually the columnar matrix with my two variables in it, x and y and that should equal my constant matrix on the other side which is 10 and 19. Now you see this multiplication is defined because this is 2 by 2 and this matrix is 2 by 1 and when I multiply those two numbers are the same so I can multiply and the answer is 2 by 1 and this is a 2 by 1 matrix over here and the way that I go about solving this is I multiply both sides by A inverse and you remember for this A I've calculated A inverse already so this is A inverse times A times x equals A inverse B and A inverse times A is the identity matrix and that tells me that the way you get x is you take A inverse times B so the way I solve for x here is I multiply A inverse times B so A inverse times B now A inverse we calculated back here is the matrix 3, negative 4, negative 2 and 3 and I'll multiply it by matrix B matrix B is the constant matrix 10, 19 and that's going to give me 30 take away 76 looks like I came up with some fairly big numbers here 30 take away 76 is negative 46 and then I have negative 20 plus 57 is 37, wow so I came up with some big values for x and y so x should be negative 46 and y should be 37 and if you substitute these numbers in you should get 10 and here you should get 19 but rather than checking that all out I think I'll move on to a different example but once you know the inverse you can solve a system of linear equations which is a unique solution by just taking the inverse times B times B as we've worked out over here okay let's go back to episode 24 and this says you should know a special formula that allows you to write quickly the inverse of an invertible 2 by 2 matrix you know I don't think I ever mentioned this in class so I'm not actually going to expect you to know this but if you want to use it you can and that is for a 2 by 2 matrix let's say for a 2 by 2 matrix of this form A, B, C, D the inverse if the inverse exists it can be calculated by first of all switching A and D so you put D up here and you put A down below and the B and C you don't move them but you change their sign negative B and negative C and then you have to multiply this by a constant and it's 1 over A, D minus B, C now I didn't justify this formula in class because actually the justification is a little bit messy it takes up a little bit of time so let me just say that if you want to use this formula you can on the exam but I won't expect you to know the formula instead if you want to find the inverse of a 2 by 2 matrix you can use matrix methods like we've just seen in the last example to find its inverse okay, let's go to episode 26 no, I'm sorry, 25 and in episode 25 you should be able to evaluate determinant of all sizes using expansion by cofactors along any row or column and then there is a simple formula for evaluating 2 by 2 determinants as well let's take some examples of both of these first of all, if you have a 2 by 2 determinant for a 2 by 2 I'll just fill in some numbers here 5, 3, negative 4 and negative 6 now to evaluate a 2 by 2 determinant you take 5 times negative 6 and then subtract off 3 times negative 4 and that'll give you the value of the determinant that's going to be negative 30 plus 12 and that's 18 now the way we explain this in class is what we're doing is we're taking 5 times its cofactor this is its cofactor and then we're taking 3 times its cofactor which requires a sign change before we add that up but the shortcut is to take this product minus this product it would be negative 18 though oh, negative 18, thank you, yeah, negative right there now for larger determinants let's take an example of a 3 by 3 suppose the first row is 4, 1, 2 and then 3, 0, 5 and then 2 and 7 and negative 4 now in this case what I can do is expand on any row or column and I think I'd want to pick the row or the column that has a 0 in it to save me a little bit of effort so let's expand along the middle row I'll just circle it and you know 3 is in a negative position so I'll just put a negative out in front and then I multiply it by the determinant 1, 2, 7, negative 4 you remember what we do is we remove the row and the column that the 3 is in then I move to the 0 0 is in a positive position but I'm not going to bother writing down the 2 by 2 determinant that goes with 0 because the product would be 0 then I'll move to the 5 5 is in a negative position and the 2 by 2 determinant is 4, 1, 2, 7 okay now I'll evaluate these by just expanding the 2 by 2 determinants that's negative 3 times a negative 4 minus 14 plus 0 minus 5 times 28 minus 2 so this gives me negative 3 times negative 18 minus 5 times 26 well I just omitted the 0 since it's irrelevant now so this gives me 54 minus 130 and that's going to be negative 76 now what would be the answer if I'd expanded along any other row or column what answer should I get? the same thing we ought to get the same thing of course so you can pick any row or column but I think you'd want to choose the one that has the most 0s to save you a little bit of work in calculating that cofactor okay back to episode 25 know the effect of each of the 3 elementary row and column operations on the value of a determinant let me just mention this quickly that if you interchange two rows or if you interchange two columns of a determinant you change the sign of the determinant if it had been negative 76 before it'll be positive 76 afterwards if you multiply a single row or a single column by a constant let's say you multiply by 3 what that does is to triple the value of the determinant and then finally if you take a multiple of one row and add it to another row surprisingly that has no effect on the value of the determinant so those are the effects of the three elementary row operations now Kramer's rule let's come back to the board here and let me just remind you how Kramer's rule works if you have a system of equations I'll make up a short one here 2x plus y is 7 and 5x minus 3y is 12 let's say this could have been a 3 by 3 or a 4 by 4 or a larger system of equations but if I want to solve this to get x and y first of all I calculate determinant d which is just the coefficient determinant 2, 1, 5, negative 3 then I calculate a determinant d sub x where you replace the x column with these constants so that's going to be 7, 12, 1, negative 3 and then for d sub y I go back to the original coefficient determinant and I replace the y column with the constants so the 2 and the 5 return and the 7 and the 12 appear over here now we actually justified this procedure using the effect of elementary row and column operations on determinants but to calculate x you take the ratio of d sub x over d this one over that one and to calculate y you take d sub y over d and if this had been say a 3 by 3 system of equations then what I would do would be to have a d sub x d sub y and d sub z where those individual columns have been replaced and I divide each of those by d to get the value now if it turns out that this problem should have no solution or infinitely many solutions so there's no one solution to generate what will happen is this determinant will be 0 and it won't allow me to divide and so I won't be able to come up with an answer so if d turns out to be 0 it tells you you have infinitely many or no solution for the problem okay, in episode 26 things you need to be able to do you should be able to solve systems of equations that include non-linear equations and you'd probably want to use the substitution or the elimination method now I'm not referring to Gaussian or Gauss-Jordano elimination but I'm referring to the typical substitution and elimination methods you've seen in math 10-10 when I say non-linear equations what I mean is something like this suppose we knew that x squared plus y squared is equal to 4 and that y is equal to x plus 2 I'm just making this up so I have no idea what the solution is but you see here's a linear equation but that's a non-linear equation because there are squares involved so I think probably the substitution method would be the way to proceed on this okay back to episode 26 you should also be able to graph the solution of a set of a system of inequalities into unknowns so you would graph those on the x-y plane you might be graphing circles, parabolas lines, vertical horizontal lines, etc but you should be able to graph systems and shade in the appropriate region and then finally you may have a problem involving linear programming and if you remember linear programming involves graphing a system of inequalities and then maximizing or minimizing the objective function or the subject function by using the vertices of the graphed solution set okay well I think that covers everything that you'll need to know in episode 26 now obviously I can't ask you all these things on this exam there's just too much some of these problems take quite a bit of time to work as you've seen but I'll be selecting from these things and to be prepared you should know how to do all of these hey best of luck to you on this exam and I'll see you afterwards for episode 28