 College Algebra. This is episode 25 and I'm Dennis Allison. You know we've been talking about matrices, how you multiply matrices, find the inverse of a matrix and some of the applications for this sort of thing. Today we look at something that's related to that. Let's go to our list of objectives. Today we're going to introduce these two by two, three by three determinants and larger determinants. We'll also look at how determinants are related to deciding whether a matrix has an inverse, so determinants and inverse matrices. Then we'll look at the three elementary operations that we can perform on determinants. These are the same elementary operations that we performed when we were reducing matrices, although now we'll be applying them to both rows and columns. And finally we'll look at one of the one of the primary applications of determinants, not the only one, but Kramer's rule. And these things are all in your text. So let's look at first of all what is a determinant. Okay let's begin by tying this in with something we've already seen. You know normally for a matrix we use a capital letter to represent the matrix for example a capital A. And if this is a two by two matrix then the entries in a two by two matrix I might call A, B, C and D. You know there's another way of writing the matrix that is you can use parentheses. In your textbook you sometimes see parentheses used A, B, C and D. But what you never see here are vertical bars because that represents something else. In fact that's what, that's the notation that we use for a determinant. Now if I want to take the determinant of A the way I'll abbreviate that is I'll put vertical bars on the A's. Looks sort of like an absolute value but it doesn't mean that. Another way to abbreviate determinant of A is to write D, E, T parentheses A to mean the determinant of A. And then when I go to actually express the two by two arrangement I'll put vertical bars out here and I'll put A, B, C and D. Now the way I evaluate a determinant is I multiply the number in the first row first column times the number in the second row in the second column. A times D minus B times C. And that would be A, D minus B, C. In other words I take this product minus that product and I come up with a real number answer. You know for a matrix you never get a real number answer for these things. One matrix can equal another matrix but a matrix doesn't ever equal a real number. Not if it's like a two by two matrix but for a determinant it has a real number value. Let me just work an example right below here. Suppose I had the determinant 4, negative 2, negative 3 and 5. And I want to find the value of this determinant. Well I take the product 4 times 5 minus the product negative 2 times negative 3. Negative 2 times negative 3. Now that's going to give me 20 minus 6 is 14. So I say 14 is the value of this determinant. You know another representation for a matrix going back up to this first notation that you haven't, I don't know that we've used this much, this notation in class is another way I could represent matrix A is I could say that the entries are, I could use a little a for every entry, a one one, a one two, a two one, a two two. Now this time I'm putting two subscripts below or just below and to the right of each of each a and this represents first row first column, first row second column and then second row second column. So in other words the two numbers that you see in the subscript represent the row number and the column number of that entry. Now that would be equivalent to saying a, b, c, and d, but this way quickly I can identify which position the number came from. So in this case, well let's see for a determinant, if I were going to take the determinant of a, I would write this as a one one times a two two minus a one two times a two one. Okay let's work another example now of a two by two determinant and then we'll look at it from a slightly different point of view. So I have, I have two ways of representing the entries of a matrix and that gives me two ways of expressing the value of this determinant. Okay so another two by two example. Suppose this time I had matrix b and the entries of matrix b this time might be zero and four and negative three and negative one and I'd like to evaluate the determinant of matrix b which I'll write as the determinant zero four negative three negative one. Now once again I'll take this product minus this product so that it'll be zero times negative one minus four times negative 13 and I get zero plus twelve is twelve. Okay so that's said to be the value of this determinant. The matrix had no value but the determinant does. Okay now that's sort of the, sort of the quick way of evaluating a determinant, a two by two determinant. Now let me show you another way, another definition for evaluating it and this definition can be expanded to larger determinants. You know one thing I haven't mentioned yet so far is that when you take the determinant of a matrix the matrix has to be square like a two by two or a three by three. So if I were to give you a two rows, three columns you couldn't take a determinant of that. Okay so going back to the two by two case suppose I want to take the determinant, this time I'll use this notation. I want to take the determinant of a matrix C and let's say the entries of the determinant this time are five, two, three and four. Five, two, three and four. Now I want to look at this from a slightly different point of view. The number in the very first position is said to be in a positive position. It has nothing to do with the fact that five is positive. We just say that the number in the first row, first column is in a positive position. The number in the first row, second column is in a negative position. So positive and negative and the signs alternate going down as well. Five's in a positive position, three's in a negative position and if three's in a negative then this is in a positive position. So I'm thinking that the signs here will be a plus, minus, minus and plus. Now so every position the number, every position within the determinant has a sign associated with it. Now the way I might expand the determinant this time, it's going to give me the same answer I would get in any other way, in the other way I expanded it, is I would take the number five and the number five is in a positive position so I'm going to put a plus in front of that and if I remove the row and the column that the five is in, I'm going to multiply it by the number four. Five times four. Okay now I come to the number two and the number two is in a negative position so I'm going to put a negative in front of it so that'll be a subtraction sign and if I remove the row and the column that the two is in, I'm left with a three, I'm going to multiply it by three. So this gives me five times four minus two times three. Now you know this time the reason I have a minus here is because I say two is in a negative position and this gives me twenty, take away six, twenty, take away six is fourteen. You notice that's the same thing as taking the product five times four minus the product two times three. Now the number four is said to be the minor, M-I-O-R-N, M-I-N-O-R, is said to be the minor of the number five and the number three is said to be the minor of the number two. That is when I remove the row and the column the two is in, I'm left with this minor right here. Now when I include that sign either the plus or minus sign with the minor then this is said to be the cofactor of two and four is said to be the cofactor of five. Let's go to a graphic right here that will summarize some of this information. Okay so you notice here for the two by two case I've indicated that the signs are positive and negative alternating on the first row. They also alternate going down plus and then minus but along the main diagonal they're all pluses along there. And so when I go to expand a two by two determinant this is the formula I think you want to use right here that's multiplying a one one times a two two that would be this product along here minus a one two times a two one. Now before I talk about the three by three case let's work an example. So let's go to the green board. Suppose I have this determinant four one negative two zero three five minus one four minus minus one again. Okay I'd like to expand this. Now this time I'm thinking that the number four on the first row first column is in a positive position and the signs alternate going across minus and the negative two is actually in a positive position even though it's a negative number. And then going down the signs alternate plus minus plus and the signs continue to alternate going across and going down. So if I were to expand this along the top row I would write down four times well let's see now four. Four is in a positive position so I'm going to emphasize that by putting a plus in front of it and now if I remove the row and the column that the four is in I'm left with this two by two determinant right here that's what I'm going to multiply the positive four by three five four negative one. Now when I go to the number one one is in a negative position so I'll put a minus in front of that I'll be subtracting and if I remove the row and the column the one is in I'm left with the two by two determinant zero five negative one negative one. And then finally when I go to the negative two negative two is in a positive position so I'll put a plus in front of that and I'll multiply by let's see the two by two determinant that remains when I remove the row and the column that the negative two is that the negative two lies in that leaves me with zero three negative one four zero three negative one four. Okay so multiplying this out I have four times now I'll go back and use my previous formula for the two by two determinant that's going to be negative three minus twenty minus one times zero minus negative five that'll make it a plus five and then minus two times zero minus negative three that's zero plus three. So what we have here is four times negative twenty negative twenty three minus five minus six so this is going to be minus ninety two minus eleven is minus one hundred and three minus one hundred and three. Okay so for the case of a three by three determinant we take the very first entry a one one times its co-factor minus a one two times this two by two determinant plus a one three times the two by two determinant that remains. Let's go to another example suppose we want to evaluate this determinant let's say we put in one two negative four zero five negative one two three two. Okay so to evaluate this I'm going to expand along the first row and I'm going to take one times its co-factor its co-factor is the determinant five negative one three two five negative one three two. Okay now I go to the two the two is in a negative position so I'll go ahead and just put a negative out in front and that negative is actually part of the co-factor and the co-factor is zero negative one two two zero negative one two two and then I go to the four so now I add on a negative four well instead of adding on a negative four I think I'll just write minus four times its co-factor now there won't be any sign change here because negative four is in a positive position so I'll just write zero five two three zero five two three. Now to continue this I have to evaluate each two by two determinant and I do that by using the shortcut rule that we just saw a moment ago I take this product minus that product one times ten minus negative three ten minus negative three minus two times zero minus negative two zero minus negative two and then minus four times zero minus ten. Well now we just have simple arithmetic here it's going to be one times ten plus three is thirteen minus two times that's going to be two minus four times negative ten so this is thirteen minus four plus forty let's see that's going to be fifty three take away four is forty nine is forty nine is the value for that you know one of the reasons that these these values are referred to as determinants is because this answer is determined once you fill all those numbers in in other words once those numbers have been those nine numbers have been established then this answer has been established even if I haven't computed yet it can't change as a matter of fact if I expand on any other row or any other column I'll get that same answer let's just try doing that see we got forty nine here let's keep that in mind I'm going to put a forty nine up here beside it and let's expand along another row or another column and see if we if we get the same result so I'll just rewrite this determinant one two negative four zero five that's a five there a little hard to read and negative one and two three two suppose we were to expand along the middle row along the middle row let me just circle that one so we see it highlighted now my first entry is zero by the way is zero in a positive or negative position negative it's in a negative position so when I write down this determinant the cofactor determinant two negative four three two I'll put a negative out in front to adjust the sign of course that product's going to be zero anyway so there's no real real effect on that next I go to the five I'll put a plus five and I'm putting a plus because fives in a positive position and it's cofactor if I delete the row in the column will be one negative four two two and finally I come to negative one now let's see negative one is in a negative position so I'll be have I'll be changing the sign of the cofactor what I'll do instead is just change the sign of the negative one make that plus one and it's cofactor will be one two two three okay well without even evaluating this determinant I can see that that product's going to be zero so let's move to the next one plus five times the value of that determinant is going to be two minus negative eight and then the next one is one times three minus four so we have five times ten plus one times negative one just negative one so that'll be fifty minus one is forty nine again yeah we got the same result again I would have gotten that same answer if I'd expanded on the bottom on the bottom row and you know you'll get the same answer if you expand along a column as well as a row let's just try doing a column expansion and verify that for one of the columns so I'm going to take out that mark let's say we expand along the last column over here I'll just use that one so I'll start off by writing down a negative four well I want to change the sign of negative four leave it alone in other words will I be putting a plus or a minus in front leave it along yeah it's a positive position so there's no sign change here and it's cofactor is zero five two three okay then I come to the negative one now the negative one is in a negative position so I'll make that a plus one and it's cofactor is one two two three and then we come to the two let's see now the two is in a is in a positive position and so we'll leave it alone plus two times its cofactor which is one two zero five okay now evaluating the two by twos we have negative four times zero minus ten zero minus ten plus one times three minus four three minus four plus two times five minus zero okay we'll keep our fingers crossed here we think the answer is forty nine let's see if it works out negative four times negative ten is forty and then one times negative one is negative one and then two times five is ten yeah there's a fifty minus one is forty nine once again you notice we get the same answer but we're not adding up exactly the same numbers to get it so as you pick a different row or different column these products and sums individually turn out to be different but the result is the same so that's one of the reasons why these are called the determinants because the answer has been determined once you put the numbers in there you know if you were going to be working this one in the future what would be a shortcut for choosing a row or a column to expand along so that you could get the so that you could come up with the answer what what shortcut would you suggest Stephen finding rows or columns that have zeros in them so yeah you can not even have to determine it because it's going to be zero at least at least some of the products you could leave out because that would make a coefficient zero you know a minute ago when I expanded along the second row here I didn't have to worry about zero times its cofactor because the product's going to be zero anyway so I think the shortest way usually is to pick the row or column that has the most zeros if you see a row or a column with say two zeros by all means go for that one because there's only one product that you'll have to actually carry out operations to make one of the another number is yours that you could yeah yeah as a matter of fact those those three elementary row operations that we used earlier to reduce augmented matrices we're going to be applying those to determinants in just a moment in fact that's that's coming up very quickly here that that's a good suggestion okay we can we can summarize this with the fact that says the value of the determinant can be found by expanding by cofactors along any row or column not merely just along the first column like we did in the in the very beginning okay now let's consider the situation of a matrix rather than a than a determinant and suppose I have a determinant or rather a matrix a and a matrix b and I multiply them together now if I take the determinant of that product that's the same thing as taking the determinant of a times the determinant of b separately and multiplying together in other words this is a number terminated a times the determinant of b gives me the determinant of the product matrix that this is a theorem in algebra that we haven't proven but I'm going to make use of and the way I'll make use of it is as follows suppose I have a matrix that has an inverse which we call a inverse that product is always an identity matrix now if I take the determinant of this product I should get the determinant of the identity matrix and you remember in the identity matrix you have ones along the main diagonal so this is going to look like a one and then a zero and a zero and then a zero and a one and a zero and etc if it were if it were a three by three it may not be a three by three but the idea is I get ones along the main diagonal now if I if I evaluate this determinant by expanding along a row or column I think you'd find out that its value is always equal to one so what we have is that the determinant of the product is always equal to one and if I use that rule that I just mentioned a moment ago the determinant of the product is the product of these two determinants separately and so what this what this tells me is that a that if a matrix has an inverse then the determinant of matrix A cannot be zero because if this had been zero there's no way I could get a one for an answer so what I can conclude from this is if matrix A has an inverse then the determinant of matrix A cannot be cannot be zero because if it were zero this product can never be one as a matter of fact you can actually get more information out of this than what we're going to use what we can actually determine is that the determinant of A inverse is equal to one over the determinant of A although I don't think we'll be using that fact in this course see what I've done is to divide by the determinant of by the by the determinant of A so the determinant of the inverse is the reciprocal of the determinant of the original matrix okay let's go to the first to the next graphic and we have a theorem that summarizes what we've just established here and that says that a square matrix A has an inverse if and only if the determinant of A is is not zero now this is another reason why determinants are named the way they are because it's with the determinant that you can determine whether a matrix has an inverse or not I've come back to the greenboard suppose we just take this suppose we take this matrix I'll think I'll call this one matrix B and let's say matrix B is 5 4 negative 3 and 2 and we ask the question is B an invertible matrix you may remember there was another word we use for this we might also ask the question which is equivalent to it is B a non-singular matrix you remember non-singular minute has an inverse so I'm asking two questions but they really ask this at the asking the same asking not the same idea well to find out if this matrix has an inverse or not I merely have to take its determinant and see if I get zero if I get zero it doesn't have an inverse if I get anything other than zero it does have an inverse so if I take the determinant of 5 4 negative 3 2 let's see that value would be 10 minus negative 12 and I get 22 now what's the most significant fact in regard to 22 at this moment it's not zero 22 isn't zero so as long as it isn't zero the answer is yes B is invertible now that means we might proceed by the methods we saw in the previous episode to calculate its inverse but but what this does is it gives us a way of deciding whether there is an inverse before we go to all the trouble of actually computing it or not okay now I'd like to show you get another way to evaluate determinants let's see what we've what we've seen is we could expand by cofactors if it's a 2 by 2 determinant there's a shortcut because you can just take the diagonal products and subtract them but now let me show you got another way that we can evaluate determinants and this is using the elementary operations that we performed on matrices earlier here's an example suppose we have the determinant 2 1 5 3 can anyone tell me the value of that determinant 1 is 1 yes because it's 6 take away 5 is 1 now look what happens if I invert 2 rows what if I switch the 2 rows and put 5 3 on top and 2 1 down below now what's the value of that determinant negative 1 is negative 1 yeah it's 5 minus 6 is negative 1 as a general rule if you have a determinant even if it's larger a 3 by 3 or larger if you switch 2 rows you will change the sign of the determinant by the way if you switch 2 columns you'll change the sign of the determinant what if I switch columns 1 and 2 I'll put 1 3 and I'll put 2 5 you notice still the product is 5 take away 6 is negative 1 so if you interchange 2 rows or interchange 2 columns you get the negative of the determinant value that you had before ok that's the first of the three elementary operations although now these are row and column operations let me take another example and let's try performing a different operation on it suppose we have 4 negative 1 3 negative 2 what's the value of that determinant at the moment negative 5 yeah it looks like it's negative 8 minus negative 3 negative 8 minus negative 3 is negative 5 ok suppose now that I double the first row if I double the first row only not the second row we have 8 negative 2 3 negative 2 this time the answer is negative 16 minus negative 6 let's write that down negative 16 minus negative 6 how much will that be negative 10 negative 10 ok so what did we do we took the original determinant I double the first row and I got double the answer by the way that would happen also if I double the second row I double the answer what do you think I'd get if I double the first row and the second row four times the answer we get four times the answer because we double it and double it again and this answer would have been negative 20 instead of negative 10 this also works if I double or triple or whatever a column let's say I multiply the first column by 5 that make it 20 and 15 going back the original determinant and but still leave the negative 1 and the negative 2 then this answer will be negative 40 minus negative 15 that's a negative 40 plus 15 is negative 25 and you see this is five times the original answer so if you take a multiple of a row or a column you will get the same multiple of the final answer ok that's the second of the elementary row and column operations ok now to me this is this next one is the most surprising one what if I take a multiple of one row and add it to another row or if I take a multiple of one column and add it to another column let's take an example let's take one four five three at the moment the value of this determinant is three take away 20 is negative 17 now I'm going to take a multiple of row 1 and add it to row 2 suppose I take 2 times row 1 and add that to row 2 let's see the way I usually write that I think is row 2 plus 2 times row 1 here's what the new determinant would look like I'm not changing row 1 but if I take 2 times row 1 and add it to row 2 what would the new entries be on row 2 7 and 11 7 and 11 ok and so we get here the value of this determinant is 11 take away 28 is negative 17 what's the effect on the determinant when you perform this operation doesn't doesn't look like it affects it at all and as a matter of fact that's the general rule if you take a multiple of one row and add it to another row or you take a multiple of one column and add it to another column you get the same answer now to me that's surprising because that seems like the most complicated of the three operations and yet it has no effect this is summarized in the next graphic on elementary operations you'll notice the title of this graphic is not elementary row operations but just elementary operations because we can perform these on rows or columns number one if we interchange two rows or columns in a determinant the value of the determinant changes signs yeah we saw that we went from a negative number to a positive number vice versa in our example number two if we multiply a row or column by a constant the value of the determinant is also multiplied by the constant and then number three if we add a multiple of one row or column to another row or correspondingly a column the value of the determinant remains the same now here's how I can use this to evaluate determinants using an alternative method let's come back to the green board and take this example suppose I have the determinant one one two zero three negative one two negative one four okay so I'm just picking these numbers at random and I'm thinking you know what if I were able to reduce this to a determinant that was in this form suppose I got zeros below the numbers along the main diagonal and above on and above the main diagonal it doesn't really matter but what if I got zeros below then if I were going to evaluate this determinant what I would do is I would take this entry right here and multiply it by its co-factor by the way this entry is in a positive position so its co-factor is just these four numbers so it'd be this entry times the value of this determinant up here I'll just put a box around it now the value of this determinant is just this product minus zero so it would be this number times this product minus zero or to sum it all up it's just the product of the numbers along the main diagonal so if I can make this determinant into this form where I have zeros below the main diagonal then the value of the determinant is just the product along the main diagonal it's this number times that determinant and that determinant is just this product right here okay so how can I make this determinant look like that well if I perform elementary row and column operations and make the appropriate the appropriate adjustments along the way using the facts that we've just seen I think we can get there so let's say I already have a zero here I'd like to get a zero there and a zero there how can I get a zero in this position well let's see I can take row three and add on negative two times row one row three plus negative two times row one okay so this is exactly what we saw with augmented matrices earlier but now we're doing this to determine it's one one two zero three negative one now here's where the change comes in this is going to be a zero this is going to be a negative three and this is going to be a zero right there okay I've got a zero a zero I'd like you to zero right here how can how can I do that add row two to row three add row two to row three okay so if I add row two to row three by the way you know when you take a multiple of one row and add it to another row you're not changing the value of the determinant so these these determinants are exactly equal one one two zero three negative one now if I add row row two to row three then I'll get zero zero negative one zero zero negative one okay so at this point I'll just take the product along the main diagonal and that's going to be one times three times negative one which is three times negative three excuse me three times negative one or negative three of the value of the determinant sometimes this is the quickest way to evaluate a determinant quicker than using cofactors to to come up with these zeros in the right positions let's take one more example where we do this and this time I want to take an example where I have where I can factor a number out or make some other change okay in this three by three determinant suppose we have zero two one five five negative ten and negative two one and four so if I can get ones along the main dot or just anything along the main diagonal but zeros below the main diagonal then I can multiply along the main diagonal and get the answer by the way you can also do this where you get zeros above the main diagonal and the answer will be the product on the main diagonal again well let's see I'd like to get a one where the zero is can anyone suggest a way to get a one there okay let me let me re let me rephrase the question give me several different ways you could get a one where the zero is what's what's one way we could do that you can change row two and row interchange row two and row one and then divide the new row one by five okay yeah so what Jeff has noticed is we have a common factor of five there so if we could get rid of the five make it a one one negative two if we interchange rows one and two I think we're on our way to getting a one in the one one position okay what's another way I could get a one there interchange column one and column three okay interchange columns one and three yeah then that would certainly put a one there okay we have to remember though of course when we interchange two rows or two columns we change the sign of the determinant so we have to account for that and David I think you mentioned one other way or no okay someone I thought mentioned that if you took negative one half of row three and added it to row one then you'd get a one right there okay so those would all work let's try interchanging two columns because we have the one already okay so I'm going to put column number three in the first position and I'll leave column two alone and of course we have to change column three but you know these two determinants are no longer equal because when you switch two columns you have the opposite sign so I'm going to put a negative in front which is going to correct the sign and bring it back to bring these back to being equal okay by the way what's the purpose of getting the one what why is my interest in getting a one there because it's easier to multiply by one well see the one is a lever and I can use the one to get zeros below whereas if this had been a three for example it would be kind of awkward to use a three to get a zero with the negative ten but you know let's do something differently here also instead of getting zeros below the diagonal let's get zeros above the diagonal one for one reason I already have a zero up there so that means I only have two positions to worry about I'd like to get a zero in this position and to get a zero there why don't we multiply column two column one by negative two and added to column two in other words this is going to be c2 plus negative two times c1 we've never written it that way before because we've never done a column operation quite like that before by the way when I take a multiple of column one and added to column two will it change the sign of the determinant no that has no effect but I don't want to forget to this negative I have to keep this negative up with me all along so I have to keep recording that negative from the column change okay column number one stays the same column number two becomes a zero twenty five twenty five and negative seven negative seven I need a little bit more room I think on that one but let me switch markers here and column number three is okay zero five and negative two now you notice in this second row I have a common factor so what I'd like to do is factor out that five and call this a negative five out in front so what I'm doing is I'm factoring out a common factor on that row this is using the second elementary operation and this becomes one zero zero negative two five one four negative seven negative two so using the second elementary operation row operation I'm now factoring out a common factor in bringing out to make these numbers smaller okay now I'd like to get a I'd like to get a zero where that one is what's one way I could get a zero there you could switch row two and row three yeah I could put a five there and a one in the middle and it'd be easier to let the five the one make the five into a zero than making the five convert the one into the zero so why don't we convert these two rows switch those two rows which is going to put yet another negative out in front so now we have a plus five and the first column remains the same but the third column moves into the second position and the second column goes into the third position and Stephen's idea was that you're getting a one in the middle I can use the one to create a zero there so let's now do that I'm going to take a negative five times column two and add it to column three so let's see that's column three plus negative five times column two I'll just write that right above column two so I have one negative two four oh I left off my five that's it's very easy to drop off those coefficients if you're not paying attention there and I almost did the same thing the second column becomes zero one negative two and the third column becomes a zero zero and what's the last entry going to be three it's going to be three yeah okay I have now made this into a matrix that has zeros above the main diagonal so if I take five times the product of the main diagonal that'll be five times one times one times three this answer is 15 now you know you might say well Dennis I don't know if this is really a very practical way to evaluate a three by three determinant this one turned out to be kind of long it could be that it would have taken some other steps it could have been a little bit shorter but actually where I want to make use of this sort of a procedure is in the next problem when we talk about Kramer's rule so let's move to to a whole new problem but I'm still going to use this idea or these elementary operations anyway to consider Kramer's rule now first of all let me tell you what the rule says and then afterwards we'll go back and establish what it's about suppose I have a system of equations of the form a x plus b y equals a constant we'll say the constant is r and then another equation c x plus d y equals a constant s this is a two by two system of linear equations now let's assume that this system for the moment has a solution though the word we use for that is we say suppose it's consistent meaning that there is a solution for this problem in exactly one solution then I can find the solution for x and for y by computing three determinants here's what I do first of all I compute a determinant that I'll just call capital D and it's the coefficient determinant that is it's the determinant made up of the coefficients a b c and d so I have a b c and d that's easy to evaluate it would be the product a d minus the product b c now there's another determinant I'm going to make up here I'm going to call d sub x and what the difference between this one and d is that I'm going to replace the numbers in the x column with the constants over here r and s I'm going to put those in place of a and c r s b and d and that would be of course the product r d minus the product b s and then there's a determinant d sub y it's a two by two determinant and if I go back to the original determinant I'm going to replace the y column with r and s that's why I call it d sub y so this will be a and c in the first column r and s in the second column now you might say well what's the what's the purpose of these determinants well I can calculate x by taking the ratio of d sub x over d and I can calculate y by taking the ratio of d sub y over d and this idea is referred to as Kramer's rule is Kramer's rule so if you set up your determinants d d sub x d sub y and then take these ratios you will get the values for x and y now the problem has to have a solution it has to be consistent if if this problem has no solution or if it has infinitely many solutions what happens is the determinant d turns out to be zero and you can't divide by zero so Kramer's rule will keep you from coming up with a particular solution because there is no particular solution in that case there there might be no solution or infinitely many so we say we want it to be consistent so that this determinant will not be will not be zero right here okay and by the way this works for larger systems of equations to if this had been a three by three system of equations I would have had three by three determinants and there would be an extra one called d sub z down here I think but let's work an example using a two by two and then as time permits at the end we'll go back and see why Kramer's rule works okay suppose we have the system of equations 3x minus y equals 12 and 2x plus 3y equals 19 and of course we want to solve for x and for y okay and the way I get x is I take d sub x and I divide by d so I'll need to find both d and d sub x and the way I get y is to take d sub y and divide by d okay so I'll need to figure out what is determinant d and then afterwards we'll get d sub x and d sub y so determinant d is that 3 negative one two and three it's the coefficient determinant and this gives me 9 plus two is 11 d sub x let's see can you tell me what would be the entries in d sub x negative one three 12 and 19 no let's see not for d sub x 12 yeah negative one 19 and 3 exactly yeah see what we're doing here Jeff is we're putting these constants in the first column where the x's were so the 12 and the 19 go down here and then we leave the y column alone let's go ahead and evaluate this one this is going to be 36 and then minus the negative 19 that's plus 19 and 36 and 19 is 55 55 okay now d sub y okay Jeff we'll give you another chance here what's going to be the determinant d sub y 12 here no doesn't it go okay negative let's see what we're going to do is we're going to put the 12 and the 19 in the y column okay and I'll leave the 3 and 2 3 12 3 and 12 2 and 19 and then 2 and 19 okay exactly okay so we've we've replaced the constants in the y column whereas for d sub x we put them in the x column so this gives me 3 times 19 what's 3 times 19 anyone 57 57 right minus okay here's another one what's 2 times 12 24 24 of course yeah okay so we get 33 okay now if I substitute these numbers in over here d sub x is 55 over 11 or 5 and y is 33 over 11 or 3 so our answer already is an ordered pair is 5 3 that's the solution to our problem let's just go back and check those answers up here I'll just put a 5 above x and a 3 above y substituting those and let's see 15 take away 3 is 12 and 10 plus 9 is 19 yeah that that is a solution in fact that's the only solution okay now I tell you what let me just talk about a 3 by 3 but I'm not going to work out all the determinants because that'll that'll take more time than we have but suppose I had a 3 by 3 system that look like this suppose we had x plus y plus z equals 4 and suppose we had x minus y plus 2z is suppose that's also 4 and suppose we had 2 y minus z equals 0 okay so we have a coefficient of x there that's 0 well this time I'd have to set up a determinant d a determinant d sub x a determinant d sub y and a determinant d sub z let's see now determinant d is just the coefficient determinant 1 1 1 1 negative 1 2 0 2 negative 1 and I'd evaluate that but I don't think I'll do that at the at this moment because it would take more time to do all four of these than we have now for d sub x I would take out the x column and I would put 4 4 0 in that column so this is the column that's going to change right here so I'll just put the 4 4 and 0 there the other columns are left intact okay now for d sub y I'm going to change the middle column so I'll put the 4 4 0 there and the first column stays the same and the last column stays the same and finally for d sub z will insert 4 4 0 in the z column and then this is still 1 1 0 that hasn't changed and 1 negative 1 2 okay then to get my answers to get x I would take d sub x and divide it by d and to get y I'd take d sub y and divide it by d and to get z of course I'll take d sub z and divide it by did by z but by d rather and that would reduce to give me my final answer okay now you know I think that's why don't we go to the next graphic and we can we can see that the two by two case outlined that was the that was the model that I had written on the board a little bit earlier for the two by two for a consistent linear system x is equal to d sub x over d and those are the determinants that should evaluate and y is equal to d sub y over d okay so let's spend our last few minutes here just looking at a justification for this you know we have if we have the system of equations a x plus b y equals r and c x plus d y equals s then the coefficient determinant what we call d is just a b c and d now suppose I were to multiply this determinant by x okay so I've x times this determined this coefficient determinant now you know when you multiply by x then you can multiply any row or any column by x so I'm going to multiply the first column by x because that's where you'd expect to see the x as I suppose so this is a x c x b and d because for determinants when you multiply by a constant you can multiply any row or any column by the constant okay now I'm going to take a multiple of column two and add it to column one so the multiple I'm going to take is y times column two and add it to column one so if I take y times column two and add it to column one this will be a x plus b y and this will be c x plus d y but then we haven't really changed column two so that remains b and c so all of this is column one and then this little bit is column two okay now let's see we have a x plus b y given my original system a x plus b y was equal to r so let's replace this with r and c x plus d y that's really s so let's just put an s there so what we have here is r b s d and I think I'll just bring this down x times this determinant a b c d is equal to the determinant over here I've in other words I've actually begun with this and I it's evolved into this other expression now you know what what this is is actually the determinant we call d we have x times determinant d and over here this is what we were calling determinant d sub x because you see I have the constants in the x column so x is equal to d sub x divided by d that's the justification for the rule Kramer's rule for x you know if I wanted to derive the rule for y equals d sub y over d how do you think this would this would have begun differently multiply by a y yeah and what do you think the next step would be how would I bring the y into the problem times it by the second column right instead of multiplying y in the first column multiply y into the second column and this remains a n c this becomes b y and d y then I'd take a multiple of a n c and I'd add it to the second column so I would begin to build a more complex expression in the second column rather than the first column then I would replace each of those expressions with r and s right here instead of here and we would have y times determinant d and this would be determinant d sub y now the only the only problem a person might only question a person might bring up I suppose at this point is Dennis can you divide by d I mean what if you're divided by zero well I don't think we've justified this but it wouldn't be that difficult to do so is that if this system of equations is consistent the coefficient determinant will not be zero but that's never been established but that is the case so if this is consistent this determinant will be zero I could divide by it and that's how we isolate the x and that's how we isolate the y for for Kramer's rule well you know if we sort of recap what we've done today we started off introducing determinants for the simplest case two by two and we said that you could just take the cross product the two diagonals and subtract those products to evaluate that determinant for three by three determinants and larger you expand by co-factors and in that case you pick any row or any column and you expand along it and you take each entry times the corresponding co-factor which is a determinant with the appropriate sign placed in front of it either a plus or a minus you know we didn't do a four by four or a five by five determinant but they're done in exactly the same way you expand along a row or column and let's say if it's a four by four the co-factors will be three by threes and then each of those co-factors have to be evaluated separately as three by three determinants then we saw that there was there was an alternative I don't know if you'd call it a shortcut and that's where you make the determinant into a diagonal and multiply along the main diagonal I'll see you next time in episode 26