A Matrix Determinant Problem

[@intellecta2686]

That feeling when you realize you’re watching a math tutorial to fall asleep easier 😅 Nice example 👍

[@mokopa]

I do the same! I watch it twice, then go to bed, trying to remember as much as I can and do the math in my head while I'm drifting off. So much better than enumerating sheep.

[@ramanunnikrishnan7354]

how can you sleep after that, it gets me so excited that sometimes my sleep gets lost

[@satishchaudhary7978]

@@ramanunnikrishnan7354 same happens with me

[@drpeyam]

*Screams in Vandermonde*

[@aashsyed1277]

how did you write in black??????????????

[@matteo23battini]

The matrix at 4:25 is the Vandermonde matrix. It has some curious facts for example the determinant it's equal to the product of (aj-ai) with 1

[@SimonClarkstone]

I never learnt these properties of the determinant formally, but they are amusingly clearly true if you consider how the determinant is the (signed) volume of a parallelepiped whose 12 edges are 4 copies of the 3 line segments given as the 3 columns of the matrix. Thanks to the channel 3 Blue 1 Brown for teaching me that.

[@drsonaligupta75]

I have another solution: We know that if a homogeneous system of linear equation has non zero solution then the determinant Of the coefficients is zero ( cramer's rule) So lets consider the system of linear equations t(x) + t^2 (y) + (1+t^3)z=0 For t= a,b,c (Like replace t by a , b, c we will get three linear equations of variables x,y,z) We want to find the condition on a,b,c given that the system of linear equation has a non zero solution ( a solution other than x=y=z=0) Which can be rephrased as: The cubic equation in t (z)t^3 + (y)t^2 + (x)t + z is satisfied by distinct numbers a,b,c for some x,y,z not all zero So a, b , c must be the three roots of this cubic as they are different Now by viete's relation product of roots = - (constant term)/(coeff. Of t^3) Which gives abc=-1 *** (The last conclusion can be said by saying the given cubic is same as z(t-a)(t-b)(t-c) as they have same roots and t^3 coefficient)

[@pravargupta6285]

well you can also use factor theorem of determinants to do easily: As at a=b, |A|=0 so (a-b) is a factor Similarly At b=c, |A|=0 so (b-c) is also a factor At c=a, |A|=0 so (c-a) is also a factor. Now by multiplying diagonals we see that the degree of resultant polynomial will be (1+2+3)=4 and it is homogenous too. So there must be another linear factor in (a,b,c). Let that factor be k(1+abc). Now substitute different values of a,b and c in |A| and find that the value of k is 1. Hence we get the result by multiplying all factors : (a-b)(b-c)(c-a)(1+abc) I am sorry I couldnot give a good explanation why and how to use this amazing method because I did not get it too! I searched up the internet but still didnot find anything. any help would be highly appreciated.!!

[@drsonaligupta75]

@@pravargupta6285 did you mean use factor theorem on individual determinants

[@pravargupta6285]

@@drsonaligupta75 The theorem actually states that if for some valuw of variable the value of determinant becomes 0 then (x-a) is a factor of the resultant polynomial

[@drsonaligupta75]

@@pravargupta6285 understood

[@AdoNir]

Writing a≠b≠c is incorrect, because it doesn’t mean a≠c, such that 7≠9≠7 is perfectly true. You should write a≠b, a≠c, and b≠c. (I lost 2 points in a my midterm in linear algebra because I wrote a≠b≠c instead of all the inequalities)

[@ffggddss]

It also works to write a≠b≠c≠a If there are 4 or more variables involved, then it's best just to say, "no two of which are equal," or, "all different." Fred

I thought the same! a≠b≠c don't implies that a≠c

[@prithujsarkar2010]

Woah , mind blown

[@davidepierrat9072]

"inequality" isn't the right term, something like 1

[@44hwxyz90]

You could just set a=c, the condition doesn't cover that lol

[@ЕгорКолов-ч5с]

🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔

[@kormosmate2]

Nope. It says a!=b!=c so a!=c also holds.

[@mokopa]

@@kormosmate2 1!=2!=1

[@user-en5vj6vr2u]

Máté Kormos the != is not transitive

[@gaurav.raj.mishra]

I always learn something new from every video you make.

[@DavesMathVideos]

A nice but useful little question!

[@ProfOmarMath]

Vandermonde 😍

[@helo3827]

you are the best! Yesterday I was just thinking when can you do a video on matrices, keep it up your videos

[@Sam_on_YouTube]

I got a 5 on BC Calculus and placed out of Calc 1 and Calc 2. My very first class in college, the professor with a hard to understand slavic accent writes a matrix on the board and does things with it I had never even considered. I though... oh no. This is going to be tough. I had a much easier time with Calc 4. I understood at least what I was supposed to be doing. My calc 3 (linear algebra) teacher was not very good and I just didn't get it. Then, when I got up to quantum mechanics the following Spring, suddenly I needed to know and use linear algebra and I remembered nothing. Where were you 20 years ago when I was in college?!

[@laurelcreek7]

You can check your result by seeing that interchanging any two of (a,b,c) you just change the sign of the det and that’s the same as interchanging two rows of the matrix.

[@fracaralho]

Why is the determinant of the original matrix equal to the sum of the determinants of the two matrices into which it was broken down?

[@eliseuantonio6652]

That's a theorem, I don't remember its name, but you can try proving it yourself by using Jacobi Theorem

[@xanthoconite4904]

Hi I'm wondering where can I buy the shirt you're wearing in this video, I can only find the cat version on your website.

[@TheBodyOnPC]

From where do we have that c minus a cannot be equal to 0?

[@pravargupta6285]

Well you can also use factor theorem of determinants to do easily: As at a=b, |A|=0 so (a-b) is a factor Similarly At b=c, |A|=0 so (b-c) is also a factor At c=a, |A|=0 so (c-a) is also a factor. Now by multiplying diagonals we see that the degree of resultant polynomial will be (1+2+3)=4 and it is homogenous too. So there must be another linear factor in (a,b,c). Let that factor be k(1+abc). Now substitute different values of a,b and c in |A| and find that the value of k is 1. Hence we get the result by multiplying all factors : (a-b)(b-c)(c-a)(1+abc) I am sorry I couldnot give a good explanation why and how to use this amazing method because I did not get it too! I searched up the internet but still didnot find anything. any help would be highly appreciated.!!

[@Liesse_SportSante]

Very good video ! Thanks

[@ramanunnikrishnan7354]

for any Indian in school, this is directly from NCERT and could come in your boards.

[@pragati9821]

That's why I found it easy 😂😂

[@dasamlan9874]

That's why i did it.

[@mohammedafridibeedi3730]

Ya it's too easy

[@aaravgavshinde]

This is the same example my teacher taught!

[@zerovibritannia6216]

I am learning Determinant at the moment and this video came up 😀😀

[@mathophile1912]

O my god I searched for the solution of such question asked in my book. I got this legendary solution

[@RyanLucroy]

When you factored out the abc, you just used the rule det(A*B)=det(A)*det(B) with A=diag(a,b,c), right? Was a bit confused and it took me way longer to figure it out than I would admit 😅

[@johannesh7610]

There are easy complex solutions. Let z ≠ 0 be an arbitrary complex number. Set a, b, c to the three complex third roots of z, or one less than those. All of these choices equate the determinant to zero, because of the last column. We can add 3 times the first two columns to it to get (1+a)^3.

[@lordmomstealer]

Sir jiii u are great ♥️❤️ from india🇮🇳

[@dbgk2342]

ur beard looks cool

[@onthecover5042]

Fun fact= If you add one to another one, you’ll get two

[@blackpenredpen]

whoa!

[@NovayaK]

Mind = blown

[@DhruvPatel-qe2yw]

What is the example of det(A+B)=/det(A) +det(B)🤔🤔🤔🤔

[@RyanLucroy]

@@DhruvPatel-qe2yw Take A=B=diag(1,1,1). Then det(A+B)=8, but det(A)+det(B)=2

[@DhruvPatel-qe2yw]

@@RyanLucroy ohh yeah thanks

[@alejrandom6592]

by a/=b/=c can we infere that a might be equal to c?

[@alexandre6881]

Awesome video!

[@EngineersCalculatorTechnique]

thank u sir very informative😍😍

[@alainrogez8485]

Can we prove the sum of determinants BpRp did in the beginning? Is it legitimate?

[@TheBodyOnPC]

Entire books have been written just about rules of determinants.

[@joacinhof3style37]

1:35 c^3 twice lol, he wrote ^3 cuz he said "third column" at the same time

[@ameersahi2168]

Why when you chnage coluom 3 by coluom 1 mulitply by -1?

[@peakpersona-yash]

Logical proofs and Blackpenredpen Both are awesome. 😎😎😎

[@ZyloSol99]

Goat.

[@Zeusbeer]

Why not just devide the (abc+1)*matrixdet = 0 by matrixdet on both sides to get abc+1 = 0 -> abc = -1 ?

[@ffggddss]

(I'm assuming you meant, a ≠ b ≠ c ≠ a.) After a bunch of reductions and factoring, I wind up with that determinant = (b - a)(c - a)(c - b)(1 + abc) Setting that = 0, with the inequality condition rules out any of the first 3 factors being 0; that leaves only abc = -1 So any three unequal values whose product is -1, solves it. @bprp: How ya doin? I heard a rumor you were hospitalized for a while. PS: Love the shirt! Fred

[@blackpenredpen]

Hi Fred, I was! I had surgery about a month ago to remove my appendix. Luckily it was just a one-day thing and I was fine after that (except for some usual pain from the wounds). Thank you. How are you? I hope you are doing well, too.

[@ffggddss]

@@blackpenredpen That was what I thought it probably was, from what I heard. Glad you came through it all right. One benefit of that, which most of the rest of us don't have, is you no longer have to worry about your appendix!! No complaints here, either. I and my housemate got tested for COVID today because someone at the shop her boyfriend works in, tested positive. Two results to come; one tomorrow, and a more definitive one in 5 days. I expect negatives. Meanwhile, I posted my first YT video! It's in the MegaFavNumbers playlist that James Grime (singingbanana) established. I'm not entirely satisfied with it; I need better equipment, and I was somewhat rushed because I was getting in so close to the deadline (Sep 2). I haven't watched all those yet; I think I saw you had one, which I will watch soon. Fred

[@sarthakjain1824]

Did same question before with a wayy longer method

[@integralof2880]

Nice example. ☺

[@priyanshsingh1753]

In that much space and time I'd solve it directly.

[@kusoae]

His beard is looking so weird 😂😂

[@خوارزميخوارزمي-ص1ص]

why not 3, 1/3,-1 and many values multiple it -1

[@MEBVishwaS]

I approached differently which gives me different answer. I maybe wrong. By manipulating the row i just got( a^3+ka^2+Ka+1) I solved it to get three different roots which a,b,c. So then the matrix becomes zero. -1 is one of the root. So divided it by X+1 to get quadratic equation. By the quadratic formula I changed k to get different values of b and c which should solve it. I don't know where I'm wrong.

[@shivansh668]

Make a vdo on the ques of abstract ALGEBRA

[@manamtiwari]

Can you make more on matrices and determinants please. 🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗

[@Dreamprism]

Looks good.

[@peakpersona-yash]

Is this possible that 1+1/n=1/n. 🤔

[@kusoae]

No

[@MEBVishwaS]

If n=0

[@kusoae]

@@MEBVishwaS n can never be zero since it would make 1/n undefined

[@MEBVishwaS]

I just makes it infinite.

[@kusoae]

@@MEBVishwaS 👍

[@mahfuzhasan576]

Woah you grew your beard out. Nice.

[@jayantverma2136]

Well i solved it a few days ago..its an standard indian ncert problem

[@Visputescooking]

You please actually start to teach precalculus till the calculus... Me and many people like me can just read the calculus in this age. Please let us see the calculus 🙏 please sir 🙏

[@blackpenredpen]

？

[@Visputescooking]

@@blackpenredpenPlease sir🙏 you can just open a brand new channel in which you teach(something).

[@sushantyadav7806]

Some different method please

[@cristianv2850]

The way I saw a≠b≠c is like this: We start out by using statements of logic. a -> ~b b -> ~c Taking the inverse of the second statement, we get ~b -> c. We can conclude that a -> c, thus A = C.

[@jumpierwolf]

Your old beard looks nicer.

[@hamiltonianpathondodecahed5236]

imma just gonna pretend dat I knew all that EDIT: And the description is killing me (≧▽≦)

[@langhuningtha8774]

A challenge for blapenredpen : √4x^3-√x=√x+192

[@Med_Alzubaidi]

❤

[@kushagrakapoor2105]

Am I the only one who feels that bprp is the quang tran of maths?

[@peakpersona-yash]

Fun fact: 2×3=6.

[@呂永志]

我記得 a不等於b不等於c 這種寫法，a可能等於c。

[@addicted7766]

tremendous

[@tomatrix7525]

Epico

[@rajeshmonu7606]

Mia ho

[@sohildoshi2176]

first

[@brunocardosodeoliveira3799]

24th :)