Thanks. You're so enlightening and wonderful, moreover, brilliant. You have heard this many, many times, I hope.

Professor Balagovic, thank you for a solid explanation on Key ideas in Linear Algebra. This is an error free video/lecture on KZread.

This video will be totally understandable once you reach the lecture 8 of prof strang.

@egor.okhterov

7 жыл бұрын

Gisari a secas I've watched only 7 lectures. Now this explains why I am so confused :)

@sounakbiswas2239

6 жыл бұрын

please provide me link to the lecture or at least title of the video

Rank(A) = number of cols - dimension of null space (explained in minute 6) = 2. So there must be 2 independent columns. C1 and C2 are multiples of b, they are not independent, the other independent column must be C1. Then it can't be a multiple of b.

@007myzorro

6 жыл бұрын

Daniel B f

Particular and special solutions? I don't remember have seen that in the second lecture.

@ohyonghao

6 жыл бұрын

I think that may be coming from Differential Equations, 18.03.

Great, thanks for the video.

how can you assume someone knows what Xp and Xs are????

Awesome video! Thank you!

great day to watch this video. thanks

2:53 In an equation which (for all we know) is only right if C=1, she substitutes b for Ax_p, which (for all we know) is only right when C=0. I can't say she explains it.

my best professor

@Ryan Hirst ... I agree ... or I'm also missing something. C1 can be arbitrary (including zero vector or multiple of b) because x1 is zero. Despite her claim about null space at the end, Matrix A can have rank 1 OR 2 and still work. In terms of the column picture, x never probes in the column 1 direction. Since x1 is zero, Range of A*x gives a (2D) plane (in 4D column space). If Range of A alone is also a (2D) plane (or 3D range in 4D space) then the earth is still turning.

@SilverArro

9 жыл бұрын

Ryan Hirst It's true that since the solution shows that x1 is 0, we can write c1 of A as any column vector that we like since it essentially disappears in the solution. This might lead to the assumption that c1 is entirely arbitrary and can be any multiple of b, but this is not the case. IF c1 was some multiple of b, then we never could have arrived at that solution to begin with. The fact that there is only one special solution indicates that our original linear system (represented by A) only has one free variable. If c1 was a multiple of b, then our system would have TWO free variables and thus, two special solutions. Try writing c1 as another multiple of b and then compute rref(A) and you'll find this to be the case. So, it's important to remember that because we're working backwards here, our solution gives the appearance of an entirely arbitrary c1, but it's important that our original c1 is such that the solution we're given IS the solution we would have calculated had we started with a defined matrix A. In fact, we can't say for sure what c1 is; we can only be assured that it is linearly independent of c2 and c3 thus allowing x1 and x2 to be pivot variables leaving a single free variable, and thus a single special solution. Ryan touched on this a bit when he said "the linear independence of c1 is enforced by its absence in the solution, not by the matrix A'. That's right! In other words, imagine that this was a traditional problem where you were given the columns of A and and asked to find the solution yourself. If c1, c2, and c3 were all multiples of one another, it would be impossible to arrive at the solution that we KNOW we get. Does that make sense?

Dr Martina Balagovic Department of Mathematics University of York Research group: Pure Mathematics Undergrad (University of Zagreb), PhD (MIT)

wow im so impressed right now

When she writes what she says is c equals 1, what kinda 1 is that? Is it due to Cyrillic? Alpha?

how do you get c2+c3 =b?! please explain

@mokzalla8170

5 жыл бұрын

I think , the dot product of A and Xp results in c2+c3=b because c1*0=0, c2*1=c2 and c3*1=c3.

Why does A have three columns?

Should have added this video later in this course. But still it was good learning.

I meant C2 and C3 are multiples of b.

@JustCurious637

7 жыл бұрын

I think Daniel is right. C2 and C3 cannot be independent because they are multiples of b. Did I miss something?

@dunga309

7 жыл бұрын

That´s right: C2 = s.b and C3 = t.b => C2 = (s/t)C3 => not linearly independent (s and t scalars)

great instructor! :)

Interesting problem.

Fascinating to watch her; did anyone else notice their vector becoming tensor?

She says A* xp = b when c=0, and after she uses it to show A*xs=0 when c=1. If c=1, A*xp is not anymore equal to b, isn't it ??!!!!!!!!!

hola amigoss

cute!

I teach myself math and I really like this Q&A format. But I think there is a slight mistake. If the complete solution is (written as transposes, sorry): x = [0, 1+2c, 1+c] then C1 is arbitrary, no? What's stopping me from writing C1 as a linear combination of b? C1 x Zero? :) The linear 'independence' of C1 is enforced by its absence in the solution, not by the matrix A. The solution allows each entry in the first column to be an arbitrary parameter. {am I missing something?}

@SilverArro

6 жыл бұрын

Ryan Hirst I see this comment is 6 years old, but I can’t resist. You are missing something indeed. C1 is not arbitrary at all. If it were linearly dependent, then Ax=0 would have another linearly independent solution. Thus, all multiples of that solution would be included in the complete solution to Ax=b. But we know the complete solution to Ax=b, and we know that it only includes multiples of ONE solution to Ax=0. Thus, if A were a rank 1 matrix, we couldn’t possibly arrive at the solution that we KNOW we get (remember, we’re working backwards here). I hope that makes sense.

Finally a legible black board. This Euro nerd does a good job explaining high school math.

damn girl are you a continuous function? because you ain't got no gaps ...

please, neatly, write things on the choke board! otherwise, viewers from other countries may get confused!