 Good morning, everyone. So this is going to be your fourth session on matrices and determinants. This would be the last session In this session, we are going to primarily deal with Product of determinants. We are also going to talk about differentiation and to a certain extent integration of determinants and then we are going to talk about the Kramer's rule That's the method to solve system of linear equations by determinant method apart from that When the time permits, we'll be taking up a little bit of concepts on functions All right So let's get started with some problem solving because last class we did not get enough chance to solve questions So I would like to begin today's session with some questions Let's take up this question Meanwhile, many people have not joined in they will be joining in It's a show that question show that if x1 x2 x3 are non zero then this determinant is given by The right-hand side. Okay, I guess everybody has joined in All right, so let me start discussion for the same question See the first thing that I will do is I will write this determinant as sum of two determinants Okay, so we'll take the left-hand side determinant left-hand side determinant X1 plus a1 b1 and we'll write this as We'll break this up as a1 b1 a1 b1 Let's do one thing. Let's keep the second and the third column as it is Let's break with respect to the first column Okay, so I'm doing this splitting with respect to the first column So let me let me like write it like this So x1 a1 b2 a1 b3 zero x2 plus a2 b2 a2 b3 again zero a3 b2 And x3 plus a3 b3. So as you can see I am trying to break it with respect to column number one So the other component will be a1 b1 a2 b1 a3 b1 These two columns would remain as they are no changes in these two columns So the first determinant you can expand with respect to column number one So the first determinant we can expand with respect to column number one because two zeros have been created over here Right So let us try to expand the first determinant Along column number one So this will become x1 times x2 plus a2 b2 x3 plus a3 b3 minus a3 b2 a2 b3 Let's open this up further This will become x2 x3 You will also have x2 a3 b3 and you'll also have x3 a2 b2 I think the other term would get cancelled off The other term would get cancelled off. So there's no need to write it So ultimately you will end up getting x1 x2 x3 x1 x3 a2 b2 and x1 x2 a3 b3 Okay, now what about the second determinant? What's about what what can you do with the second determinant? First thing can I do can I take b1 uh common from the first column? Let's take the first column So let's let's do do the change here itself so that we don't have to write it again b1 b1 b1 Now as you can see there is an a1 present over here and there's an a1 b2 and there's an a1 b3 Can we do anything to get it cancelled? Can we do this operation? Can we do c2 as c2 minus b2 c1 And in the same way can I do c3 as c3 minus b3 c1 if I do that operation This determinant only I'm writing it it becomes b1 a1 a2 a3 You'll be left with a zero here zero here This will be x2 This will be zero again. This will be zero again and this will be x3 Okay, I think this is uh pretty easy for us to expand we can expand with respect to uh, let's say the first row itself So this will become a1 b1 x2 x3 a1 b1 x2 x3, okay now all together if you see All together if you see if you combine If you combine this expression And this expression Your expression will be x1 x2 x3 It will be x1 x3 a2 b2 It will be x1 x2 a3 b3 And it will be x2 x3 a1 b1 So you can take now x1 x2 x3 common and that will give you one plus a1 b1 by x1 as you can see this term It'll give you a1 b1 by x1 This term will give you a2 b2 by x2 And the final this term will give you a3 b3 by x3 That's your RHS. That's what you wanted to prove hence proved. Okay No problem Richard. Do you want me to explain this again? Voice is breaking like anything So watch the recording afterwards. Oh, no problem. Okay. Is this fine? Let's take up the next question Let's take up the next question Yeah, you can take this one If they have given you an identity the best way to solve this is put some dummy values for a b and c This should be easy. I think you should be able to get this easily That's correct. But um, uh, because I think some sign mistake has happened. Correct. Myth Okay, so I'll do it in two ways. One is by use of a rigorous method. So let's take up a rigorous method first The first operation that I would do here is uh, first I will change r2 By using this operation r2 minus r3 um When I do that operation I'll end up getting a 4a over here. Okay So take 4 out common Take a 4 out common Okay uh, next operation that I would do is uh, I will do r3 as r3 minus r1 Plus two r2 Okay, so minus r1 because I want to get rid of a square term Plus two r2 because I want to get rid of a term. So that will leave you only with one one one, right? So this will become a square b square c square a b c one one one Now if you just swap the position of these two You will end up getting four one one one a b c a square b square c square but with a minus sign over here Okay, what is i here? Cement k K no no k is not a 8. It's minus 4 Okay, now this is This is a well-known determinant for us It's one of the determinants which I asked you to keep in your mind. So it's a minus b b minus c c minus a If you compare it with the given expression this term is going to be your k Okay, that's one way to do it Another way to do it is uh, take a as zero Take a as zero Okay, take b as one Okay, and take c as minus one Okay, uh, make sure you don't take any one of them to be equal else What will happen zero will become equal to zero and k will be lost somewhere Right, so you're not able to get the value of k So when you do that a is zero and b is one so it'll become uh zero one one This is one. Uh, this is four This is again a zero This is one This is zero and this is four This you're comparing with with k times a minus b a minus b is minus one b minus c b minus c is two And c minus a c minus a is again a minus one Okay, so you can expand this if you want, uh, let's expand with respect to, uh The last column plus minus plus So it's a zero minus four Plus four Minus one So you'll get two k is equal to this that means minus eight is equal to two k. So k is equal to minus six Minus four, sorry Simpler way to do it. Is that fine any questions here? Oh, you did a sign mistake. No worries Let's take another one. I hope you can read the question here slightly darkened. It has become on this side, but I think so if summation alpha n From n equal to one till n is a n square plus b n a n b are constants And alpha one alpha two alpha three belong to okay And 25 alpha one 37 alpha two 49 alpha three b three date numbers then prove that this is equal to zero So it's five seven nine red five seven nine So in the determinant like the marker is on if you know that there is a summation which is giving you a quadratic in n without a constant term I hope you remember we had a discussion on this in the series sequence progression chapter last year This can only happen if your Sequence alpha n is an ap Only in an ap you would see that you'll end up getting a quadratic and that too without a constant term if you just try to check it with the Expression itself you would realize that this expression would always be a Quadratic without any constant term. I mean, there will not be a c kind of a term They'll always be n square and n containing term correct Done ruchir. Very good Okay Now look at this expression Look at this expression First of all, I can comment alpha one alpha two alpha three They're they're claiming it to be single digit numbers Okay, very good Can I also say that alpha one alpha two and alpha three will be in ap Okay, since the entire series in the entire sequence is in ap Alpha one alpha two alpha three will also be in ap How to use this information that these are three digit numbers Because I can see five five seven seven nine nine And alpha three basically what they are trying to hint as hint at if you do this operation r3 as r3 minus 10 times r2 minus r1 See Basically this this number would be like I just take a guess. Let's say it is 252. I'm just taking an example Okay, because alpha one is a single dated number If you do this operation that means if you're subtracting 50 minus 2 it will leave you with 200 Isn't it So when I do this operation The entire last row The entire last row will become 200 This will become 300 And this will become a 400 correct Yes or no Now when such a thing arises, uh, it's a very big sigh of relief because you can do It's that I didn't get this. Why does 49 83 have to be bigger than 400? Sorry, sorry So why does 49 alpha 3 have to be more than 400? It's not a product. It's not a product. These are numbers Oh, yeah, it is just a proof of alpha one there. It is not like 25 into alpha. It is not like that I thought it is okay, but it is not product. Yeah So first of all, let's take 100 common from the Last column. I think it won't make much of a difference to us But just for this thing and then we can do this operation C2 as C2 minus C1 plus C3 by 2 So when I do that, remember since they are in ap alpha 1 plus alpha 3 Is equal to 2 alpha 2 Okay, so you are subtracting you are doing this operation in the middle row You're doing alpha 2 minus alpha 1 Alpha 1 plus alpha 3 by 2 that will give you a zero, isn't it? So alpha 1 alpha 3, there'll be a zero similarly 7 minus 14 by 2 will also be zero. This will also be zero So I think the entire column number 2 has vanished That means this determinant even though multiplied with 100 it will give you zero into 100 at C Is the idea clear? Any questions anybody here? Only that the way they represented it got confusing No, but they have written no b3 digit. Oh, okay. Yeah, some of you thought that even after making a product It is a three-digit number. Yeah I agree with you. They could have written Okay, next we are going to talk about multiplication of determinants product of determinants Okay, one question normally comes in several of the exams Let me ask you this question. Find the largest value find the Largest value Of an order three determinant Of an order three determinant Whose elements are only zero and one whose elements Can only be zero and one I've seen this question coming in many competitive exams I would like you to try this out Yes, mehul. Yes, correct correct here correct ruchir correct chaitanya correct adhrik, okay So most of you are giving the right answers to this correct Dhruv Dhruv, how are you now? How are you feeling now? Things under control Okay, don't worry. You are Yeah, you are strong enough Okay, so if you just take correct Varun If you take basically any let's say I take any determinant in general, which is of order three. Let's say I take this So if I take this determinant and if I expand it I'm sure you would have done the saris method also. Let me expand this with respect to saris method So a1, uh, okay, let's write here itself without wasting much time a1 a2 a3 b1 b2 b3 Okay, so if you use the saris method What you will have you will have on the right side r is equal to a1 b2 c3 plus a3 b1 c2 plus a2 b3 c1 And the left side would be Left side would be these terms Let me choose green color So this will be a3 b2 c1 Plus a1 b3 c2 Plus a2 b1 c3 And your determinant is actually found by doing this operation r minus it r minus l Okay Now if you want this guy to have a maximum value, can I say All those ones in the r Should be one am I correct? Yes, sir Correct And all those In the l we should try to have it as a zero so if you If you Take a1 b2 c3 and a3 b1 c2 and a2 b3 c1 all as one That means each one of them has to be one Yes So the elements in these positions And What about these they should all be zero, right? One of those one of any of those elements would be zero Sorry So like for each of them to be zero either one or two or all three should be zero Correct So what we can instead of you is just make one of these three guys zero Make a2 b3 c1 zero by making all three zero then all of your left hands become zero. I didn't get you So we can sacrifice one of the right hand guys to make all the left zero Yeah, that's what I'm trying to see. So if I make all of them as one Then I'll have such kind of scenario that a1 I'll just try to write it down Let's say a1 b1 c1 a2 b2 c2, I think everything will be getting covered Yeah Right and yes, this will give me a zero determinant So what I have to do is I have to sacrifice one of them. Let's say if I sacrifice a1 and make it zero Okay, let's say make this as zero Okay So this will become a zero Okay, and this will become a zero And let's say I make b2 as zero So b2 is zero is not coming up over here. So this will become a zero Okay And let's say I make c3 as zero c3 is also not coming here. So this will become one one zero That means the largest possible value of the determinant that you can obtain from this is actually let's expand this So it's minus one Minus one plus one One so that means two is the largest value you can get Okay Is this fine? So this is a question that normally comes in many competitive exams So this is how we have to deal with it Next concept that we are going to talk about Is product of determinants Is product of determinants or you can say multiplication of determinants of the same order Now the first question that arises. Why do we need to learn product of determinants? Aren't these two simple numbers which we can multiply anyhow? So why do we need to learn a rule for finding the product of determinants? Uh, there are certain questions that you would get that would be solved only when you try to break up a determinant as product of two determinants It's not it's not that this process is important It's actually this process which is important Breaking a determinant. Let's say breaking a determinant d3 as product of d1 and d2 So sometimes breaking a complicated determinant Sometimes breaking a complicated determinant into simpler determinants Which can be easily be evaluated That is more desired rather than learning how to multiply two determinants to get a third one So why we are learning this concept is primarily because of this Now, how do we multiply determinant? I hope you remember how do we multiply matrices? How do we multiply matrices? Let's say I talk about two matrices which are of the same order. Let's say two square matrices How do we multiply it? We normally take the row of this And multiply it with the column of this. Okay. This is the symbol that we follow So it's basically row into column multiplication that we do Okay, so this is true for a matrices. This is true for a matrices But for a determinant, there is no such rule like this You can do row with row. You can do row with column as you do it in matrices You can do column with row. You can do column with column End result is going to be the same anyhow Let me take an example to illustrate this Let's say I take one two three four Okay, I want to multiply this determinant. Let's say with two one um zero Uh, let's say three Okay now We all know that the answer coming from here should be minus two into six. Isn't it answer should be minus 12 Now I'll show you that I'll obtain a determinant Whose value will be minus 12 irrespective of whether I do row with row row with column column with row column with column Now see let's say I do this operation row with row So let me take a row with row operation So when you do row with row operation you take the first row And you multiply it with the first row of the second determinant So one will multiply to two two will multiply to one and that would come in the First row first A column position Okay, then you take the first row Multiply it with the second row and that you can write either in this position or in this position. That's your call Okay, so let's say I I write it in this position Here So this will become a zero Plus six which is six Similarly, this row multiplies with this will give you a 10 Correct me if I'm wrong and this so multiply this will give you a 12 Okay, if you evaluate this you'll get 48 minus 60. That's nothing but a minus 12 That's what was expected if you do row with column The result would be no different. Let's see Row with column means the same way as you do the Matrix multiplication. So this with this so that will give you two Okay, this with this will give you a seven. I believe This with this will give you a six This with this will give you a 15 So this is going to be 30 minus 42 again a minus 12 Okay, no change in the answer whatsoever If you do column with row That also I'll show you So column with row means take the column and multiply with the row Okay, so this column multiply with first row will give you a five if I'm not wrong Okay this with five and This with this will give you a nine that you can write here also. No worries This with this will give you eight correct And this with this will give you a 12 Am I right? So this will give you 60 minus 72 which is again a minus 12 Okay, if you do column with column also nothing will change So this column with the first column will give you a two This column with this column will give you a 10 I believe This column with this column will give you a four And this column with this column will give you a 14 So it's 28 minus 40 again a minus 12 Okay So in in in a determinant multiplication You can follow any of these four rules without affecting your result Okay Now where is it used? How is it used? We'll come to know when we take few questions here when we take few questions over here Let us say Let us say somebody gives you a question like this There's a determinant There's a determinant given by a1 alpha 1 b1 beta 1 a1 alpha 2 b2 beta 2 Think this should be b1 beta 2. I think I think because it's not following the trend And then it's a1 alpha 3 b1 beta 3 This will be a2 alpha 1 b2 beta 1 Let's say a2 alpha 2 b2 beta 2 a2 alpha 3 b2 beta 3 And finally you have a3 alpha 1 b3 beta 1 a3 alpha 2 b3 beta 2 a3 alpha 3 b3 beta 3 My question is Prove that this determinant is zero Prove that this determinant is zero, okay Now let me solve this question for you because then you'll get an idea about how to approach in these kind of problems Okay, but I was already done. Okay, but I'm good So what I will do is I will in in massage this determinant as product of two determinants So I will try to in massage this as a product of two determinants and let me tell you this is not easy Um, if at all a student feels problem in determinants, it is this concept where he feels the determinant he he tries to Try all possible ways and then later he realizes that Oh, nothing is working. So probably I should try giving it Try breaking it as a product of two determinants. So it doesn't come very easy Okay, this topic of product of determinants is not easy. Okay Now if you look at these trends, you'll always see a1 b1 a1 b1 a1 b1. So what is happening? Let me take a1 b1 over here Okay Now this guy Is multiplying with some determinant to produce this. So if alpha 1 beta 1 has to come there has to be alpha 1 beta 1 Okay Now what about this element And this element Okay, as of now, I'm not sure about any one of them, but at least one of them has to be zero because Because let's I take this as zero Because when you multiply when you're multiplying only a1 alpha 1 b1 beta 1, let's say I'm following row with row multiplication I'm following row with row multiplication So only a1 alpha 1 plus b1 beta 1 is appearing that means the other one is getting Zero because of one of them being zero or at least one of them being zero Okay, so as of now, I'll take one of them as zero. I may have to come back and change it See, I'm ready to do that because it's just trying to work it in a reverse direction. I'm doing reverse engineering over here This may be something I don't know but one of them has to be zero. So I've taken a leap of faith and I've called this as zero Again, I'm repeating it. I may have to come back and change it Okay Next I have a1 b1 And alpha 2 beta 2 Multiplied to it. That means this guy has to be alpha 2 beta 2 for sure Because then only a1 alpha 2 plus b1 beta 2 will appear and this could be anything Okay, this could be anything Okay, similarly for this it will be alpha 3 beta 3 and this could be anything Okay What about this element this element will only come when there's an a2 b2 and there is a Zero now any one of them could be zero actually, but let's say this is zero for us Because since we have already taken a zero over here, no need to waste time taking a zero here Similarly a2 alpha 2 b2 beta 2 Zero will come that is this element Similarly, this element will come from a2 a alpha 3 b2 beta 3 and zero again Similarly, this will be a3 b3 and zero because then only I'll get a3 alpha 1 and b3 beta 1 and same for this Now I'm not saying these question marks Yeah I'm not saying these question marks are all the same. They may be different also They may be zero also. I mean, you don't know But I don't care about it because one of the columns here is completely zero Right if it is completely zero that means this entire determinant will collapse And whatever is the other determinant Whatever is the other determinant it will give you a zero itself. That means your result is zero But here what if like One of this one of the zeros I can replace with the two question marks. Okay. That's that's one of the question that arises Let's check whether that can happen Okay, if let's say Okay, now try try multiplying this guy with this guy That means focus on this term There should be an answer coming from this and this Correct, so there should be an extra term coming from here there But there is no extra term Can't put it anywhere else. Yes. So that violates the fact that There cannot be a switching off one of them has to be having all the columns as zero I'm just replacing it back else Is that fine Let's take more questions. I think as we solve more and more questions the concept would be uh cleared I'm sorry. This this slightly looks like deviated This is your first column This is your first column. This is your second column and this is your third column I think there was no space. That's why they wrote it below But I hope you can read this out, right? Prove that this determinant is a zero Okay, Richard very good Both the riches Okay, let's let's try to predict. Let's try to predict which two determinants Which two determinants would give you this determinant? So let's try to predict that Okay So see you have a two here Okay, and you have alpha plus beta plus gamma plus delta and you have alpha beta gamma delta Okay So let's try to predict. See I'm I can just take a leap of faith. I may be wrong Okay, let me tell you I there may be errors in it. So I may have to come back and revisit my my working also. Okay So, uh Okay, let's take this as one one Okay Let's take this also as one one Now since there is no other that means let's say this is zero and this may be something I don't know. What is it? So at least I can put a two over here Okay Next is alpha beta gamma delta. I cannot have alpha beta gamma delta because it's a three by three determinant. That means Alpha plus beta gamma Yeah, and I can see these kind of pairing ups are also happening Okay, there's a pairing up of these terms happening. So let's say I take this as alpha plus beta And I take this as gamma plus delta Gamma plus delta Okay, and this may be anything I don't care because the zero here will take care and hence I have Got this term also For this term. Similarly, I can say alpha beta gamma delta and I don't care whatever is here Okay Now let's choose the element of the second row for the first determinant Remember second row multiple. This is also taken care second row first row here will give you this element Okay, so can I let's say let's say I take alpha plus beta gamma plus delta Okay, and this will be zero We will check if things go wrong. We'll check Ah, yes Ruchi is absolutely correct because when you multiply this with the second row I should get alpha plus beta whole square gamma plus delta whole square, but this is not the element which is sitting over here So yes, you are absolutely correct. It should be the other way round. See this is what I was saying You have you may have to revisit it It's not that you have put and it'll be Some writing in stone Same thing here will be gamma delta alpha beta And this With this will give you this this will give you this and zero. So that's correct Okay So this element is also taken care of no worries with this Now here, let's say I take alpha beta and gamma delta. Then yes That will create a problem. Yeah so let's take Gamma delta and alpha beta and a zero So this with this This with this and zero with this will give you this element This with this this with this and this with a zero will give you this element And this will give you to alpha beta gamma delta This will Yeah, so this is the right way to do it and since one of them is completely having uh, the third column as zero So it's a zero column. So everything will become a zero here zero into anything will give you a zero Got the point Any questions? Next Show that this determinant is always non negative Where is this determinant zero? So like So So like this method of like splitting the determinant can we like we can also use it to like normally solve determinants, right? see If if it's like integration by parts If you can normally solve it do it right If you if you think You know, it cannot be solved by any of the method And then you figure out, okay It's not like if you if you're not able to solve it by any method your product will Of determinants will work definitely no not like that It may work. So you will give that a try probably that may work So there's no like rule that if anything doesn't work if you have to apply product of determinants You may feel that easy determinants can also be written as a product But as you can see it is not easy. You have to do some reverse. You have to do a guesswork It's a reverse engineering that is happening. Yes Yeah, this this is going to be a simple integration actually I'm sorry simple determinants actually Okay, let's say Let's say I want to write this as a product of two determinants. Let's say I can solve it other way also But let's say I want to write it. Can I write it? Let's try to check Can you write it as a product of two determinants? Uh, I see a square b square c square Okay, so let's have a b c a b c Okay, that will definitely generate Correct here that will generate definitely generate a square b square c square Next I want b c c a b. So there's a c here So I can have a b So a b will be generated C b c will be generated and a Okay Similarly this I can write it as c a b also. No problem will happen. Same term will be generated. Okay For this I can have again b c a check it out Okay, and I think everything else will work fine See this element can also This element can also be obtained when you multiply this with this Okay, so if you want this element to happen, that means there has to be c a b here also And just check if every element is fine. Basically you're writing the cyclic determinant whole square All right, you're writing the cyclic determinant whole square So you are writing a b c b c a c a b the whole square So this can never be Negative it is always a greater than equal to zero quantity Now we have also learned that the expansion of this determinant is this If you want this to be zero If you want this to be zero remember the factorization of this was half a plus b plus c A minus b whole square b minus c whole square and c minus a whole square Okay, so under two situations it will give you zero one Your a plus b plus c should be zero And to your a b c Could all be equal to each other Like this correct So they will all be equal then also it will be zero Need not be all zero actually Because it need not be all zero So if they're equal Or their sum is zero Then only they will be zero got it Any questions here? Okay, let's take more questions Okay, I'll I'll help you out in this Uh Just one minute. Okay Let me at least Expand this and write so that we can This is b c b plus c x x square Okay Yeah Can we envisage this as a product of The first could we start by like subtracting some of the rows Let's come up the rows Okay, a good idea. So like you can Yeah Minus Okay, so what I plan is uh, basically param when I look at this right side expression. Okay These this expression looks very familiar to me because I have seen this in One of these determinants, which basically I keep on using one one one a b c A square b square c square. This is what I I know I think this will you know help me to get it Same with this guy Okay, I I plan it to break it like this so that you know directly these two determinants which I am encircling with yellow and green They come directly I think when you do that subtraction you will end up getting few factors directly out Which I can do but I'm avoiding it right now Okay, let's see. I may have to you know come back and do whatever you are saying But let's try to see what best we can do Okay, so let's say I want to get this element Okay, and in the back of the mind those two determinants are also running in uh here So can I do one thing? I can definitely see one x and x square So let's let's keep this one x and x square and let me keep this as b c b plus c and a one Let's not put brackets unnecessarily Okay, right Correct Now for this element let's choose ac a plus c and one over here And for this element, let's choose a b a plus b one. I may be wrong. I may have to come back and change it Okay, so I'm just taking a leap of faith as of now So I think these three elements have been taken care of So for this I need to multiply the first row with the second row here So it has to have one one y square And thankfully not only this this is also falling in place when you multiply second row of the first determinant with the second row of the second And when you multiply third row with the first determinant with the second row of the second so these two are falling in place So Now this multiplied with one z and z square will do the wonders So this will be obtained when the first row multiplies with the third Then again second row multiplies with the third. I'll get this again third row multiplies with this. I'll get this Okay, so everything is falling in place Uh, and thankfully this determinant is known to be x minus y y minus z z minus x So at least I got that A green circle part which I was trying to obtain I hope I'm correct Okay, what about this? Uh, what about this? What about this? What about this? Uh, uh, uh, yeah, can we do Can we do a simplification of this? I think we can take one of the rows subtract it from the other two rows Or would you like to split this up and do it? Yeah, we can split this up and do it. So let's try that because it's it, uh Won't be that difficult. So let's say bc ac ab b a Oh splitting in order to be that great. Okay. Let's let's expand it So let's do this operation r one as r one minus r three and r two also as r two minus r three So this will be, uh bc minus ab I think c minus a will be left zero ac minus ab c minus b will be left zero ab a plus b one I think c minus a can be taken out from here and c minus b can be taken out from here So c minus a c minus b So if you expand it now with respect to the third column This should give you just b minus a So you'll end up getting b minus a let me write it as a minus b And switch this position b minus c and c minus a and you already had this term Let's copy it So your final answer would be this into x minus y y minus the z minus x Done any questions here clear abiram Abiram nowadays is not answering. What happened abiram? You've gone very quite All well abiram. You're there. Okay I think the frequency of your answering has reduced Uh, this we have already done we'll take one more. Uh, let me just take up one more Uh, let's take this question prove that Prove that a minus x whole square a minus y whole square a minus z whole square b minus x whole square b minus y whole square b minus z whole square c minus x whole square c-y whole square c-z whole square is equal to 2 times b-c-c-a a-b is actually slightly wrong times x-y y-z z-x. So could you show the right side? Done. Chetanya also done. Very good. Most of you have been able to do this. So if you expand this and write, I think I need more space, it's not be sufficient. So the property the b-c a-b b-c-c-a that works only for a particular order. Which I didn't get you. Like in the determinant which we generally expand it's like 1 1 1 x y z and x square y square z does it work only for that? Yeah, it works only for that. Okay. Or if you are able to substitute something with y z x y z. Yeah, so this is b square minus 2 b x plus x square b square minus 2 b y plus y square. This is b square minus 2 b z plus z square. Now let's try to write this as a product of two determinants. So let's do this element. Can I just take a square minus 2 a and 1 and keep 1 x x square here? Similarly, I can keep 1 y y square here and 1 z z square here. I think this determinant will take care of the last three factors that we have on the right hand side. Okay, what about this? This has to be b square minus 2 b 1. This has to be c square minus 2 c 1. I think everything will fall in place when you multiply these two determinants. Just take a minus 2 out. Just take a minus 2. We already have the result for the determinant. Not exactly. We have to switch. This this result is known to us. We have to one switch will give a minus sign. We have to switch this first. So when we switch this first, we'll end up getting, let me just change in the next step to 1 1 1 a b c a square b square c square. I think this term would be the one responsible for the first three factors that we have on our right hand side. Okay. So yes, I think post this, we already know the result. This is a minus b b minus c c minus a. This is x minus y y minus z z minus x. Done. Okay. Is that fine? Any questions here? Can you explain how do you remove the minus sign again? Which one? Minus sign? Yes, sir. How I removed it? I swapped the position of c 3 and c 1. No. Okay. You can only write this result when it is 1 1 1 a b c a square b square c square in the same order either row wise or column wise. Yes, sir. But here it was slight deviation in the pattern. That's why I flipped the position and removed this negative sign. Yes, sir. Great guys. Then I think we are good to go with differentiation of a determinant enough questions we have taken on this differentiation of a determinant. Differentiation of a determinant.