 Hello. Countest. Hello guys. Am I audible now? Can you see the screen? Great. So we'll begin with this question. So this is going to be a session on matrices and determinants and I think since we are half an hour late because of the technical issue, I may extend by a few minutes on the other side of the session as well. So please cooperate. So let us begin with the first question. So here we have a question which says that alpha is a root of the equation x to the power 4 equal to 1 with negative principal argument. With negative principal argument. Then we have to find the principal argument of this determinant. So again, I'll wait for just three people to respond and then I'll start discussing the solution for this. So let's see who's the first one to answer this correctly. Just a second. Yes, anyone? So if we just take the a to the power n common from the second row and 1 by a to the power n common from the third row, we will get this expression as a to the power n by a to the power n and we'll have 1, 1, 1, 1, 1, a. In fact, this expression is not a, it's going to be alpha, I guess, because there's no a in the expression. So this is going to be just alpha expression. Okay. So treat this as your alpha values. So this is b alpha. This will be alpha. This will be cancelled. This will be alpha cubed. And this will be 1 by alpha, 1 and 0. By the way, it's very clear that alpha in this case would be the one with the negative principal argument. So the roots of this equation will be a plus minus one and plus minus I. Right. So I'm talking about the one which has got the negative principal argument. Negative principal argument means only minus I has a negative principal argument. Right. That's equal to minus of pi by 2. Right. So argument of minus I is negative pi by 2. Okay. All right. Let's discuss what's the answer in this case. So let me expand now with respect to the first row itself. So it will be getting 1 times 0 minus alpha cubed. Again, minus 1 times 0 minus alpha squared. Okay. And the third row will be 1 minus 1, which is 0 itself. So it's going to give me minus of alpha cubed plus alpha squared. And when we put the value of alpha is minus of I, we get minus I cubed plus minus I squared. Right. That's going to be plus I cubed and this is going to be plus I squared. That's going to be minus I minus 1. Now, minus I minus 1 happens to lie in the third quadrant. Okay. So minus I minus 1, if you represent it like a point, it happens to lie in the third quadrant. And the argument for this would be minus of 3 pi by 4. So option number 2 is going to be correct in this case. Is that fine? Purvik Rohan, did you realize your mistakes? Okay. Any question with respect to this? So easy question to begin with. So I'll be moving on to the next question now. So this is the second question for you. If determinant 1 or delta 1 is given by this determinant, delta 2 is given by this determinant, then which of the following option is correct? Okay. So Amog has given the first response here. All right. So Purvik also feeds the same Rohan, Sai, everybody feeds the same. Okay. So this is a pretty simple question. So if you evaluate delta 1, okay, I'll be using Saras method to evaluate it. So let me write the determinant in this fashion. And let me repeat the first two columns after this. Yeah. So basically we write the product of this. We write the product of this. We write the product of this over here. That's going to be x cube plus b square a plus b a square. And again, we write the product of this. We write the product of this and we write the product of this over here. Right. So that gives me x a b, x a b, x a b. So 3 x a b. So your total function delta 1 is going to be x cube plus b square a plus b a square minus 3 x a b. Okay. And when you find the determinant delta 2, it's going to be x square minus a b. So I'm sure the first option is not going to be correct. Now let's try to differentiate delta 1. So delta 1 derivative will be 3 x square minus 3 a b. That looks like 3 times x square minus a b, which is actually 3 times delta 2. And yes, option number 2 is the right option in this case. So the first one to answer this correctly was amok. Okay. Well done. Simple question to begin with. So let me now move on to the next one. Yeah. So third question. If a square plus b square plus c square plus a b plus b c plus c a is less than equal to 0. For all a b c belonging to real numbers, then the value of this determinant is. All right. The public has given the answer here first. I'm not saying it's right or wrong. Let's wait for a few more people to respond. All right. So I've got the response from other people as well. So let's discuss this guys. So here we have been given a condition that a square plus b square plus c square plus a b plus b c plus c a is less than equal to 0. Right. Let's multiply both sides with 2. So multiplying with 2 both sides. I will get 2 a square 2 b square 2 c square 2 a b plus 2 b c plus 2 c a is less than equal to 0. Now this is as good as saying a plus b the whole square b plus c the whole square and c plus a the whole square is less than equal to 0, which implies that the only possibility is each of these terms are 0. Which means each of these terms a plus b b plus c c plus a all are 0, which clearly means that a b c all of these terms are actually equal to 0. Correct. So this particular determinant actually simplifies to 4 0 1 1 4 0 0 1 4. Right and we can expand it with respect to let's say the first row. So it will become a 4 times 16 minus 0 and plus 1 1 this becomes 64 plus 1 which is 65, which means option number one is correct in this case. Okay. So almost everybody who responded gave the right answer for this question. Let's now move on to the next one. So let's move on to the fourth question. The product of the roots of this equation is so this is basically if you simplify this it will result into a polynomial equation and they are primarily asking you the product of the roots of that equation. Okay. So I'm going to note it down your response. Let's wait for two more people to answer. So guys, how are your practical exams going on? I think I see has already started. What about CBC? Oh, you have came practical tomorrow. TBC Janta, when is your practical starting? Oh, both practicals are over. Oh, great. Great. All right. So guys, discussing this now to get the product of the roots, we know that the degree of this particular quad polynomial equation would be decided by the degree of the product of these terms. So it's going to be a fourth degree term, right? It's going to be something like this. Okay. Right. So the sum of the root is minus B by the sum of the product to at a time is C by then minus D by then E by. So if the roots are alpha one, alpha two, alpha three, alpha four, its product will be E by. So for that, we need the constant term and we need the coefficient of X to the power four. The constant term is nothing but when you put X is zero into this determinant, the answer that you get will be a constant term, right? So just put X as zero. So you get this expression, correct? Let's evaluate this. So it'll be one time, two minus three, which is minus one, then minus one time, two minus zero, then plus one, two minus zero again. So that's going to be minus one, minus two, plus two, that's going to be minus of one, correct? Now, coefficient of X to the power four can be easily obtained when you see that the term 2X, X and minus X square will generate X to the power four, right? So this A term can actually be generated by multiplying 2X, X and minus X square, which is actually going to give you minus of two. So your E by A is going to be minus one by minus two, which is going to be half, right? So you don't actually have to evaluate the determinant. You can directly get the constant term by putting the variable X as zero and you can get the coefficient of X to the power four by directly looking at the diagonal expressions, right? And the diagonal expressions, you would note that 2X to the power four, X to the power four will be generated by the product of 2X, X and minus X square. Is that fine? So option number one is going to be correct. So let's move on now to the next question. Initially, you'll find some very easy questions. So moving on to the fifth question, in a triangle A, B, C, if A, B, C are the sides opposite to angle A, B, C respectively, then find the value of this determinant, okay? So I've got a response from Aditya and Atmesh. Guys, it's a super easy problem. What you have to do is just do C3 as C3 plus C1. And when you do that, you end up getting this. C cos B plus B cos C and here you get A cos C plus C cos A and here you get B cos A plus A cos B, okay? And we know from the projection formula that this expression, this expression here is going to be, this expression here is going to be small A. This expression here is going to be small B. So this is small B and this is small C. So as a result, what happens? Column number C3 and C2 become the same and hence your determinant collapses and it becomes zero. So absolutely correct. So as I said, initially you will find very easy questions. Just to keep you motivated. It's not like I'll be demotivating you later on. Yeah, so let's now go to the sixth question now. So A is a series which goes like 1 plus 2 plus 4 up to n terms and B goes like 1 plus 3 plus 9 up to n terms and C goes like 1 plus 5 plus 25 up to n terms. Then the determinant delta is equal to, okay? So yeah, guys. So we'll discuss this quickly. Again, if you see this, this is a GP and then A will be nothing but 1 into 2 to the power n minus 1 by 2 minus 1. That's nothing but 2 to the power n minus 1. Similarly, B will be nothing but 1 into 3 to the power n minus 1 by 2 and C will be nothing but 5 to the power n minus 1 by 4, right? So I can write this delta as A which is 2 to the power n minus 1, 2B will be 3 to the power n minus 1 and 4C will be 5 to the power n minus 1. Now let's take a 2 common from the second row. So if you just take a 2 common from this row, okay? And do this operation, R3 is R3 minus R2. Then you get delta as twice of 2 to the power n minus 1, 3 to the power n minus 1, 5 to the power n minus 1, 1, 1, 1. Again, 2 to the power n minus 1, 3 to the power n minus 1, 5 to the power n minus 1. And since row number 1 and row number 3 become the same, your determinant value will become 0. So option number 3 is going to be the correct option, okay? Again, a very simple question. So let's now move on to the next one. Question number 7. If a1, a2, a3, 5, 4, a6, a7, a8, a9 are in harmonic progression, then the value of this determinant is D. Find the value of 21 D by 10. Yes, guys, anyone? So let us discuss this. For example, here I know what is my fourth and fifth term, right? So first of all, if I say a1, a2, a3, 5, 4, a6, a7, a8, a9 are in HP, I can say 1 by a1, 1 by a2, 1 by a3, 1 by 5, 1 by 4, 1 by a6, 1 by a7, 1 by a8, 1 by a9, etc., are in AP, okay? And since we know these two terms, I can know the common difference as well. Common difference will be 1 by 4 minus 1 by 5, which is going to be 1 by 20, right? And I can find the first term as well. So 1 by a plus, let's say this is your fourth term. So 3D is equal to 1 by 5. So I can say 1 by a1 is 1 by 5 minus 3 by 20, which is again 1 by 20, okay? So the terms a1 is going to be 20, correct? a2 is going to be 2 by 20, which is 20 by 2, which is 10. a3 is going to be 3 by 20, which is 20 by 3, okay? a4 is going to be 5, a4 is going to be 5, a5 is 4 given, okay? a6 will be 20 by 6, a7 will be 20 by 7, a8 will be 20 by 8, and a9 will be 20 by 9, okay? So once we know these values of a1, a2, a3, a4, a5, a6, I can actually write the determinant back again as 20, 20 by 2, 20 by 3, 5, 4, 20 by 6, 20 by 7, 20 by 8, and 20 by 9. You can take 20, 20, 20 common from each of the given rows. So you can actually write it as, let me just clear this part of the board, okay? So once we have figured this out. So I can write 20 cube, and I can have one, half, one-third, one. We can have four-fifth, then, sorry, one-fifth, not four-fifth. So here we can have one-fourth, one-fifth, one-sixth, one-seventh, one-eighth, one-ninth, right? Now we can take 1 by 4 common from the second row and 1 by 7 column from the third row. So if I do that, I get a 4 and I get a 7. And this I can write it as 1, 1. And this will have 4, this will have 4, this will have 7, this will have 7, okay? The reason I did this was to create 1, 1, 1 in the first column. Now let's do this activity, R2 as R2 minus R1 and R3 as R3 minus R1. So take the first row and subtract it from the others. So when you do that, you get a 1, 0, 0, half, one-third. And this is going to be four-fifth minus half. That's going to be minus 3 by 10, okay? And this is going to be two-third minus one-third, which is again going to be one-third. Here also I can subtract seven-eighth minus half. That's going to be 3 by 8 and 7 by 9 minus 1 by 3, which is going to be 4 by 9. So now let us expand this. So when you expand this with respect to the first column, you will get 1. And you have minus 12 by 90 minus 1 by 8. And outside we have 20 cube by 28 waiting. So that's going to give you, again, make some space for myself. So this is going to give you minus 96 minus 186 by 90 into 8. Just check whether all the calculations are proper or it's plus 3 by 10. Okay, this is plus 3 by 10. Okay, so this will become plus 4. Yeah, thanks for that. So in this case, my answer will now change over here. So it will become 96 by 6, 690 into 8 into 20 into 20 into 20 by 4 into 7. So this is going to be my determinant and they have asked the value of 21 by 10 times this. So I think we can cancel this by a factor of 3 and this goes by a factor of 5. One of the zeros will go off. Okay, so there are too many expressions over here. So this is going to be 10, 200 into 6 into 3 by 10 into 9 into 8. So this is going to be gone by a factor of 12. Yeah, so this is gone by a factor of 4. So yes, so it's gone by a factor of 5. So option number 2 becomes correct in this case. Okay, so not a difficult problem, but yes, a bit of calculation involved towards the end. So the right answer is option 2 and let's see who has given the right answer. Oh, none of you so far. Anyways, one more opportunity is there with the next question. So let's move on to the next question now. Question number 8. So the value of this determinant is dependent on A, B, C dependent on A dependent on B. On actually dependent on B dependent independent of A, B and C is a small change in the question. This term over here, this term over here, change it with A plus 2B plus C by AC. Sorry about that typo, but make a change in the term. That term is not one by that is A plus 2B plus C by AC. All right, so let me discuss this now. So first we do one thing. First you multiply the first column C1 with A. Okay. So when you multiply with A, we have to divide outside with A. Correct. So let me make the changes by writing it separately. So 1 by A and I will have the first column as A by C minus B plus C by A and minus B B plus C by AC. Okay. Let me multiply the second column with B. So I will have 1 by B outside. So it will be B by C, B by A. And as I already discussed with you, this will be B times A plus 2B plus C by AC. And let me multiply this column number 3 with C. So I will have 1 by ABC and you have minus A plus B by C. Then you have C by A and then you have B times A plus B. In fact, minus B by A plus B by AC. By AC. Okay. Now having done this, we can add everything to the first column. So column 1 is column C1 plus C2 plus C3. So delta will become 1 by ABC. So when you add everything to the first column, you would realize A by C, B by C will get cancelled with minus of A plus B by C. So this will become 0. In fact, this will also become 0. And this will also become 0. That means the first entire column has now collapsed. So I don't care about the other elements. So the answer is going to be 0 in this case. So it is independent of A, B and C. So this could be a potential question for you in the board exams as well. Is that fine? Any question? Please type in your chat box if you have any question, any concern with respect to this solution. All right. So moving on to the next question, which is question number nine. Find the number of positive, sorry, if the number of positive integral solution of U plus V plus W is N, be denoted by PN. Then the absolute value of this particular determinant is, I think the word absolute here is not important. So just ignore this word. Just find the value of this determinant. First of all, what is the positive integral solution of this particular equation? So when I say positive integral solution means U, V, W, all of them should be actually be greater than equal to 1. So what I do now, I call U as capital U plus 1, small V as capital V plus 1 and small W as capital W plus 1. So since small U, small V and small W are greater than equal to 1, I can say that capital U, capital V and capital W will be greater than equal to 0. Correct. So when I put this into this particular expression, I get capital U, capital V plus capital W is N minus 3. Okay. And now you have no restrictions on U, V, W. They're all greater than equal to 0. So it is like distribution of identical objects into three groups. So you are distributing N minus 3 identical objects. So it is like distribution of N minus 3 identical objects among three groups says that the groups can be empty as well. So it is like, you know, you're trying to separate out N minus 3 identical balls by using two sticks. Okay. So wherever you place your sticks, okay, let's say these are the balls. So wherever you place your sticks, you know, the three groups will automatically get formed. Right. So the number of ways in which you can arrange N minus 3 plus 2 objects out of which N minus 3 are identical of type 1. So out of this N minus 3 are like, let's say balls, which are identical. And let's say two are sticks. Again, these six are identical. So this can be done in N minus 1 by N minus 3, sorry, N minus 1 factorial by N minus 3 factorial 2 factorial. Correct. Which you can actually also write it as N minus 1 C2. So this is your value of PN. Okay. So once I know PN value, I can find out the value of PN plus 1, PN plus 2, etc. Just by changing N with N plus 1, N plus 2, like that. So the first expression is N minus 1 C2. Okay. The next expression is going to be N C2, then N plus 1 C2. Again, N C2. Again N plus 1 C2. Again N plus 2 C2. Again N plus 1 C2. Again N plus 2 C2. Again N plus 1 C2. Let me just erase this part. This is going to be N plus 2 C2 and this is going to be N plus 3 C2. Okay. So you just have to find the value of this expression. Now, since your answer over here is independent of N. Okay. You can actually choose a suitable value of N over here. For example, I can choose N value as, let's say, since the maximum is 2, I can choose N value as 3. Okay. Let's try choosing N value as 3. Let's see what happens. So 3 minus 1 C2 is 2 C2, which is 1. Okay. This is 3 C2, which is 3. 4 C2, which is going to be 6. Again, this will be 3, 6. 5 C2 is 10. And here I will have 6 C2. 6 C2 is going to be 15. Correct. So let me expand this. Let me just take out some terms or we can directly expand it. Also, it's not a big expression. So one times this is going to be 90 minus 100 minus 3. This is going to be 45 minus 60. And this is going to be 30 minus 36. That's minus 10. And here I will get plus 45. Okay. And here I will get minus 36. So that gives me the answer as minus of 1. Which is option number 1 is correct in this case. Guys, just make your life simple because your answer doesn't depend on N actually. Right. So you can choose a suitable value of N to make your life easy. And the only person to answer this correctly was Amogh. Okay. So the idea that was tested over here is a distribution of identical objects in two groups. Nothing else. Moving on to the next problem now, question number 10. So f of x, h of x are polynomials of degree 4. And this determinant over here is equal to mx to the power 4 and x cube rx square sx plus t. Okay. So this is an identity in x. That means these two expressions are equal for all values of x. It's not an equation. Then we have to find the value of this determinant. Please note that these two terms are together. Here these two terms are together. These two terms are together. So f triple dash 0 minus f double dash 0. F, g triple dash 0 minus g double dash 0. H triple dash 0 minus h double dash 0. Yeah, it's triple dash. Triple dash means the third derivative. All right. So I got the response from Aditya Amogh. All right. So let me begin the discussion for this. So guys, if you differentiate this determinant on both the sides of the identity, since row number two and row number three are all constants, the derivative would be f dash x, g dash x, h dash x, and you'll have a, b, c, p, q, r, and all the elements would be 0. Okay. And that's going to be 4 m x cube, 3 n, 3 n x square, and 2 r x plus s. If you differentiate this once more, you're going to get f double dash x, g double dash x, h double dash x. And here you have a, b, c, again, p, q, r. And that's going to be 12 m x square, 6 n x plus 2 r. Okay. Triple derivative of the determinant will give me f triple dash, g triple dash, h triple dash. Okay. I need not write the other elements because they will be 0 anyways. Okay. And this will give me 24 m x plus 6 n. Okay. Now what I'll do is in this, let's say I call this as the first, second, and third. So in the first, in the second expression, I'll put x as 0. Okay. So when I put x as 0 over here, this gives me f double dash 0, g double dash 0, h double dash 0, and you have a, b, c, p, q, r. And on the right-hand side, you'll get 2 r. On the right-hand side, we end up getting 2 r. Here also, when we put as 0, we get, let me put here itself, 0, 0, 0. And this will give me 6 n. Right? So when I subtract these two, when I subtract these two expressions, only the first row is going to get subtracted. Row number 2, row number 3 are going to be the same. So it will become nothing but 6 n minus 2 r. So you can always take a 2 common, so it becomes 3 n minus r, which is nothing, but your option number 1 is correct in this case. So the first one to answer this correctly was Aditya. Okay. Easy question. So let's move on to the 11th one now. So in this question, we have to find the coefficient of x in delta x. Where delta x is a determinant, which is again a function of x. So we have to find the coefficient of x in this. Okay. So Aditya has given the response. Okay. I'll move things, different answer. Okay. Guys, we all know that, first of all, this is going to be 1, 2, 3, 6 degree term. Right? Okay. And the coefficient of x is going to be nothing, but the derivative of this function at 0. Correct? So if you differentiate once, what happens? The constant term goes off. For example, if you write this as a 0 plus a 1x plus a 2x square all the way till, let's say, a 6x to the power 6, when you differentiate it once and put 0, you only are left with a 1. So all we need to do is differentiate this and put x as 0. Let me start with differentiating the first column. So when you differentiate the first column, you get 1, 1, 1 and put 0 in all the elements. So this is going to be 1, 1, 1. And again, this is going to be, sorry, this is going to be 0. And this is going to be 0, 1 and 2q, which is going to be 8. Okay. Next is minus 2, minus 1, 0. The derivative of this will be 2 times x minus 1. So that's going to be minus of 2. This will be 2x. That will be 0 itself. And this will again be 2. Okay. And here I will get a 0, 1 and 8. When I am differentiating the third column, again, let me put 0 in the first. So it will be minus 2, minus 1, 0. And this will be 101. And this is going to be again 0, 3 and 12. Okay. So let us expand this. So the first determinant will give me 1, 0 of minus 1. And you have a minus 1. So you have got 8. Sorry, 8 minus 1. Yeah. In the second determinant, you can write minus 2 times 0, minus 2, plus 2 times minus 8. And this you can forget again. This will become minus 2, 0, minus 3, minus 1, minus 12. Okay. So here my answer is going to be minus of 8. From here, my answer is going to be plus 4, minus 16, which is minus of 12. And from here, my answer is going to be 6 plus 12, which is 18. So that is going to be 10 minus 2, which is minus of 2. That is option number 2 is correct. Okay. And the first one to answer this correctly was Aditya. In fact, he is the only one to answer this correctly. So option number 2 is the right option. So Amok, Purveks, I, please check if you're wrong somewhere. Guys, can you see the screen? Am I audible? Simple, guys. I'm expecting the answer to come within one minute for this question. It's a super, super easy question. Okay. So option 4 is what Amok says. So yeah, when you're dividing this function by x square, you divide it in such a way that you end up giving 1x to the first column. That's sin x by x, x square and 2. And you can give 1x to the third column. That's sin x by x, 1, 1. Okay. And when you take the limit at both the sides as x ending to 0, then this converts itself to 1, 0, 2, 1, 0, 1, 1, 1. And you can always expand this. You can expand this with respect to the second column. That's my second row. So it's plus, minus, plus, minus. So it's minus 1. So just... So it's 1 minus of 2. That's going to be 1. That's option number 4 is correct. Yeah. Let's move on to the 13th question. If ABC are non-zeros, then the system of equation has a non-trivial solution if which of the following option should be correct. I think it's option number 4. So guys, non-trivial solution means what? Non-trivial solution means the determinant formed by the coefficients of these variables. This should be equal to 0. Now, in order to simplify this, I can make these transformations. Let's say row 2 is row 2 minus row 1. And let's say row 3 is row 3 minus row 1. Okay. So alpha plus a, alpha, alpha, this will become minus a, minus a. This will become b. This will become 0. This will become 0. This will become c. Okay. And this should be equal to 0. So when you expand it, let's expand this with respect to the first row itself. So you'll get alpha plus a times bc minus alpha. This will be minus of ac and alpha and this will be ab. Okay. So you can take alpha common, you'll get ab plus bc plus ca plus abc equal to 0. So I can say 1 by alpha could be written as minus ab plus bc plus ca by abc. Okay. So this means alpha inverse is going to be negative a inverse, b inverse, c inverse, which is obviously option number 4 is correct. Okay. So the first one to answer this correctly was Amogh. Amogh was the first one to answer this correctly. So guys, let's now move on. Let's now move on to question number 14 for the day. So the value of theta and lambda for which the following equation sin theta x minus cos theta y sin theta is in brackets cos theta is in brackets y plus lambda plus 1z equal to 0. Again, cos theta x sin theta y minus lambda z is 0 has non-trivial solution. Yes, again, the same process has to be followed over here for again non-trivial solution. So this is the determinant which is formed by sin theta minus cos theta lambda plus 1 again cos theta sin theta minus lambda and lambda lambda plus 1 cos theta this should be equal to 0. Right. So it's a system of homogeneous equation and if you want non-trivial solution means you want at least solutions where x, y, z all are not 0. Then the determinant should be 0. In that case, the determinant made from the coefficients of the variable must be 0. Okay. Now we need to simplify this expression. Let us expand with respect to the first column. Sorry, first row. Sine of theta and here we have sin theta cos theta plus lambda square plus lambda. Then we have plus cos theta. So we have cos square theta plus lambda square and lambda plus 1. We have lambda plus 1 cos theta minus lambda sin theta. So let us try to expand it. So from here I can get sin square theta cos theta and from here I can get cos theta cos square theta. Okay. And then we will have lambda plus 1 whole square cos theta and the sin theta term will get cancelled. So here this term I can actually write it as lambda lambda plus 1. Okay. So this term here and this term over here will get cancelled. So I will get one more cos theta which is lambda square cos theta term. Okay. Now here if I take a cos theta common I get sin square theta plus cos square theta in the brackets and here I will get cos theta and this will give me lambda plus 1 whole square plus lambda square equal to 0. Now this term over here this is 1 actually so we can take a cos theta common throughout and write it as lambda square lambda plus 1 square plus 1. Okay. And if this is 0 so it implies it implies cos theta has to be 0 because this term cannot be 0 because it is made up of all squares of positive term. This cannot be 0. This cannot be 0 so cos theta has to be 0 which means theta can be an odd multiple of pi by 2. Okay. So n belonging to integer theta can be odd multiple of pi by 2 and lambda can be any real number. So option number d matches with this. So as you can see theta is odd multiple of pi by 2 and lambda could be any real value. So option number d becomes the right option in this case. So aditya no psi is the first one to answer this correctly. Next question number 15. If the system of equation has at least one solution then which of the following option is correct has at least one solution. Okay. So I've got a response from psi. So at least one solution means it can have either exactly one solution or infinitely many solution. So let us first check the determinant formed by the coefficients of these variables. So let us expand this. So it's going to be 1 minus 3 plus 6 which is 3. Then we have minus of minus 2 which is plus 2 minus 6 plus 2 and you have 6 minus 1. So that's going to be 3 minus 8 plus 5. That's going to be 0 anyways. So it cannot have exactly one solution. So this implies it should have infinitely many solutions. If it has infinitely many solutions that means all the determinants that you make from remember Kramer's rule. So if d is 0 in order to have infinitely many solutions d1, d2, d3 etc. They should all be 0. Okay. So let us make d1 0. d1 is basically a determinant formed by the first column from these elements. So you have minus 2, 1, 1 minus 2, 3 minus 3. This should be equal to 0. Okay. So let us expand this. So it is 8 times minus 3 minus of minus 6 is plus 3. Okay. So we have a plus 2. We have minus 3b plus 2c and you have 3b minus c equal to 0 which actually gives you 3a minus 6b plus 4c plus 3b minus c equal to 0 which is 3a minus 3b plus 3c equal to 0 which actually results into the condition a minus b plus c equal to 0 which actually is number 4. Okay. And since it has to be one option correct so you can just stop here. You don't have to do d2 equal to 0 and d3 equal to 0 but however if you want to do that you will end up getting the same condition. So don't waste your time. Once you got one of the options correct and you know it's single option correct. So move on. So option number 4 is going to be the right option in this case. Oh I'm sorry. Option number 2 is going to be the right option. My bad. Sorry. Option number 2. So the first person to answer this correctly was Sai. Well done. Let's move on to the 16th question. So a, b, c are angles of a triangle and the system of equations sine a, x plus y plus z equal to cos a. So this is one system. Another system is this and another one is this. So this has unique solution, no solution, infinitely many solution, etc. Whatever the following option will be correct. Okay. So Aditya's things finitely many solution. So let's move on first to find out the determinant obtained from the coefficient of these variables. So we have sine a, 1, 1. Then you have 1 sine b, 1. You have a 1, 1 sine c. Okay. Let's try to figure out what is this determinant. Let me do a simple operation. Let me first do c2 as c2 minus c1 and c3 as c3 minus c1. So let's take the third column and subtract it. Sorry, first column and subtract it from the second and the third. So it will give you sine a, 1, 1 and this will give you 1 minus sine a. Okay. And again we have 1 minus sine a. Okay. And then we have sine b minus 1, 0, 0, sine c minus 1. Right. So if you expand this with respect to let's say the first row itself, we will have sine a and this will be sine b minus 1, sine c minus 1. And again we have minus 1 minus sine a and this will become sine c minus 1. And then again we have 1 minus sine a and we have a negative of, negative of sine b minus 1. Now since we're talking about a triangle, right, you cannot have all of them as 0 because if you want to have all of them as 0. Okay. Then you need to have your, you know, a. Okay. For example, let's say b is 90 degree. Then you may have the first and the third one as 0. Okay. And 1 minus sine a and sine c minus 1 can never be 0 in that case. Okay. So this entire expression will not be 0. If it is not 0, you'll always find there will be a unique solution existing. Right. So your option number 2 will become correct in this case. So both of your answers are wrong. So there will be a unique solution existing. That's option number 2 is correct. So now let's move on to the 17th question. So this question says determinant of a minus b is not 0 and a to the power 4 is equal to b to the power 4 c cube a is equal to c cube b. Now a, b, c, etc. They're all matrices. Then we need to find out the determinant of a cube plus b cube plus c cube. Okay. So Moog says the third option is correct. Let's see what others have to say. Okay. 17, 2. All right. So let's do a simple operation first. Let's try to multiply a cube plus b cube plus c cube with a minus of b. When you do that, you get a to the power 4 minus a cube b. Then you get b cube a minus b to the power 4 plus c cube a minus c cube b. Okay. Now, since a to the power 4 is b to the power 4, these two will get cancelled. Okay. c cube a, c cube b. So these two will get cancelled. And a cube b is b cube a. So these two will get cancelled leaving you with a null matrix. Okay. So a cube plus b cube plus c cube times a minus b is equal to a null matrix. That means the determinant of this should also be equal to determinant of a null matrix, which is zero, which means a cube plus b cube plus c cube. Okay. Assuming that they all are square matrices, this should be equal to zero. And since this is not equal to zero, implies the determinant a cube plus b cube plus c cube has to be zero, which means option number one is going to be correct. Guys, it was a simple question. I don't know why you messed it up. Okay. I'll give you another one based on the same concept. Again, if a b plus b a is a null matrix, then which of the following is equivalent to a cube minus b cube. Okay. So third option is what wishes says and so does Simon. So let's check this out. So here, of course, we have this information given to us that a b is equal to negative b a. Correct. Now here the best way to figure it out whether which is equivalent to a cube minus b cube is to, you know, try expanding these terms and see which of them works. Since you all have given the answer as three, I will start with expanding the third one. So a plus b a square minus a b minus b square. So when I expand it, I get a cube minus a square b minus a b square. Then we get b a square minus b a b minus b cube. So I can take a cube minus b cube together. Now we have minus a square b. This term I can write it as b a into a, which is actually minus minus a b into a. Okay. So this term I can write it as minus a b a. Okay. And this term I can write it as a b square. So hope I've not missed out on a term. Just one second. I'll just revisit these terms. So these two terms are taken care of. Good. So minus a square b I'll write as such. Okay. In fact, minus a b square also I would write it as such. Okay. Now we have b a square b a square. I can write it as minus a b a. And we can write minus b a b as plus a b square. Okay. So these two get cancelled. Okay. Now here again, this b a could be written as this, which is again getting cancelled with this term. And yes, you guys are correct. Option number three becomes the right option in this case. Again, the first one to answer this was wishes followed by Simeer. All right. Let's move on to the next question. ABC are three matrices of the same order such that two are symmetric and the third one is Q symmetric. So we have been given that X is ABC plus CBA and Y is ABC minus CBA. Then XY transpose is which of the following option. Okay. So it says it should be a symmetric matrix. It should be a skew symmetric. All right. So let's discuss this. So we have been fine. We have been asked to find out XY whole transpose, which is actually Y transpose X transpose. Correct. So let's find out Y transpose. Y transpose means transpose of ABC minus transpose of CBA. Correct. Which is C transpose, B transpose, A transpose minus A transpose, B transpose, C transpose. Okay. Now we know two of them are symmetric and one, the third one is Q symmetric. So this will result into minus CBA. Okay. And this would result into ABC, which is actually nothing but your Y itself. Correct. In a similar way, X transpose would be ABC transpose, CBA transpose, which is C transpose, B transpose, A transpose, then A transpose, B transpose, C transpose. Again, this will be minus CBA and this will be minus ABC, which is negative of X. So Y transpose X transpose is nothing but Y into negative X, which is minus XY. So option number four is correct. So again, none of you are correct. Okay. So now let's move on to the... Okay. So let's now look into the 20th question. Okay. What about others? So guys, this is a smaller version of the question we just noted. So whenever these two informations are given, it's better to start with the product of these two, which actually gives you A cube minus A square P plus P square A minus P cube. Now it's already given as A cube is equal to P cube and A square P is equal to P square A, so everything gets cancelled, so it becomes a null matrix. So the determinant of this is going to be zero, which is nothing but determinant of this into determinant of this is going to be zero, which implies determinant of A square plus P square has to be zero since this is not equal to zero. Again, option number two becomes the right option in this case. So Puruvik was correct in this case. Quickly moving on to the 21st question. It's not as easy as it looks. Determinant of AB and determinant of BA are same if they were square matrices. So Aditya says option four. In other case, you can just take an example. Let's say we have a 2 by 3 matrix A, which is let's say A1, A2, A3. Let's say this is A4, A5, A6. And we have a 3 by 2 matrix B, which is let's say B1, B2, B3, B4, B5, B6. Now, let us multiply BA. When you multiply BA, that means you multiply a 3 by 2 with a 2 by 3. You end up getting a 3 by 3. So let us write down the elements of BA. So row with column. So that's going to give you B1, A1 plus B2, A4. And then we have B1, A2 plus B2, A5. Then you have B1, A3 plus B2, A6. And when you take the second row and then start multiplying, you get B3, A1 plus B4, A4. Then B3, A2 plus B4, A5. And B3, A3 plus B4, A6. Similarly, we have B5, A1, B6, A4, B3, A2 plus B6, A5. And then we'll have B5, A3 plus B6, A6. Now, if you see this closely as a determinant, if you treat this as a determinant, let me make this as a determinant over here. You would realize that this determinant could have been obtained by multiplying A1, A2, A3, A4, A5, A6 with the third row as 0. That means you can actually treat this as coming from two square matrices determinants, which is actually the same thing but you have added an extra row down over here and an extra column over here. So if you do the column into row multiplication, that means if you take this column and this row, you will end up getting this element. If you take this column into this row, you end up getting this. If you take this column into this row, you end up getting this. Same with second column, first row, second column, second row, second column, third row. Then third column, first row, third column, second row, third column, third row. You'll keep on getting all these elements. Now, actually it is nothing but determinant of 0 is multiplied with 0. So your answer is going to be 0, which is option number 4 is going to be the right option. It's a good question. It can be slightly tricky. So Amogh is the one who gave the right answer. Sorry, Aditya Mishra is the first one who gave the right answer for this. Well done, Aditya. This is a slightly tricky question. So here, this information is actually useless. This is of no use for us. Let's move on to the next one. That's question number 22 for the day. So mostly everybody is going with option number 2. The best way to arrange all this problem is to make a situation where you have a 3 by 3 square matrix. The sum of elements of each row should be 1. So the best option would be choose our identity matrix. So the sum of elements of each row over here is 111. So this adds to 1, this adds to 1, this adds to 1. So a square will also be identity matrix. Now the sum of all the elements of a square, of course all elements add up to give you 3. So option number 2 becomes the right option in this case. So most of you have answered this correctly. Let's move on to the next question. I think that's question number 23 for the day. If a and b are 3 by 3 matrices such that a b plus a plus b is equal to 0, then which of the following option is correct? So let's look into this. a b plus a plus b is equal to a null matrix. So do one thing, add a i to both the sides. So it becomes a b plus a plus b plus i is equal to i. If you look at it carefully, it is actually nothing but it's the product of a plus i and b plus i. What does it mean? It means that a plus i and b plus i are inverses of each other. a plus i and b plus i are inverses of each other. And if they are inverses of each other means you can also write a plus i into b plus i as b plus i into a plus i. It doesn't make any difference. In other words, you can write a b as b a because when you expand it, you will get a b plus a plus b plus i is equal to b a plus a plus b plus i, which means a b is equal to b a. If a b is equal to b a, then from the first expression, which you can write it as a plus b into a plus b, which will be nothing but a square plus a b plus b a plus b square. You can actually add, you can actually make these two terms same because a b is equal to b a, so it actually gives you a square plus 2 a b plus b square. That means option number one is going to be definitely correct in this case. And since only one option is correct, I'm not going to look any further. These conclusions cannot be drawn. I think two is not evident. We can only say a b is equal to b a. And of course, three cannot be said. So Aditya is the first one to give this answer correctly, followed by Atmesh. Is that fine guys? So let's move on to the next question, which is question number 24. Okay, so our move says fourth option. Guys, it's very simple again here. When you expand this, you get a square plus b square plus a b plus b a equal to a square plus b square, which clearly means a b is negative of b a. Okay, so determinant of a b is determinant of minus b a. Now, since both are odd order matrices, can I say it's going to be just negative of determinant of b determinant of a. Okay, now since determinant is not equal to zero. The only possibility is determinant of b has to be zero, because then only determinant b could be negative of determinant b. So which means option number four is correct. Guys, it was a simple question. Let's move on to the next one. That's a 25th question for you. So a is a square matrix given to you as follows and b is given as a joint of a. And c is five a, then find the value of determinant of a joint of b by determinant of c. A joint of b means what? Determinant of a joint of b means determinant of a joint of a, right? Which actually is nothing but determinant a to the power of n minus one square. So n is the order of it. So n here is three, so it will be determinant a to the power of two square, which is determinant a to the power of four, correct? And c is five a, so determinant is determinant of five a, which is actually five to the power of three determinant of a. That's 125 determinant a. So I can write this as a joint of b determinant by determinant of c as a to the power four determinant of a to the power four and 125 determinant of a. So one of them will get cancelled. So it becomes determinant a cube by 125. Now, what is determinant of a itself? Let us expand with respect to the first column. So let's use the first column for expansion. One times you will have three and plus two times you will have one. So that's going to be five. So it's going to be five cube by 125. That's going to be one itself. That means option number three is going to be correct. Yes. So Si give the right answers. I was the only one to answer this correctly. So it's not a waste time finding the adjoint and all you have to use the properties of determinant of adjoint. Remember, I had also told you that if you have adjoint of adjoint of adjoint of any matrix a. Let's say this is m times and a is a matrix of order n. Then your answer will be determinant a to the power of n minus one whole to the power of m. So just remember this is a very useful isn't so here we took adjoint two times. So you'll have n minus one whole square had I taken adjoint three times I would have got n minus one cube like that. If I take m times it will become determinant a to the power of n minus one to the power of m. All right. So moving on to the 26th question. Let NBB to non singular square matrices says that B is not equal to they should have written I over here. And AB square is equal to BA. If a cube is B inverse a cube B to the power n, what is the value of N? Okay. So Ram Prasad says option one. Let's see what others have to say. Yes. Anyone else? Okay guys. Let's let's look into this situation. BA is given to us as AB square. Correct. Right. Okay. Let's see what is A first. So if I pre multiply with B inverse, can I say A is going to be B inverse AB square. Correct. So if I multiply pre facto with B inverse, I'm going to get this. Right. Okay. Now what will be A square? A square, can I say it will be the same thing multiplied to itself? So let me simplify this. B inverse A and B square and B inverse will make a B over here. Okay. Now this BA can again be replaced with AB square. So if I do that, it will be again replaced with AB square. So it will become B inverse A square B to the power 4. So this is your A square. Right. Now if you see the pattern, when there is just a single A, you get B inverse AB square. When there is A square, this becomes A square B to the power 4. So by the same logic without, you know, repeating the step once again, I can say A cube will be nothing but B inverse A cube B to the power 6. Okay. I can see the pattern over here. Right. So the next pattern has to be A cube B to the power 6. So N value here, if you compare these two, N value has to be 6, which is nothing but option number 3 is going to be correct in this case. So be watchful guys. You can actually save your time and effort. You can see a pattern being built up. It's very important to observe pattern in mathematics. Okay. One of the greatest mathematicians, Srinivasan Ramanujan was great because he could see pattern in things. Alright. So we'll move on to the last question of the day here because since we started half an hour late, I'm extending it by 20 more minutes. So question number 27 we are working on. A is an idempotent matrix. Guys, let me tell you idempotent matrix is when A square is equal to A. So any square matrix is idempotent when A square is equal to A. Where I is a unit matrix of the same order as A, then find the value of alpha. Alright. So I'm getting response from Ram as 1, Vishis says 3, Sai Meer also says 3. Shallow will work this out guys. So first of all, since this is the inverse of this, I can directly write the very first statement that identity matrix will be I minus 0.4A into I minus alpha A. Okay. Let me expand it. So I minus alpha A and this will be minus 0.4A plus 0.4 alpha A square. A square again will be A because, you know, A square is A because it's an idempotent matrix. So I can write I as this. So I will get cancelled. So null matrix is going to be, you can take A common. So it's minus alpha minus 0.4 plus 0.4 alpha. Hope I'm not missing out anything. Minus 4 alpha plus 0.4 alpha. Right. Oh yeah. That's correct. Okay. Now if this has to be a null matrix, which means minus alpha minus 0.4 plus 0.4 alpha should be equal to 0. Right. That means 0.4 is equal to minus 0.6 alpha. That means alpha is equal to minus of 2 by 3, which is clearly option number three that is going to be correct. So since we have just one problem left off, we can complete that as well. Question number 28. If A and B are two non-singular matrices which commute, commute means AB is equal to BA. Okay. Then we need to simplify A times A plus B inverse, B again inverse AB. Okay. So I'm getting the response as the first option. So let us look into this. First of all here, there's a product of three matrices and we are inverting it so I can follow the law of reversal. So it will become B inverse A plus B A inverse. Okay. And here we have AB outside. So let it be like that. So here when you multiply, you get B inverse A plus I times A inverse. And of course here we have AB outside waiting. So we can write this as B inverse A plus I and this will become B again. So this will become B inverse AB plus B. And since they are commutative, you can write AB over here as BA. So it becomes B inverse BA plus B and B inverse into B will be I. So that will become A plus B again. So option number one is correct. So absolutely easy question. So wishes for the first one to get this right. So all right guys. So we'll end up this session. And those who have chemistry, practicals tomorrow, best of luck. Keep your cool. Okay. Bye bye over and out from Centrum Academy. Those who have chemistry, practicals tomorrow, best of luck. Keep your cool. Okay. Have a good day. Sorry. Have a good night. Bye bye.