 So next concept that we are going to talk about is differentiation of a determinant not Determinant if you see I just take a very small example of a 2 by 2 Let's say if I have F1 G1 F2 G2 as our determinant Okay Let's say I call this as netx. So here F1 and And G1 are functions of X or can be constants also. Okay now when we expand this determinant When we expand this determinant we get F1 G2 minus F2 G1 If somebody says differentiate this determinant with respect to X What do I do? I'll apply product rule individually to these two, isn't it? So if you look at the first one it becomes F1 G2 dash plus F1 dash G2 minus minus This becomes let me put back its this becomes F2 dash G1 plus F2 G1 dash Correct. Okay. Now, what I'll do here is I will restructure it like this F1 G2 F1 G2 minus F2 G1 dash F1 dash G2 so I've taken this term and I have clubbed up with this Okay plus the non-tick mark terms that is F1 G2 dash Minus F2 dash G1. I'll club up like this Okay, if you see these two terms very very carefully, they themselves are actually determinants Having this as the first row This as the second row for the first one. So for this one, I have written it in the form of a determinant Please confirm whether this is fine and for the second one That is for this guy You can also write it as a determinant F1 G1 F2 dash G2 dash correct So what does this tell us? Sorry, this one is slightly above so it has to be Yeah, what does this tell us when you're trying to differentiate a determinant? We can do the differentiation Like this as you have done it here row wise But let me tell you you can also perform it column wise Okay, so if you do a different type of a pairing over here You'll end up getting a column wise for the same. That means you can also do it like this I'll let it down over here F1 dash F2 dash G1 G2 plus F1 F2 G1 dash G2 dash that means either you do it row wise or you do column wise The result is going to be the same as you can see here F1 dash G2 minus F2 dash G1 Right, it is basically clubbing this term with this term See it F1 dash G2 minus F2 dash G1 so it is clubbing these two terms Okay, so if you make a pair of these two and these two you'll end up getting this form. Is that clear? Yes, sir. Okay, so Let me expand this or let me extend this to even a three by three so you can follow the same thing Similar approach can be followed even for even for higher order determinants Okay Now it is your call you want to differentiate it row wise But remember when you're differentiating it by any one of the approach you can only differentiate a row or a column One by one You can't differentiate everything at the same time. So as you can see I've differentiated first row Kept the second column as as it is then plus keep the first row as it is Then differentiate the second column a second row Okay, same with column wise operation also So if you're doing it do first column keep the second column as it is plus keep the first column as it is differentiate the second column, okay Now again, many people ask me, sir, why do we need it? We can expand it and differentiate it. No What is the big need for this particular approach again? As I told you there may be some complicated determinants I know you you cannot expand it and then differentiate It's better to differentiate it before you expand it. So in such cases The differentiation of determinant method would become very very handy will become very very convenient for you Okay, let's take a few questions. Then we'll understand this approach in a much deeper way. Let's say Let's say I take this question Question is find f dash zero. Now as you can see if you expand this determinant and then try to differentiate It would be too tricky. It would be too lengthy for you No, Richard. It's not that correct Richard Parik correct Advik Richard Singh it is not the answer that you have stated No, it is not even that Sure. Sure. Take your time. I know Abhiram. That's not correct Check your working again correct Chitanya No, but I'm That's not right Okay, guys, let's let's discuss this Let's discuss this See when you're differentiating the first row Second row elements, you can directly substitute zero because you'll have three determinants and ultimately you'll have to substitute zero in place of x So this will become a zero. This will become a one. This will become a zero. This will become a zero This will become a zero and this will become a zero So anything that comes here, I don't care because the last row is zero. So this will be completely zero This will be completely zero next if Let's say I differentiate the second row then also I'll get completely zero because the third row will still be zero Okay, so this zero will come again The only respite that we'll get is when we are differentiating the third row Okay, now see for third row I'll put zero in the first and the second row. So this will be one zero minus one zero one zero Derivative of this will be two sine two x that will give me a two This is zero and this will be zero Okay, let's expand this with respect to the last row So two Zero minus minus one plus one. So this will become f dash zero. That's nothing but two So guys one thing I would like to tell you don't Unnecessarily solve those determinants where you realize one of the rows is going or one of the columns is going to become a zero So as you can see column number three will be zero when you put x as zero, right? post differentiation Post differentiation when you're putting x as zero Column number three will be completely zero at least for the first and the second one, right? So why to even expand it? But I'm you need to practice differentiation first Okay, let's take another one FG and HR three polynomials of degree two Prove that this is a constant polynomial easy Okay, so this is a very easy question see if you differentiate it Let's say if I do five dash X When you differentiate the first row keeping the other two same you would realize that the first and the second will become the same, right? So it'll happen something like this. Okay, so this guy will become a zero Okay When you differentiate the second row keeping the first same I think the first second and the third will become the same. So this will become a zero and When you differentiate the third row since it's a degree two polynomial We can say that the last row would be all zero zero zero Because the third derivative of a second degree polynomial has to be zero That means this is also zero. So if your five dash X is zero, that means this polynomial is a constant polynomial simple as that No problem with this no issues Let's take the next question G of X is given as this F of X is a polynomial of degree less than three Prove that the derivative of G will be This determinant divided by this is a division symbol Okay, this is not a easy question actually it requires a bit of thinking So they're saying that the derivative of this term can be expressed like this determinant divided by this determinant Okay Now first of all first of all Can we can we break G of X into partial fractions now? Is it a candidate for partial fractions for that the degree of the numerator must be lesser than degree of the denominator? Is it that way? Yes, because they have mentioned that it is degree less than three Okay, so I'm going to break this up as a partial fraction. So how I'm going to break this as a partial fraction So let's let's check it out So this term this term I'll break it up as a by X minus a B by X minus B and C by X minus C. Okay Now if you take the LCM and all If you take the LCM and all this will become a times X minus B X minus C B times X minus a X minus C and C times X minus a X minus B X minus B Okay, and you're comparing it with f of X Okay, so if you want to get an a the obvious substitution is put X as a Correct. I hope you have exposure to partial fractions, right? We have done this in integration as well So if I put X as a this will become f a whereas these two terms will vanish and I will end up getting a times a minus B a minus C as f of a that means your a which is your The cof the numerator part of X minus a Will be f of a by a minus B a minus C. Am I right? Similarly, I'll not waste too much time. I can see a trend over here Similarly, I can say B will be f of B by B minus a B minus C and similarly and Similarly, C will be f of C by by C minus a C minus B. Correct. Any questions so far? Okay, Richard. Okay. So when you put it over here When you put it over here, you would realize you have written something like this f a X minus a a minus B a minus C plus B will be f of B X minus B B minus a B minus C and C will be f C by X minus C A minus B a minus C. Oh, sorry. I was wondering I wrote something wrong Yeah, C minus a C minus B. I looked at the top one by mistake. Anyways Can I take one by a minus B a minus? Sorry one by a minus B B minus C and C minus a common Okay, if I do that if I do that correct me if I'm wrong The first term will become f of a into C minus B By X minus a correct. So from this term if you take One by a minus B B minus C C minus a remember there was an a minus C here So if you're taking a C minus a this will become up Ulta and go on the numerator, right? Okay, similarly what happened to this term? So B minus a has been taken out as a minus B Okay, and B minus C is B minus C. So this will go and multiply as a minus C on the top, right? Similarly here f C you have taken This has Ulta and this has same so B minus a will go and multiply correct me if I'm wrong guys Guys and girls both. What does this remind you of? The one one one ABC Determinant and right and they have written it here, but slightly in a deviated fashion. Probably they have written negative like Interchange the corner. They're interchanged. So let me do one thing as of now. Let me park it as it is We can do something with this We can do something with this Can I write this as a determinant? Which has got one a Basically, I'm taking this as one term and C minus B is another term. So f of a by x minus a And this must have a C minus B. So there will be a one here There will be a B here and there will be a C here and they'll be a one here. So that's why this term is obtained C. I'll explain you what I did This term is obtained When you are expanding with respect to let's say the third column and you are multiplying this guy with C minus B. That's its co-factor, right? So that's how you end up getting the The first term similarly, this will become f B by x minus B So when you're expanding it, this will become this into C minus a with a negative sign that will give you a minus So this last Element of row 3 will be f C by x minus C Is there anybody who is feeling trouble in understanding this? Let me know if at all you're feeling trouble just Expand this determinant with respect to column 3 Expand this determinant with respect to column 3 you'll end up getting the whole this part Check one. So now when you differentiate two will give zero and the last one will give a minus sign Absolutely. It was a planned move when I look at this expression x minus a whole square Okay, so I thought of this. It is not like I'm doing something in in the dark I was actually having an eye on what we want to get So this is a g of x finally. So when you differentiate g of x as Ruchi rightly said Since the first C constant will not be affected constant will remain constant. It will not be changing. Okay But what will happen when you're differentiating it? Let's differentiate it column wise the first column will be completely zero So I don't need to write it in the second determinant The second column will be completely zero. I don't need to write it in third column. It's on in third determinant you will have one one one and You will have a b c and The derivative of this will be minus f of a x minus a whole square minus f of b x minus b whole square and minus f of c x minus c whole square and now we understand why they have switched the order of the Denominator over here. So basically they have removed They have removed a negative sign from here Okay, and they have placed it over here and Ideally this guy should have been Ideally this guy should have been one by determinant How they have written it they have written it column wise. Oh, yeah column wise so Ideally it should have been one one one a b c a square b square c square But now that they want to get rid of this negative sign They swap the position of the first column and the third column So let's drop this negative sign. Just swap these two positions and there you go This is what you wanted to prove done Okay, not a easy question. I would say this will be probably a J advance kind of a question Why I'm writing prove that I should write hence fruit Is that fine any question any concerns, please do let me know Any step that you want me to revisit again do let me know Let's take one more question and then we'll move on to integration of determinants That's not a very Happening concept integration of determinants, but still I'll surprise you off Question is There is a determinant There's a determinant like this Okay, so we have a determinant like this Okay question is find the coefficient of find the Coefficient of X in this determinant. Let me call it as delta X Find the coefficient of X in this determinant No other way that's not the right answer See basically this is going to result into a polynomial only something like this Correct. Let's say let's say it's an nth degree polynomial Okay, if you want to get coefficient of X How will you get it? Can I say coefficient of X will be nothing but it will be let's say I call this determinant as delta X so a one will be Derivative of delta X at zero isn't it? Because when you differentiate a not goes for a toss You'll have a one remaining and then other terms will have an X in it So if you put a zero all the other terms containing X will go off So what is the question actually asking you the question is actually asking you? Delta dash zero, that's it Yes, which is that's correct Correct Chaitanya. It's a easy question guys. Very easy question. I mean you can do it verbally also correct Richard Park Chalo base If you differentiate it and put X as zero remember, let's say if you're differentiating the first row Okay, second and third row will not be affected Right, but when you're putting zero won't it become one one one one one one That means row number two and one number three will add in become identical. Correct. So when you're differentiating it, I'll just I Just try to write it Something will come over here. I mean if I want I can write it as a one be one A one be two a one be three, but I see a one one one one one one coming over here So anyway, this will kill the whole determinant Okay, similarly for the second one you will have one one one right something will come over here a two be one a Two be two a two be three and again one one one So row number one and row number three are identical that will kill the whole determinant Okay, and finally finally you will have One one one one one one one and you have a three be one a three be two and a three be three So that will still give you zero because no number two and no number one identical. That means That means delta dash X or delta dash zero is zero which is actually a one equal to zero So they'll be zero coefficient along with X. Is this clear everybody? Is this clear? Yes Okay, now let's take up the concept of integration of determinants Integration of determinants this concept is a very very narrow one because This works only in certain cases So if let's say you have a determinant which is a function of X, but only one row or any one column okay, if Multiple rows and multiple columns have functions of X then this method will not work Okay, so this theory says that if you have a Determinant which is having any one row or any one column as Functions of X and if you're integrating that determinant with respect to X from let's say limit a to b Okay, then it is as good as integrating that row which has got functions of X in it Or that column which has got functions of X in it Okay, rest other rows and columns would remain as it is Rest other rows and columns would remain as it is now a Good question arises here. What will we do? What will we do if multiple rows or multiple columns have functions of X in them in that case? You need to expand the determinant and integrate Okay, so please note if more than one row slash column are expressed as functions of X as functions of X then Delta X Must be expanded and then integrated now Why does this happen is because let's say if you have only one row as a function of X and you expand it you would realize that You'd realize that in such a case you will always end up getting f of X along with some constant G of X along with some constant Correct, I hope you know the constants for f of X the constant will be qn minus mr Okay, but g of X the constants will be RL minus pn like that and h of X will also have a constant Now when you're integrating it the constants in any ways not participate So you'll end up ending integrating f g and h only which is as good as saying this Right that means the other two rows are not getting affected But if you have multiple rows or columns written in functions of X then you'll have a product of two functions of X coming up and for that the integration has to be done in a Different way you either have to apply integration by parts or any other method that you know off to do it So this will not be as simple as We have it in the case where any one row or any one column is made as a function of X Okay Because of this restriction it doesn't find its use you know many questions are not asked on this So it's a very very you can say a simple look concept Let's take a simple example for the same and then we'll close the session let's say Have a determinant Very simple question. I'll give you just to apply the formula. Okay. Let's say this is a function of X Evaluate Integral of Delta X from zero to one our dear all those who are following SK Goyal from Arihant Let me tell you they have given something called Wally's formula and all under integration under determinants, I don't know whether you have the new addition with you But I will not be doing Wally's formula in different determinants. I'll be doing it under definite integrals Okay, very good, which is Rahul Chetanya Mahit good Yeah, that's correct. That's correct. It's very simple. Let's let's do that So if you have to integrate remember the other rows which are made up of constants will not be affected This will become X square by two from zero to one X cube by three from zero to one X four by four from zero to one. So that will give you a B. C a B. C six four three Half one third one fourth Half one third one four. So let's multiply The row number three with a 12 and we can write it like this 12 A B C 6 4 3 6 4 3 column 2 and column sorry row 2 and row 3 have become identical So this will be collapsing and the answer is 0 And the answer is is that fine Now since it involves integration having Limits will not be able to take many questions on this before we do definite integrals So we'll start with different integrals in the next class then probably we can no solve many more questions under this Okay Now I'll take a break because we have officially Done the determinants and we have to start with Kramer's rule We have officially done with the differentiation and integration and we'll have to start with Kramer's rule So after the break we'll talk about Kramer's rule that will take another hour or so