JEE Advanced 2024 | Matrices And Determinants | Must Do Questions 🔥| Namrata Ma'am

SC hello Ninja Warriors things are getting pretty intense and J Advance is in quite a few days right and we are going to solve top questions from the high weighted chapter that is matrices and determinants right but before I start before I dive into the session please let me know whether I'm perfectly Audible and visible and also please bring your notebooks because J Advan mathematics questions you won't be able to solve without pen and paper right okay so yes I can see Aditya I can see Manoj I can see Janna okay okay like the class and share with your fellow J Advan aspirant so that we can have have maximum interaction and doubt solving over here okay okay good evening rajeshwari keep joining keep coming and keep liking the session okay so I'm here with the very amazing questions from matrices and determinant chapters that will actually not only give you concept Clarity but also boost your confidence and that will make you believe that yes ma'am we can solve J Advan questions okay all right so yes today session is espcially going to be confidence booster session why because the process or the approach we are going to use in J Advance question and you'll be amazed that okay can we solve such difficult questions just by option elimination or just by like uh uh using some tips and hacks from mathematics okay so the session is going to be super duper intense fun and uh you will get maximum takeaway from this session okay not only in terms of Concepts but also in terms of some smartness because sometimes when you're not able to get to the correct answer then what happens is what you can do is you can just use a little bit hacks or shortcuts to just eliminate such some options to get the final final answer right are we ready yes okay okay okay so time is so it session is going to be one uh 1 Hour 1 hour 10 minutes 15 minutes long okay all right so because we are having only one chapter matrices and determinants all right all right let us move on and talk about myself so over here I H I have over 10 years of teaching experience in both online and offline so I have taught many students who were preparing for J Main and advance and I have uh got top selections in various colleges you can see over here okay all right yes so that's about it and one more thing so if if you are actually looking for some great problems in mathematics especially if you are looking for the pattern-based problems because you are aware in J Advance you can see multi- select questions you can see Matrix type questions you can see numerical based questions as well okay so these books will get you some amazing questions from all the chapters over here this is J Advance rank accelerator books link is given in the description box and trust me for concept building for confid Ence boosting these books are must must have uh for your J Advance preparation okay all right yes yes yes okay so all right all right all right so let us start let us start with the very first question and the very first question is from J Advance 2023 yes this is from J Advance 2023 okay that is last year 2023 3 2023 okay all right yes yes yes clear okay let me just fix my mic over here let me just uh keep with myself in my hand okay so let us understand let us read the question carefully you know what I always say this in mathematics the very first few minutes of reading the question is very crucial right so if you read the question in rush or in hurry what you end up will what you will end up doing is you have to reread that question to understand it better right so the better way is to give some time for the question reading read the question nicely because anyway you are going to end up uh wasting your time in reading the question twice so read the question carefully slowly slowly Okay so let m is a i j where I comma J is 1A 2A 3 be a 3 cross 3 Matrix okay so basically m is a matrix from here also we can just understand or extract this information that m is a 3 cross3 Matrix right they don't have to give it but J Advance being modest they are giving this as well okay so m is a 3 cross3 Matrix you understand I how I read the question please there are just don't go ahead and just don't jump into and just looking for the solution only no no no just understand that how I approach the question how slowly I read the question obviously this is a class so I'm going to read it very slowly but still the question reading should be actually very very uh attentive and focused okay so m is a three cross 3 Matrix a i j is 1 if J + 1 is div divisible by 1 otherwise AIG J is zero okay so the first line the moment I read first line what I understood from the first and second half of the first line is first line is only telling me that m is a 3 cross3 Matrix okay and over here by reading this I can sense that okay M has only two elements either one or zero depending upon this condition so what I can do is I can actually form the M Matrix 3 cross 3 Matrix just by this information okay all right so I'll be doing that but let's proceed and understand what they are asking then which of the following statements are true so this is a multi- select question please understand this is msq multi- Select question and trust me multi- select questions are the most difficult ones right multi- select questions are the most difficult ones difficult than your Matrix match as well so I will highly recommend that please Focus first on single correct then Matrix then numerical and then multi- select questions right one or two questions may it may happen that question nature is easy but but the pattern wise if I talk about pattern wise multi- select questions are the most difficult ones why because even if you get uh like two correct uh answers but if you're not about sure about third you might end up losing marks right so that's why it is very tricky okay all right so now what they are asking all four options are correct so basically multi- select questions are nothing but four different questions over here but this question I would mark it as an easy question not a very difficult question question wise it is easy but pattern wise it is also difficult okay so now first question is m is invertible or not so we know that invertible uh are those uh Matrix are invertible whose determinant is non zero if determinant m is non zero then option A is true okay and if you carefully observe M minus 2 I these A and D are very easy J main level questions A and D right so you have to form the M and M minus 2 a is invertible right that means what they are saying is M minus 2 I determinant is non zero if it is non zero then we can find the inverse otherwise inverse is not possible Right getting it very first question very first question A and D are absolutely easy okay we can check A and D so let us quickly form M determinant okay M Matrix so what they are saying is a i j if if J + 1 is divisible by I so if I and J both are 1 then what will happen J is 1 so two is divisible by 1 definitely two is div divisible by one okay so 2 is divisible by 1 that means this will be 1 okay if J is so i a i j okay a i j so A1 2 so three is also divisible by 1 and four is also divisible by 1 so first row will be having 1 1 1 now what about this particular element so this is going to be A2 1 right so over here I is 2 and J is 1 so I is two clearly and J is 1 so if J is 1 two is divisible by 1 definitely two is divisible by one so that's why here also we'll be having one okay this is your one three is divisible by two no no no otherwise it is zero four is divisible by 2 1 right okay so this is how you're going to form your M Matrix okay so M Matrix 101 clear and last we are going to have what 1 0 1 0 0 1 0 how are we going to get so this is your a31 a32 a33 so your I is three now now 2 is divisible by 3 no because J is 1 so J is 1 so 2 is divisible by three no okay then then uh three is divisible by three yes and four is divisible by 4 is zero so this is going to be zero so that's how your m m is going to be like this this is your M understood how I figured out the uh value of M Matrix very easy formation of M is is very easy okay now determinant M we need to figure out quickly so the moment I find M what do I get 1 this is going to be minus1 and then minus one this is going to be uh okay + one and then one this is going to be okay all right one second the determinant value can be easily calculated by this so yes this is a 32 right so this is going to be min-1 and this is going to be 1 - 1 so this is going to be zero oh oh oh so determinant m is coming out as zero if determinant m is coming out as zero what does that mean that a is definitely incorrect m is not invertible okay m is not invertible all right clear okay m is not invertible that is the first thing I'm going to get now m is 1 1 1 1 0 1 0 1 0 let us check the option D so for that M - 2 I will be equals to what so you have to actually keep uh rest of the elements same except the diagonal elements so that that is going to be 1 one then 1 1 and then 0 1 right so 1 1 then we have 1 1 then we have 0 1 right and the diagonal elements what we have to do is we have to subtract one from the diagonal elements so this is going to be 0 - 1 - 1 so 0 -1 - one right so what did we do is we did nothing just oh sorry min-2 we have to subtract right so -2 if I'm going to subtract m- 2 I so this is going to be Min -2 then -2 then -2 so this is -1 -2 -2 yes -1 -2 -2 right so M minus 2 I so we are going to write M over here minus twice of I I is what one one 1 1 rest everything is 0 0 0 right so if we perform this we are going to get this now what we need to do is we need to just expand this so 0 0 will be 0 minus 1 then we are going to have Min -2 and + 1 we are going to have 1 that's it so we are going to have + 2 so three okay let me check it is is it coming out as 3 m- 2 I let me check m- 2 is -1 - 2 - 2 1 1 1 1 0 1 this is min -1 I made a mistake over here in copying yes this is minus one so this is going to be - 1 4 - 1 so - 3 will be there okay first element would be Min - 3 yes so this is going to be zero okay okay okay M minus 2 I is also yes sorry sorry but because of copying I wrote incorrect okay M minus 2 I very easy to calculate but this is it and you are going to get a zero over here that means D is also incorrect can you see determinant of this determinant of this is coming out as zero what does that mean that D is also incorrect D is also incorrect why because determinant of M minus 2 is coming out as zero only that's why it is not invertible but D option is saying it is invertible I hope that this is clear okay now B and C one by one we will understand B and C so what they are saying in B option is there exists a nonzero column Matrix nonzero column Matrix such that M * this column Matrix is equals to this m * this column Matrix is equals to this okay so we need to actually solve this particular equation let us do that option b so we are having 1 1 1 1 0 1 0 1 0 1 1 1 1 0 1 0 1 0 correct this is our M Matrix this is what is our M Matrix now we have to multiply M Matrix with what this A1 A2 A3 to get this so M * m * 1 0 1 0 1 0 m * A1 A2 A3 will give me what - A1 - A2 - A3 and this this Matrix should be non Z Matrix so there exist a nonzero Matrix such that this is possible so let us solve it okay let us check whether it is non zero or not that's what we need to check so row number one and column number one so A1 + A2 + A3 this is the ele first element of this particular Matrix and this is going to be equal to minus A1 so what do I get 2 A1 + A2 + A3 is equal to 0 so first element I'm going to forg get this second element now row number two column number one so A1 + A3 A1 + A3 will give me zero and what about this and this A1 + A3 will give me minus A2 sorry so this is going to be minus A2 so what we are going to get is A1 + A2 + A3 is equal 0 this is the second thing now what is third thing so this multiply this so clearly A2 right this is zero this is A2 this is A3 clearly A2 is what zero so a ni from here I'm going to get A2 = to minus A2 A2 = to- A2 this implies 2 A2 isal 0 so A2 is actually coming out as zero the moment I see A2 0 what do I get if I see A2 0 what do I get right if A2 is 0 from here if A2 is 0 2 A1 + A3 is 0 and A1 + A3 is also 0 so 2 A1 + A3 is = 0 and A1 + A3 is coming out as zero If I subtract what do I get A1 is also zero okay X second A2 is 0er and uh yeah A1 + A3 is 0 correct so what we are going to get is A1 + A3 is 0 so what we are going to get is A1 is equal to - A3 yes A1 is equal to - A3 is what we are going to get okay A1 is equal to - A3 what do what does it mean right so A2 is coming out as zero and A1 and A3 are having opposite time what does that mean that means that this column Matrix so one Matrix I can see this is the definitely zero if I mark it one and if it is minus one then I can get the answer over here yes this could be one this could be basically these two should be opposite typ sign then only this thing will be correct okay so what we are going to get is option b is absolutely correct there exist a nonzero column Matrix such that this will be equals to this option b is absolutely right okay and now what about option C please understand this option C is saying that set X belongs to R Cube what is r Cube all real values of X right all real values of x where so the set X MX is equals to 0 is not equal to Zer okay so can you see this set X is not equals to zero right the set X is non zero set okay the set X is non zero set what is this this is MX = 0 so this is basically a homogeneous equation so how do we solve a homogeneous equation MX = to 0 so if determinant m is equals to 0 then we are going to get a unique solution sorry non zero per unique solution and if determinant m is equals to Z equals to Z then what do we get infinite solution there is no scope of no solution in case of homogeneous so if determinant m is zero then we are looking for no solution solution for what this x so this x will be having sorry infinite now so X will be infinite solution right so this x Matrix will be having infinite values okay this x Matrix can be X1 comma y x comma y comma Z so X will be having infinite values can you see option C is absolutely right it is non zero option C is absolutely correct okay all right yes ruk that's what I wrote now ma'am A2 is equal to minus A3 a A2 is- A3 No A1 is equal to minus A3 correct see this is your M and this is your that Matrix if you multiply this A1 + A2 + A3 will be equals to minus A1 this multiply this will be A1 + A3 so A1 + A2 + A3 will be zero and this multiply by this A2 is equal to minus A2 so that's why A2 is coming definitely out as zero right okay clear okay this means A1 and minus A3 are equal so this is noninvertible okay because M minus 2 I is coming out as zero all right okay so that means this is option C B and C both are correct yes B and C both are correct all right clear okay so this is what your first question is but I always say that multi- select questions are the most difficult ones because even if you crack two options now there are two more to go so basically this multi- select question is combination of four questions okay all right yes so that's what I wanted to make you aware about so so that you Target only those pattern based questions which you confident on now we are having another question from J Advance 2023 and trust me this question will boost your confidence so this question looks like that so many things are happening matrices um Matrix match type uh match match the column type question is there but again uh match the column type question actually gives you some Liberty in eliminating the options okay all right yes so first of all let us understand what type of question so system of linear equations do you understand system of linear equations shall I revise tasa okay shall we revise system of linear equations so yes so for nonhomogeneous that means for such type of uh uh nonhomogenous for nonhomogeneous what do we say so for nonhomogeneous that non homogeneous mean you will see the constants involved over here right so XY Z you will see three equations x y z but on the right hand side constants will be there so non homogeneous so Delta if it is not equal to zero then you are going to get unique solution okay and if Delta is equals to zero then there are two possibilities either no solution or infinite solution so for no solution at least one of the rest of the Deltas that is either Delta X or Y or Z at least one should be non zero but for infinite solution all should be zero in uh ex uh including this okay so Delta equal to Z sorry for nonzero unique solution for equal to zero either no solution or infinite solution and in infinite solution all the remaining determinant should be equal to zero but for no solution at least one should be non zero okay this is what your overall uh concept involved over here is okay all right okay now let me just remove it and let me just uh Target let me just calculate Delta from here so Delta will be what 1 2 1 then 1 Alpha please calculate Delta carefully because there is no term related to y That's why I have written zero then 2 - 3 and beta let us check what is happening if it is zero or non zero depending upon we'll talk about this later okay so 1 0 so this is 3 Alpha - 2 Beta - 2 Alpha then + 1 we are going to have minus 3 and this is going to be zero okay so clearly 3 Alpha + 4 Alpha 7 Alpha - 2 Beta is = 3 or 7 Alpha - 3 1X 2 will give me beta okay all right all right so now we have understood can you see the uh list over here beta is equals to half of 7 Alpha minus 3 beta not equal to half only these two things are written okay so we can just clearly just segregate over here so if beta is equals to these that means Delta will be equal to Z Delta 0 so either no or infinite solution okay so if beta is equal then either no solution p and Q will go with no or infinite Solution please understand and R and S will go with definitely unique solution unique solution so let us just understand and talk about option elimination over here so yes R and S unique solution will match with r and s both will match with either one okay and uh yes so 2 and three cannot match with r then four or five can also 1 14 five R is also 145 and S is also 145 p and Q are no or infinite solutions so two or three right so 2 comma 3 and uh yes 2 comma 3 and also 4 five can also happen in case of p and Q okay four five can can also happen now let us have a look at the options so p is either three or two right p is either three or two so if I go with the p p is either three or two okay p is either three or two okay so no need of writing four or five so p is either three or two okay what about Q let us go to the Q so Q over here is 2 or 1 Q is in the options they are giving Q is either two or one let let me go back to one one means unique solution okay so youi no no no no this means Delta equals to Z it can't be unique solution so Q is in the options they are giving one or two so Q is definitely two right Q is definitely two Q is matching with two for sure right that we have observed by option elimination right understood yes okay all right so if I go two Q is matching with two what does that mean c is wrong C is wrong because Q cannot match with one and this is also wrong so only this and this are correct okay and in both the options s is four so we don't have to check four right p is already three this we have done Q is already two this we have done s is four this we have done only thing which will give me my answer is either R is 1 or five we need to check R is matching with one or five that we need to check r r r where is R so yes one or five one means a unique solution definitely uh it is going to give me a unique solution okay all right R is one definitely one is the answer what about five so for five we have to check now it is very obvious that R will match with one because it is definitely a unique solution okay so uh by without even soling in I can Mark a as the correct answer right a as the correct answer to make sure that five is not the answer and uh five is not the this is wrong R is not equal to five let me just show you one more thing over here so r r says Alpha is equals to 1 what is r trying to say is Alpha is one right Alpha is 1 so let us put Alpha equals to 1 and uh let us check for this these values x = -15 if I put very first equation so -5 Y is = -2 so - 4 -5 - 4 + 0 + 0 is equal to 7 is it correct is it correct so -5 - 4 and Z is 0 no it is not correct it is not satisfying that's why r five R matching with five is incorrect R cannot match with with five so that's why one is the correct answer by option elimination we have done but only by option elimination can you see only we have only uh solved one thing which is what this Delta that's it right just by option elimination we did it okay but but but I will show you the complete solution what if that because of the nature of the options we are able to do that but sometimes what happens is we are not able to reach to to the answer by looking at the option depending upon the nature of the option okay all right I hope that this is clear now and now we will solve the question nicely and we will understand that how to tackle such questions okay so I'll show you first thing okay I'll show you the solution okay so first of all we have calculated Delta and we have realized that Delta equal to zero so option p is this if B is equals to this that means Delta is zero okay and Gamma is equal to 28 in the very first okay gamma 28 is also we can see over here what does gamma 28 mean so if I talk about Gamma 28 so let us calculate d z let is calculate d z for d z what we need to do is these two things will be same 1 1 2 2 0 - 3 and this is 7 11 and Gamma let us calculate this so 1 0 and then so this is 33 - 2 then we are going to get gamma - 22 then + 7 - 3 is equal to 0 so 33 + 44 - 2 GMA is = 21 so we are going to get 77 - 21 is = 2 GMA so gamma is coming out as 28 okay what does that mean that d z will be equal to 0 when gamma is 28 okay so one more information we have received that from here d z is zero d z is zero d z is z that means Z equals to zero is the Sol one it can be the one of the solution and if this is so here also can you see D z0 and this is d z non zero okay so if d z non zero and this is also d z non Zer okay can you see Del Z d0 means what infinite or no Solutions okay Q if I talk about q d z equals to Z that means infinite or no solution Del no zero no solution because at least one of them is non zero if Delta is zero and even one determinant is coming out as non zero then we can clearly say that this is no solution so that's why clearly Q is matching with two Q is matching with two and if Q is matching with two P will match with three infinite solutions right okay so p and Q are sorted and this will actually be both unique solution we just need to check that either this or this or this is the solution for R and S what you have to do is you have to just so Z both Z is zero right so this is z non zero this is z non zero so definitely R will match with one how now please understand this one by one one by one okay so this is d0 and d z0 d0 means no solution or infinite solution d z 0 also indicate no or infinite solution right this is p now what about Q D equal to 0 d z non Zer that means definitely Q will match with two right this is definitely no solution what about this D not equals to z and d z which is what d z non zero Del Z non zero this implies Zed is also not equals to zero that means R is what unique solution but four and five are incorrect for R so only R will match with one okay all right and this is what yes so clearly this is Del not equals to Z that will give you unique solution and if so here d z is z this implies Z is z for this you have to check for this you have to check that four and five either four or five can be correct okay so five if you check so clearly the moment you put X = -5 and Y = 4 X = -5 y 4 Z 0 okay first equation is going to satisfy what about second equation so this is -5 and 0 so second equation is not going to satisfy okay right so this are not satisfying this is is still getting satisfied okay all right so first equation is not satisfying by this okay so four or five both are incorrect for R and S so s over here will match with what s sorry four is correct for yes four is correct for your uh s correct how so four Alpha equals to one you just put four let us check four for S four and S checking right so X = to 11 then we have - 4 is = 7 yes first is getting satisfied what about second 11 + 0 is = 11 second is also getting s what about third so 22 + 6 Plus beta Z okay plus beta Z so this is anyway going to be zero and this is going to be gamma gamma which is 28 yes definitely it is matching so s is matching with four okay all right clear so we did we did this question in two ways one way simply option elimination another way one by one we are just directly hitting that exactly what is matching with P right and we are reaching to the answer what is matching with Q what is matching with r what is matching with s okay exact thing we are approaching okay all right I have explained now uh system of linear equation already just before the uh question I have explained yes we give you schedule uh Ragu okay I have seen your query okay so a option is absolutely correct for this particular question now we are having again another question from J Advance 2023 only yes this is from J Advance 2023 okay all right okay this is also not a very uh difficult question not a very difficult question okay so let us understand determinant M denotes the yes denotes the determinant of square Matrix M first line is very clear okay now G is a function whose domain is first quadrant and co-domain is all real and G is defined by this F of theta F ofun by2 Theta minus okay what is f of theta the moment I see F of theta I am to are you are you scared too F of theta the moment I look two determinants Theta Theta Theta and once we calculate F of theta we have to plug in the value of f of theta and G of theta oh my God but there are some questions which look giant bigger right and also once we solve it can you see now they are attaching matrices with the quadratic equation concept But please understand this question is not at all difficult even in fact you will realize this is an alpha question okay if You observe carefully can you see this is sin Pi sin Pi is zero minus cos Pi by 2 oh my God this is also zero T Pi is also zero so this is useless can you see J Advan is playing with your brains they are just making the question bigger right nothing else so if they are adding like let's make question bigger that means question is not having any like power or strength the question is already done decoded bye-bye Tata right because they have to make this question tough the question is actually not that tough obviously they have pressure because they can't keep any every question tough they will have to keep some easy questions as well so this is what is an easy question I would say matrices determinant I will highly recommend this is a scoring chapter in J Advance okay understood all right now can you see this is zero so F of theta is only this can I calculate F of theta let me just do it over here F of theta very easy question F of theta is 1 by 2 and we are having 1 1 + sin sare Theta okay all right and now we are having minus sin Theta then we have minus sin theta plus sin Theta okay so this is going to give me zero over here and last we are going to have 1 sin s Theta + 1 so + 1 + sin s Theta so F of theta is going to be 1X 2 and 2 1 + sin s Theta oh my goodness F of theta is nothing but 1 + sin Square Theta okay F of theta is 1 plus very easy that means this whole picture is a scam J Advance has done a scam right just to like uh make question look bigger they have just did this else what they have they could have done is directly they have written F of theta is 1 + sin Square Theta are you understanding my point this is not making any sense this is only the uh like Gad Mauri we call in Hindi Gad Mauri Gad Mauri is like um it you are still doing work but it is not making making any sense okay but they are just making you work so this F of theta is coming out as 1 + sin Square Theta understood now the moment I write F of < by 2us Theta what would that be F of < by 2 2us Theta so that will be 1 + cos s Theta correct now if I plug in the values over here what would be the G of theta so 1 + sin Square Theta can you see 1 + sin s Theta - 1 + 1 + cos² Theta - 1 can you see 1 1 1 1 getting canceled out so mod sin Theta + mod cos Theta is your G of theta what is g of theta mod sin plus mod cos this is your G of theta okay so let us just understand the question now so this is G of theta now we are having actually two questions involved one from matrices second from quadratic and there is actually a bridge between those two questions okay PX be a quadratic polinomial whose roots are the maximum and minimum values of G of theta okay so the other half of the question is talking about a quadratic equation whose root are maximum and minimum value of G of theta okay so G of theta is clearly mod of sin theta plus please do not forget the modulus otherwise you will do absolutely incorrect okay so maximum value of this equ this particular thing will happen when Theta is equal to < by 4 so G of theta max value will give give you < tk2 1 by < tk2 + 1 byun2 2 byun2 that is < tk2 now G of theta minan for G of theta minan you have to put Theta in first quadrant only can you see because beginning may they have shown first quadrant so either you put 0 or Pi by2 then only we will get so either zero either Theta is 0° or < by2 in at both the points G Theta minan value is 1 now we are having a quadratic equation whose roots are one and whose first root is one second root is < tk2 so P of X is decoded and also what they are giving is p of 2 is 2 - < tk2 so P of 2 is 2 - Ro < tk2 let us use it to find the value of a p of 2 so P of 2 is a 2 - 1 1 2 - < tk2 and this value they are giving is 2 - root2 so can you see a is coming out as 1 a is 1 so P of X is is what x -1 and x - < tk2 now let us check the options what are they trying to ask okay so they are giving us some values they are giving us some values and on these values they are asking whether your function is positive or negative okay so what we can do is we can use method of intervals very easy so one root is 1 another root is < tk2 okay this is 1.4 something okay so my P of X will remain because leading coefficient is positive so it is positive negative and positive so if the value of x lies which is greater than 1.4 so the the value will be positive P of x value will be positive if the value of x lies between 1 and 1.4 then negative and Below one then it is what positive again so yes so we need to calculate the range over here 3 + < tk22 by 4 so 3 + 1 .4 by 4 so 4.4x 4 3 + < tk2 by 4 correct yes so obviously this is 1.1 1.1 actually if you see method of intervals Lies over here so that means negative so option A is correct multi- select question please understand this is a multi- select question which of the following is are true then what about this so yes if I calculate 1 + 3 < tk2 that means 3 into 1 4 so 3 4 are 12 and 4.2 by 4 so this is actually 5.2x 4 so this is 1 Point 1.43 1.3 again 1.3 actually lies where 1.3 Lies Over Here below like to towards the left of one by between these two value one and 1.4 so again it is it will give us negative so but the question option b is incorrect because this because it is showing positive okay what about this 5 < tk2 so 5 into 1.4 - 1 by 4 5 4 are 20 so 7 so 7 by so 6 by 4 that is approximately 3x2 greater than < tk2 this value is greater than root2 so positive yes option C should be positive because if my value lies greater than root2 now so it will show me positive value okay and last let us check the last thing over here 5 - < tk2 by 4 5 - 1.4 by 4 so this will give us what 3.6 by 4 so this is 0 something that is before one before one value will be positive so D is wrong can you see before one all the values will be positive so it is wrong so now we can say that A and C are the only correct option b and d are absolutely wrong okay a d c are correct d how D is correct 5 - < tk2 so the value is 5 3.6 but no no no D can't be correct okay only can you see yes so yes this is your P of 2 p of X is this so clearly A and C are correct okay all right yes clear okay now let us move on to the next problem next problem are we ready okay so we are having again Matrix type and this is from by the way J Advance 2022 they have actually repeated some pattern can we can you see that we have solved one question system of linear equations it was actually important for your J in examination as well and it is important for from your J Advanced perspective why because in 2023 they are giving a question from system of linear equations and in 2022 also they are giving a system of linear equations okay all right understanding this is actually very important point over here so yes first of all you can see that this is again nonhomogeneous type of equation in nonhomogeneous you can see some constant although this is zero this is zero but this is one so if all three constants are zero then we will call it as homogeneous if anyone is non zero then we will call it as nonhomogeneous this is nonhomogeneous okay what they are giving is pqr being nonzero real numbers respectively 10th 100th and thousandth term of harmonic progression okay yes so what they are telling us is that uh 10th 100 and thousandth term of a harmonic progression are given by PQ and R respectively consider system of linear equations over here so this is first this is second this is third okay and now what they are giving is three types of of things are giving Q by R is equal to 10 then what will happen P by R not equal to 100 P by Q not equal to 10 P by Q equal to 10 okay so first of all let me just write this is first first is what PQ by R = to 10 Q by R = 10 second P by R not 100 Third third is what P by Q not 10 and fourth will be P by Q = to 10 these are the four cases over here okay and also one more thing this what about this if I just take this into consideration so first equation is pretty clear second is pretty clear third if I divide the whole equation by pqr what do I get so X upon p + y upon Q + z upon R is equals to Z this is your third equation okay so this is actually better looking equation we have just simplified this now this is first this is second and this one is third you can just forget this okay all right now p is 10th term of HP so p is 10th term so if p is 10th term so overall P will be what 1 Upon A + 90 and Q will be 1 Upon A + because this is 100th term so 99d harmonic progression and R will be equal to 1 Upon A + 9 D okay all right so Q by R is equal to 10 Q will be what 1 Upon A + 9d and R will be a + Tri 9 d 1 upon right this is equals to 10 let us solve it out a +9 D will be equal to 10 a + 99 into 0 D so this is going to be 9 a and this is going to be uh uh uh 90 okay so a is coming out as equals to D in first Matrix type okay so this is a equal to D second similarly let us solve the second part P by R is not 100 p is what 1 upon so this is 1 Upon A + 9 D and R will be a + Tri 9 D that is 100 okay let's us cross multiply a + 9 D will be equals to 100 a + 900 D so this is going to be 99 a is equal to 9 9 D not equal sorry not equal so that means a is not equals to D A Part Two is saying a not equals to D what about part three p by P by Q not 10 P by Q not 10 what is p 1 Upon A + 9 D and what is q a + 99d is equal to 10 not equal to so a + 99 D is = 10 a + 90 d right so you going to get 9 a is equal to 9 D not equal not equal so a not equals to d a third is saying a not equal to D can you see third and fourth are same so this will say if this is not equal this will say a is equals to D okay all right all right so this is a not equal to D again and this is a equals to D we have decoded the equations at least by now okay not a very difficult task what do you think was it difficult to get these condition from this okay and now this x by P Y by Q Z by R because can you see first and second equations are sorted third equation has a problem because third equation is involving 1 by P 1X Q 1 by R so can I just solve the third equation in terms of a and d right so X by P + y by Q + z by R is the third equation PQ R let us put the value so this is 10th term a + 90 * x + a + 100th term 99 d y and a + Tri 9 d z equals to 0 also one more thing x + y + z is given as 1 and 10 x + 10 y sorry 100 y + 1,000 Z is zero and third is this equation right if I carefully observe over here so I'm going to get I'm going to get 9 D 99 D and tri9 D okay 9 999 so if I take D common if I talk about this if I take D common x + y + z okay plus 9 if I take D common not 9 if I take D common hang if I take D common so let's take D common huh D common if I take what do I get 9x + 99 y +9 Z is equals to Z right can I take help from the second + x this is 99 y + 9 Z and + x + y + z is equal to 0 so basically 9 x + 99 y + 9 9 Z is equals to - x - y - Z = 0 and this is a x + y + z okay so if I take x + y + z common what do I get a minus D and x + y + z = 0 okay all right so this is equation oh sorry sorry sorry huh stuck stuck stuck yes yes yes I am back I am back okay so let me just show you what I did till now what we have done over here so far not a very difficult question so yes first of all there are two things going on over here first is this all the options are Q by Ral 10 P by R not 100 p q not 10 P by Q equal to 10 so let us decode this one by one by assuming p q and R over here so this is your P because it is what 10th term of HP 100th term of HP and th term of HP just plug in the value solve it so basically what you're going to get is first and fourth are having same thing a is equals to d a is equals to D second and third are a is not equal to D are you getting my point yes okay and now the third thing is this the third equation over here so the third equation is so we have to divide the third equation by pqr so that you can get 1X P 1X Q 1X R now 1X P can be replaced by 1 upon uh a + 90 and similarly q and R so if I replace 1 by P I will get this is 1x P 1X Q 1 by R now ax a y a z let us take a side and 9x 99 y 99 y9 z d if I take common so if this is our thing over here so can I just see the second equation second equation is what 10 x 100 y 1,000 Z can I take help from this so this is 9x 99 y9 Z + x + y + z so 9x 99 y9 Z can be written down as - x - y- Z and if I take x + y plus that common so this is your third equation so modification of third equation is what you are having over here rest everything will be fine let me just show you yes so this is nothing but a minus D and x + y + z = 0 okay now we are looking for whether this now H so clearly if a is equals to D then what will happen is the coefficient of the third will be zero are you getting my point if you if you uh make a determinant Delta a a is equals to D then coefficient of x y and z will become zero so Delta will become zero and if Delta becomes zero then what will happen either no solution or infinite many solution okay all right so if Delta is coming out as zero then what will happen this will be zero and x + y + z is equals to 0 right so if I go by the options let us look at have a look at the options so one and two sorry A and B one either t or Q is the correct answer can you see by the options one option one is having either t or Q option four is T or R let me go back option one is having either t or Q so t or q and option four is having what option four let us talk about option four t or R okay T or r so t or R yes so by looking at the first and fourth we are actually not able to get any particular answer because everything can be right if a is equals to D no solution can be true infinite solution can be true so if no and infinite solution can be true so for infinite pqrst so basically for option one p QR St everything could be correct and same as this is pqr St so basically first and fourth we are not able to make certain uh certain um uh conclusion over here but what about second and third so second and third definitely we'll have what if a is not equal to D understand if a is not equal to D then then x + y + z has to be zero so in both the cases what will happen is x + y + z has to be zero and can you see the first equation first equation is x + y + that equal to 1 so second and third will definitely have no solution so there is no other option S and S is correct for second and third both right this is very strict over here and which option is giving me second and third as s is only B option so clearly B option is correct this is from option elimination method we have actually done nothing just eliminated the options over here are you getting my point yes but now what if what if they give some options where in second s third s is given in any of the other as well another option as well so we have to be very clear conceptually as well right so over here yes so over here what we can do is we can just uh yes check can you see we can just check for this q and we can check for p as well right check for p and check for Q because this is anywh zero so x + y + z will be 1 so only these two equations will confirm us whether we are correct or not okay so if we put Z is equals to 0 what do we get if Z is 0 then x + y should be 1 and 100 x so can you see 100 by 9 and Y is - 1 by 9 no no no no no yes - 1 by 9 will be 0 okay Z is equal to 0 okay and if x is0 then 10 by 9 okay all right so yes now we what we need to do is we need to solve this x + y + z is = to 1 and we are having 10 x + 10 x + 100 y + th000 z is equal to 0 let us put Z is equals to Lambda let us put Z is equals to Lambda the moment I put Z is equals to Lambda I'm going to get x + y = 1 - Lambda and 10 x + 100 y = -000 Lambda put Z is equal to Lambda and now what we need to do is we need to get the value of X and Y in terms of Lambda only so 10 x ++ so can I multiply 10 over here or 100 over here yes let us multiply 100 in the first equation and subtract so what we are going to get is 100 x - 10 x is what 90x 100 100 gone and this will be equals to 100 - 100 Lambda + 1,000 Lambda so this is 100 minus you are going to get so can you see 1,000 - 100 is what plus 900 Lambda this is 90 x 0 0 0 cancel so X will be equals to 10 upon 9 + 10 Lambda correct 9 and this 10 Lambda this is your X right so if your Zed is Lambda if your Zed is Zer X should be 10 by 9 right so for option one if I talk about option one Q can be the correct answer okay but Q cannot be the correct answer why because if a is equals to the D then either infinite or no either infinite or zero solution for this we have a unique solution right so format is not matching that's why we have already concluded over here that this option is absolutely correct first Q fourth R first Q fourth R first Q first Q sorry first Q fourth R okay okay yes first Q is matching and fourth R is also matching infinite okay there is minus in two equation minus he n NE second equation which second equation plus plus orus and x + y = 1 - Lambda correct to okay and what we are doing is subtracting so 100 x - 10 x so 90x 100 - 100 Lambda minus- th000 correct th000 so this is plus correct absolutely correct okay so if I talk about first first answer Q yes first l l this is correct okay ma'am option two would be no solution right option two would be no solution you are seeing uh no no no no no you are getting confused no one more time let me just make it uh clear to you so a equal to D that means infinite or no solution for this particular what what all can be true over here is infinite or no solution right so infinite so P can also be correct Q can also be correct r s so for option one p q rst or R st all are open and the same goes with this pqrst please understand but second and third will actually have what no solution so S and S this is fixed this is absolutely fixed second and third will have no solution right so yes that's what you're saying sorry sorry second and third is having no solution why it is having no solution because a is not equal to D so if a is not equal to D then x + y + z is equal to 0 if this is zero and this is one so can you see at both at the same time x + y+ plus Z cannot be equal to 1 and Zer at the same time these are two parallel planes if I If You observe that's why these two are no solution and we have to stop the question right away because the only option which is satisfying second and third s is this okay sorry Kushi ha that's what you were saying Cho now let us move on to the last uh topic last question over here from the matrices I hope that this is clear now last question from the matrices m is given this then which of the following matrices is equal to M to^ 2022 okay so this is actually absolutely easy peasy question from J Advance 2022 yes this is from Advance 2022 very easy absolutely easy okay so what they are saying over here is m is this and M raed to 20 022 so for this is pattern based question so m is this let me write m is 5x2 3x 2 - 3x 2 - 1 by 2 clear what about m² so m² will be 5x 2 3x2 - 3x 2 - 1 by 2 and one more time let us write 5x2 3x2 - 3x2 - 1 by2 shall we proceed Row 1 column 1 so 25 by 4 - 9 by 4 25 - 9 is 16 by 4 so 4 okay now Row one and uh Row 1 column 1 Row 1 column 2 5 3 are 15 minus 3 15 - 3 is 12x 4 12x 4 is 3 right now Row 2 column 1 -5 and + 3 -5 + 3 is -12 - 12 by 4 is what -3 okay all right last but not the least Row 2 column 2 - 9 and we are having + 1 - 9 + 1 is what - 8 - 8 by 4 is -2 okay one thing if you realize over here is okay so M square m Square m² m² is nothing but so this is m² so can you see this and this are exactly reverse of each other and here also 3x2 so - 3x2 right so can you see value is equal but the this is positive this is negative now what m to^ 3 m to^ 3 will be 4 3 - 3 - 2 okay and we are going to have 5x2 3x2 - 3x 2 - 1 by 2 so Row 1 column 1 4 5 are sorry this is 3 H 4 3 is are 12 12 by 2 and - 9 by 2 12 - 9 is what 3 so this is going to be 3x 2 okay sorry 4 3 is are 12 and 3 3 are 9 so 12 by 2 - 9 by 2 is it 5x2 sorry it is 5 5 4 5 are 20 - 9 20 minus 3 3 are 9 correct 4 5 are 20 and [Music] 3 3 is are 9 M Cube value let me check 11 by 2 yes so we are getting 11 by2 4 5 are 20 - 9 11 by2 then 4 3 is are 12 - 3 so 9 by 2 and this is going to be - 9 by 2 and what about the last so this is going to be - 9 and -2 so - 11 by 2 is it correct M Cube value let me check M Cube value is - 7 by2 over here I have done something wrong so uh this is going to be - 9 + 2 - 9 + 2 is 7 yes so yes seven seven now we need to understand this is a pattern based question so we need to understand can you see the very first term over here so 5x2 is the first term and four and then we are having 11 by2 so the first term if I talk about 5x2 then we have four then we have have 11 by2 so over here if we check so we have to add + 3x2 to get 4 and + 3x2 to get 11 by2 so first term is 5x2 Right a n will be 5 by 2 + N - 1 and D is what 3x2 so this is going to be 5 - 3 so 1 + 3 n by 2 so basically for m to^ n this is going to to be 3 n by 2 + 1 okay 3 n by 2 + 1 so the moment you put what you have to figure out 2022 so the moment you put 2022 over here so 3x2 into 2022 + 1 that is going to give me 3033 + 1 3034 so this could be the correct answer this could be the correct answer C and D are wrong and also one more thing these two should be same yes so out of A and B clearly option A is correct why because can you see the pattern says this is positive this is positive so this element throughout is positive this element throughout is positive 9 by2 so that's why this is correct but what you can do is if you want to form a pattern so you can do that as well 3x2 right second term 3x2 3 then 9 by2 clearly by looking at this this is 3x2 into n so this is 3x2 into n and this is - 3x2 into n right and this is 1 - 3 n by 2 correct sorted right so the moment you put 20 22 what do you get M to^ 2022 is actually matching with option a easy peasy yes clear clear this was in J Advance 2022 now you just tell me one thing I want you to solve each and every problem one more time to get a better Clarity and let me know that what do you think because these are not some previous year question these are recent year questions 2022 2023 what do you think matrices can be do you think it is scoring do you think you would be able to solve questions from uh matrices and determinant this is an important chapter by the way and can you see how many questions how many questions in 2023 so many questions right three four questions in 202 three three four questions in uh 2022 right so this is a again High weighted chapter you will see at least two to three questions in your uh uh J advance so this is an important chapter and also this is a scoring chapter as well I agree there are certain questions which might be challenging right but overall if I talk about in general sense then this is a scoring chapter exception can be there it can be there that out of 10 years you will get that paper where mat is determinant they are giving tough questions right but usually if I go by the previous year Trend analysis it is likely that the questions will be this level only and this is not a difficult level question right yes okay so that's about it about today's session and I know that you are talking about algebra and calculus so we'll be uh doing the questions from uh calculus and algebra as well right pattern based questions we will do do not worry about that now I will take your leave this is meam signing off I'll see you all super soon

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Id: UUhgMzEHs8U

Length: 75min 1sec (4501 seconds)

Published: Tue May 14 2024