Linear algebra question paper tough aaaa⁉️yaru sona very easy .....

[00:00:01]Hi guys. Welcome back 3GB Max.

[00:00:05]So remember the last couple of months ago come back with the call for the personal reasons that I So I put my videos portal by the way. So any killing algebra question paper by and get tough on the money that you have to be basic thinking and I need to get a theory that I have.

[00:00:20]So I will either be easy and a question paper by the way get away money that I have. So let's see.

[00:00:27]So first question paper by the way front page is it done?

[00:00:31]So in the what you pattern change when I tell a one marks, three marks and 12 marks.

[00:00:35]So I will first semester in the method that I get the problem. Just what we have to see. Which of the following span R squared? So in case of the sums that I put in the span R squared values in the pattern that I put in the problem. 1 0 0 1 So in the end of the number sums that I put in the problem. So in the sum of the answers that I get it.

[00:00:57]So next dimension of the subspace. So in the dimension that I put in the problem the same as the problem that I put in the problem. So X equal to X equal to X minus 2 Y plus Z equal to 0.

[00:01:14]So in the problem that I put in the problem the dimension so in the problem that I put in the problem the variables that I put in the problem. So in the unknown variables that I put in the problem so in the problem that I put in the problem the equal to T that I put in the problem.

[00:01:27]So in the problem that I put in the problem so in the problem that I put in the problem. X equal to 2 S plus T so therefore X Y Z equal to X value 2 S plus T Y value S and Z equal to T So in the problem that I put in the problem the value that I put in the problem the problem that I put in the problem. So in the problem that I put in the problem the problem that I put So in the problem that I put in the problem the problem that I put in the problem the problem that So in the problem that I put in the problem the problem that I put in the problem the problem that I put in the problem the problem that I put plus next T coefficient. So T is good. T is the coefficient one.

[00:02:05]So second term T is zero. So third term one.

[00:02:09]So now the dimension by T and now so this is a one dimension this is a two dimension. So now the dimension therefore dimension two. So now upper answer one B one two. Answer B.

[00:02:22]Next so already the formulas by T and rank of T plus nullity of T equal to dimension domain. So this is a formula.

[00:02:31]So next the definition of the answer is a matrix A is a diagonalizable. So A is a diagonalizable A has n linearly independent eigen vectors.

[00:02:42]Next so now if we find the norm so if we X and term norm of X equal to So square one square one two square two and two square two. So total add a nine root nine value three.

[00:02:58]So if the inner product so it will be definition U V equal to zero vectors are orthogonal. Next so now positive definiteness one two.

[00:03:10]Or definition two marks important definition. So now definition and three marks concentration full and full one marks concentration and again.

[00:03:19]So positive definition and again again and again and again.

[00:03:23]Positive so this is the definition and again and again and again so again and again and again and again and again and again and again and again and again and again and again and again and again and again and again and again.

[00:03:33]Next QR decomposition of A gives. So now QR decomposition QR decomposition so what is mean by Q and what is mean by R and again and again and again and again.

[00:03:43]So Q is orthogonal so and R is upper triangular matrix upper triangular okay.

[00:03:50]So now we are going to number again and again and again and again and again and again and again and again.

[00:04:01]>> Next.

[00:04:04]To compute rank in MATLAB.

[00:04:08]So, to compute rank in MATLAB.

[00:04:18]Next, V and D. Equal to eigen year returns.

[00:04:24]vectors D equal to eigen values of diagonals.

[00:04:32]nine nine one marks may easier than you can do. So, nine maximum seven eight seven eight nine blue heart comment and then check whether is equal to is a linearly independent.

[00:05:08]So, matrix method put on the one So, matrix method when you put on the one one two three I'm sorry, one two three and zero one two and two three and then four. So, number matrix answer when the zero one is that is a linearly dependent. Not equal to zero and that is a linearly independent.

[00:05:41]You know A1 A2 A3 which is number one form one new So, A1 or the A2 or the value and A3 or the value find one more.

[00:05:50]So, other the value one there and you only get killer more than my zero and the that is the other the opposite in the zero one that only get in the answer on the linearly dependent and another so in the zero and the other than that only get in the linearly independent one.

[00:06:02]Next.

[00:06:04]So, number in the one that only get equal to zero and the so other than that only get linearly independent one linearly dependent one more. So, find a basis for the sub sub space W equal to XY Z X plus Y plus Z equal to zero in R cube. So, number in the every find one more.

[00:06:21]Second one choose every find one more same other than that one more number in the one more second the 12th one or choose one more. So, number in the every one more so in the one more values one more. So, other the values one more number in the one more basis dimensions one more. So, the number basis and other the bracket one more that is the other than that is the basis.

[00:06:37]So, that is the value three marks every one simple one more than that one more number in the one more. Find the matrix of the so number in the in the term one more matrix find one more. So, the other than that steps one more so number in the one more T1 E1 one more in the one more straight time in the one more in the straight answer one more. 3 minus 1 1 2 in the one more steps in the one more in the one more in the one more two marks in the one more in the one more.

[00:07:00]General 14 one more in the one more in the one more find eigen values and eigen vectors. Okay.

[00:07:06]Next find the angle between. General one more in the one more in the one more in the one more in the one more in the one more in the one more angle between one more in the one more in the one more in the one more.

[00:07:14]So, the one more. General one more in the one more in the one more in the one more in the one more in the one more so formula in the one more theta equal to cos inverse UV and norm U and norm V.

[00:07:28]So, number in the one more number in the UV in the one more in the one more in the one more.

[00:07:31]So, U equal to 1 comma minus 1 V equal to and 2 comma 3.

[00:07:38]So, general one more in the one more in the one more UV find one more so multiple in the one more in the one more.

[00:07:44]1 2 0 one two are one three are three minus UV value minus one norm U norm V square by UV law So theta equal to theta means angle between cos inverse UV value minus one norm UV norm root two norm three three are nine four 13 So root two root 13 equal to cos inverse minus one by root 26 So then you can divide by cos inverse put it by again.

[00:08:27]So I'm going to order values So I'm going to linear algebra I'm going to use calculator use one more So you are going to tell okay now you are going to find by the way again.

[00:08:37]Next one is 16th one.

[00:08:40]So verify Schwarz inequality for U equal to two comma minus one V equal to one comma two So UV Schwarz inequality the formula modulus U comma V less than or equal to norm U and norm V So I'm going to modulus UV two one are two minus one into two two minus two zero So I'm going to zero less than value root five root five so I'm going to zero value root five root five one day five one more So higher than that greater than that again It Schwarz inequality So you are already find least square solution normal question either next orthogonal diagonal is the new the matrix.

[00:09:58]Write any one software code to check if vectors linearly linearly independent or dependent It will be the answer correct. Three marks. Okay, well. So, apply Gram-Schmidt. So, when you apply Gram-Schmidt in the sum part of A using any software So, I will try my best to try to learn.

[00:10:29]20 very, very easy So, all the 10 of them are in the exam.

[00:10:35]You know what I'm saying? book like a question. Now, so I'm going to attend my exam. I don't know one two three four So, maximum many people will not do marks in the exam. So, it will be four.

[00:10:49]And then So, Cauchy-Schwarz inequality formula in the exam okay. Find least square phi.

[00:10:56]So, linear independent try to learn in the exam. Six. So, maximum six of them will be in the exam. So, the total of seven one marks And then six three marks So, you will get a mark in the exam.

[00:11:14]25 marks.

[00:11:16]So, everyone 12 marks very easy ma'am said it in the morning. So, very, very easy.

[00:11:23]The set of all vectors So, I am going to learn in the exam vector space.

[00:11:29]What are the eight conditions? Okay. So, I am going to learn in the exam. But, I will learn the eight conditions.

[00:11:38]>> And then basis and dimension.

[00:11:50]Six terms look at some length one and 12 marks.

[00:11:58]A blow easy on a question. And then basis it will be basis find on there.

[00:12:07]Diagonalize the matrix.

[00:12:10]So 12 marks up a confirm 2 12 marks easy one again.

[00:12:16]Inner product of space off here.

[00:12:19]Condition but you can I can tell me easy for four conditions other satisfied in the same next in the same again. So in the identity number prove next. So in the same first term in the same one easy done. So in the same one again.

[00:12:36]Apply the Gram-Schmidt basis vector into an orthogonal basis. And the orthonormal basis. So it is find on there. So you can again and again third unit one again question again and again it's okay. So in the same in the same first 12 marks easy second 12 marks also.

[00:12:54]And then find QR decomposition very very easy some formulas mistake one again and again very. Very very easy confirm 12 marks. And then it will be direct question least square solution.

[00:13:06]It will be 12 marks up again and again four four 12 marks again. Three 12 marks confirm again and again. So 25th question again and again just see. So same define the vectors and then Gram-Schmidt process.

[00:13:21]So in the one same easy again and again back option same easy question and again and again and again and again. Again.

[00:13:28]Create the symmetric matrix.

[00:13:44]diagonal matrix So, this is your final review of the linear algebra.

[00:14:08]So, 4 * 12 = 48.

[00:14:16]So, easy and I think the average students got 73.

[00:14:21]Okay, guys. Remember to follow up.

[00:14:23]Thank you.