I am trying! It's been over 14 years that I haven't done any of it, lol. So I am only starting with the computational part first and then I will get into the more conceptual part.
You have the best timing!!! I literally learned this topic a few days back and always complained about how long it takes! Your third method is awesome! Thank you!
The first way is the more "theorically understandable", but the third way is the coolest to perform. Once you have computed the adjugate, you can also ignore the last raw and column and use the centre to compute de determinant (if not previously done). So, an "all in one" method !
I have no idea about the pretty 3rd method !!! Thank you 🙏!! I’ll give it to my students next monday !! Very nice !! (Like the D.I. Integrate method 😉)
@@wduandy so a 3×3 matrix has a characteristic polynomial like this A^3+a_1*A^2+a_2*A+a_3*i=0 ,multiply by A^(-1) and we get A^2+a_1*A+a_2*i+a_3*A^(-1)=0 and from there you get A^(-1)
@@SimonClarkstone you compute the characteristic polynomial,det(A-x*i),but since A is a 3×3 -a1 is the trace(since the sum of the eigenvalues is the trace),-a3 is the determinant(product of eigenvalues),all those observations come from vieta's formula,a2 is a bit more tricky,is the sum of all 2nd degree diagonal minors,or just compute det(A-x*i) and those a_i will come naturaly
Another way to find inverse is by using Cayley-Hamilton Theorem, which gives |A -λ I | = 0 , where I is a unit matrix. When we evaluate this determinant we get an equation of degree n , where n is the order of A. The equation is in terms of λ, so replace it with A. Voila! we get an equation with variables being the matrix and constant is the unit matrix. Multiply by A inverse and get simplify the rest o the terms to evaluate A inverse
I just finished this topic in school , finding the inverse of 3x3 is such a pain for me because I always make stupid arithmetic blunders. Just got to be careful
There's a guy with a goat beard who holds a pokeball, has the Picasso painting The Scream and is talking about matrices... Pure Excellence! Greets from Greece!
This technique applies to any sized matrix with an inverse. It is the matrix algebra equivalent of doing simultaneous equations as usually taught to students before they meet matrices.
Fourth way: apply the BPRB technique but the second matrix is not the unit matrix but this: Det 0 0 0 Det 0 0 0 Det This gives you the transpose matrix in the second example. Remember to divide the integer matrix you get by the determinant of the original. You can either divide each element, or just write a scalar multiple of (1/Det A) in front, depending what you are about to use the matrix for. This offers an insight about why there is no inverse when Det = 0 because you'd be dividing by zero... I prefer this fourth way
a_{n}=1 a_{m}=-1/(n-m)(sum(a[j+m]tr(A^{j}),j=1..n-m)) This will give you characteristic polynomial and from Cayley Hamilton we will get the inverse This is not as fast as elimination but faster than cofactor method
Great! but in the second method, you could use Cofactor Matrix to evaluate Determinant easily! so I think the second method is much more faster than the first one.
I’ve a little "improve", making the T operation over the A matrix at the first, and then work with it. You'll avoid the final arrangement for making the T. I'm based on the property Adj(A^T) = (Adj(A))^T. That's only a suggest !! ;-)
i remember being good at matrix in college. i remember doing that second method. this was more than 5 years ago. the only chapter that gave me hope of being good at math xD
thank you so much sir, you made it look easier! I just want to ask a question, regarding the 3rd way 24:30, can I use it still when solving for determinants with 4 x4 or more matrix?
Unfortunately, I knew this before u could upload this, but it is always love to see you. PS: You and Quang Tran look alike And I love you both. One for Maths One for Mukbangs
Great video...actually Today i was trying to find some more ways to calculate the inverse of a matrix and you helped me a lot. thank you...but now I'm wondering how to compute inverse of a (n by n) matrix where, n is any unknown positive integer Please share how to do this
The last method is very beautiful for optimazing the inverse. I really want to use it in an exam, but i think that i need to demostrate it. Could you please help me, please? Thx
I have a doubt on characteristic equation of a matrix..For a 3x3 matrix A , we know that sum of eigenvalues = trace of A(sum of diagonal elements of A), and product of eigenvalues= determinant of A..For a 3x3 matrix,is there any significance of sum of product of eigenvalues taken 2 at a time? (i.e. (coeff of A) )
Can the 3rd method be extended for higher order matrices as well? That is, copy first 3 columns and rows for 4×4 instead of 2 which is for 3×3. And then take determinant for each 3×3 matrices formed inside
I think that will work for sizes of an odd number (but not even number) because when a column in a square matrix of size of an odd number is shifted to the opposite end the determinant doesn't change sign.
If you have a prime determinant, you usually end up with that as the denominator of a fraction in at least one row. If you have a compound determinant, you could have that as the denominator in one row, or you could choose to have fractions in different rows whose denominators multiply to that number. There are occasional exceptions to both the above.
What if you have a 4 x 4 or 5 x 5 or anything like and n x n where n is greater than 3. Do we still add the two columns and rows to expand the matrix for the shortcut method ?
3rd method is great. To me, second method is way worse then the first one because it is a lot harder to compute and it is harder to understand why it gives you the correct result in the first place
Can you please show a way to turn an infinite product into an infinite sum...... Basically I wanna know a relation between f(x) and g(x) so that (∑ [g(n)] as n=k to ∞)=(∏ [f(q)] as q=c to ∞)
@@SimonClarkstone no that would not work .......you can try yourself .. In order for g(n)=log(f(n)) to work You have to take the log of both sides which isn't shown in the equation.
Ah, you're right; my idea would require an extra log outside the Π. Sorry about that. There are an infinite variety of functions g(x) that could satisfy it for any f(x). For example: g(n) = [ the correct product when n = k [ 0 for all other n
@@SimonClarkstone at first I also thought that g(n)=log(f(n)) would be a solution to my question...then later I realized that it is not. Actually I want an expression for g(x) in terms of f(x) ..so that if I have a product and I need to turn it into a sum,I can just substitute the expression.....and samely of I need to turn a sum into a product ..I can just substitute the expression for f(x) in terms of g(x).
I like the fact that you are bringing linear algebra to your channel but I hope you will focus more on theory, definitions and theorems (with proofs) and not just doing pretty standard exercises
Gaussian elimination sucks, it’s a bit trial and error and if you take the wrong route you go into a black whole and can’t go out of it
It's fun that you're embracing linear algebra
I am trying! It's been over 14 years that I haven't done any of it, lol. So I am only starting with the computational part first and then I will get into the more conceptual part.
@Leonhard Euler I am a real big fan of you Mr. Euler.
But I cannot subscribe your channel.
Because you are faking
@@ranjitsarkar3126 who?
The last method is Gold.
Thanks so much.
You have the best timing!!! I literally learned this topic a few days back and always complained about how long it takes! Your third method is awesome! Thank you!
The first way is the more "theorically understandable", but the third way is the coolest to perform. Once you have computed the adjugate, you can also ignore the last raw and column and use the centre to compute de determinant (if not previously done). So, an "all in one" method !
Sorry do you mean removing the first column and last row? Because that works out, whereas what you mentioned doesn't work. ??
When you drew that big matrix for C that face you made was hilarious knowing we all are suffering from the inverse hahaha
lol, imagine if it was a 4x4 matrix
@@blackpenredpen The third way could work for 4x4 matrix as well?
@@puneetmishra4726 I guess it will work when n is odd. When n is even, the plus minus sign would mess up.
I have no idea about the pretty 3rd method !!! Thank you 🙏!!
I’ll give it to my students next monday !! Very nice !! (Like the D.I. Integrate method 😉)
Thanks for liking! Cheers!
My teacher already taught me these methods 😎😎😎😎
I like the 3rd method the most too!
the last method is the life saver!!! :))
Another method would be from the characteristic polynomial,by writting A^(-1) as a linear combination of A and A^2
How?
@@wduandy so a 3×3 matrix has a characteristic polynomial like this A^3+a_1*A^2+a_2*A+a_3*i=0 ,multiply by A^(-1) and we get A^2+a_1*A+a_2*i+a_3*A^(-1)=0 and from there you get A^(-1)
@@MA-bm9jz How do you find a_1, a_2, a_3?
@@SimonClarkstone you compute the characteristic polynomial,det(A-x*i),but since A is a 3×3 -a1 is the trace(since the sum of the eigenvalues is the trace),-a3 is the determinant(product of eigenvalues),all those observations come from vieta's formula,a2 is a bit more tricky,is the sum of all 2nd degree diagonal minors,or just compute det(A-x*i) and those a_i will come naturaly
@@MA-bm9jz I don't know enough linear algebra to understand that unfortunately.
I really appreciated the 3rd version! Many thanks for that!
Szerintem is ez a nyerő.
The second method is best. You won't need to calculate the determinant separately if you don't have it.
That 3rd method is actually very useful, thank you for showing that
Another way to find inverse is by using Cayley-Hamilton Theorem, which gives |A -λ I | = 0 , where I is a unit matrix. When we evaluate this determinant we get an equation of degree n , where n is the order of A. The equation is in terms of λ, so replace it with A. Voila! we get an equation with variables being the matrix and constant is the unit matrix. Multiply by A inverse and get simplify the rest o the terms to evaluate A inverse
i love the last one u make it look really easy i will try writting a CPP code to compute the inverse using that algorithm
That last trick is so cool! I wish my lecturer taught me about it!
Peyam - Funniest math teacher. Bprp - Coolest math teacher.
I just finished this topic in school , finding the inverse of 3x3 is such a pain for me because I always make stupid arithmetic blunders. Just got to be careful
Watching this while doing my linear algebra homework on inverse matrices
Ditto
I have an Algebra exam next week, really appreciate these videos you are uploading.
Greetings from Chile!
There's a guy with a goat beard who holds a pokeball, has the Picasso painting The Scream and is talking about matrices... Pure Excellence!
Greets from Greece!
And runs Marathons 🇬🇷
This technique applies to any sized matrix with an inverse.
It is the matrix algebra equivalent of doing simultaneous equations as usually taught to students before they meet matrices.
I took Linear Algebra over the summer (and passed!) but I’ve never seen the 3rd way! Very useful and would have saved me a lot of time
One more reason to start watching bprp is that he is now making Linear Algebra videos
As a HS math tutor, you are very entertaining!
Fourth way: apply the BPRB technique but the second matrix is not the unit matrix but this:
Det 0 0
0 Det 0
0 0 Det
This gives you the transpose matrix in the second example.
Remember to divide the integer matrix you get by the determinant of the original. You can either divide each element, or just write a scalar multiple of (1/Det A) in front, depending what you are about to use the matrix for.
This offers an insight about why there is no inverse when Det = 0 because you'd be dividing by zero...
I prefer this fourth way
a_{n}=1
a_{m}=-1/(n-m)(sum(a[j+m]tr(A^{j}),j=1..n-m))
This will give you characteristic polynomial
and from Cayley Hamilton we will get the inverse
This is not as fast as elimination but faster than cofactor method
Sir I found 3'rd method the best .Congratulations for it .DrRahul Rohtak Haryana India
3nd way is very clever, thanks Steve!
This video is so good, now I'm ready for tomorrow's exam, thx a lot
please note i do not think the last method applies to matrices greater than 3 x 3
I always used to use the second matrix. Thx for this
The adjugate is good for rings without inverses because it always works. Though it might not be a real inverse, but good enough in a lot of places.
Great!
but in the second method, you could use Cofactor Matrix to evaluate Determinant easily!
so I think the second method is much more faster than the first one.
Woohoo, I’m inverse ready 😇
do you play the sims ?
I am actually looping 9:40,13:42,15:40 those charming laughter
I know all of the method.
But, i like your way of teaching ♥️
I think I like method #3 the best for manual longhand calculation, but #2 as the easiest to program..
All the method i was knowing 😅 but i love...i taught u might have other shortcut .....the 3rd is my favourite i use it every time its easy
Gracias por compartir sus conocimientos maestro redpen 💪🙌
I’ve a little "improve", making the T operation over the A matrix at the first, and then work with it. You'll avoid the final arrangement for making the T. I'm based on the property Adj(A^T) = (Adj(A))^T. That's only a suggest !! ;-)
The method used in the thumbnail was already taught by my teacher last year
i remember being good at matrix in college. i remember doing that second method. this was more than 5 years ago. the only chapter that gave me hope of being good at math xD
Thank you Very much!
The 1st method that you've done is Gauss-Jordan method
24:25: "It's not a new way"
Title: "Inverse of a 3 by 3 matrix (3 ways)"
This is AMUUUSING! thank you! i love this trick
thank you so much sir, you made it look easier! I just want to ask a question, regarding the 3rd way 24:30, can I use it still when solving for determinants with 4 x4 or more matrix?
Unfortunately, I knew this before u could upload this, but it is always love to see you.
PS: You and Quang Tran look alike
And I love you both.
One for Maths
One for Mukbangs
thanks to this video I have coded an nxn inverse calculator in python. The determinant function was the trickiest, since it's recursive.
matrices bless you man, thanks for this dead cool video. Much appreciated
Great video...actually Today i was trying to find some more ways to calculate the inverse of a matrix and you helped me a lot. thank you...but now I'm wondering how to compute inverse of a (n by n) matrix
where, n is any unknown positive integer
Please share how to do this
Thank you for you comment.
I think, unfortunately, once we get a bigger matrix, we have to use either method 1 or method 2..
You should cover pseudo-inverses!
Fun fact: Inverses can be found vertically using column operations.
The third way is excellent. May I teach my students this method?
I don't even know linear algebra but I am watching this because it seems smart.
Congratulations from Brazil.
For the 3rd method, the crossed out -4 is only used for the det. now?
And can you actually start anywhere but 1,1 (where the -4 is) is easiest?
Can you do proof for second method? Thank you
Superb video brother
I've always loved method two.
The last method is very beautiful for optimazing the inverse. I really want to use it in an exam, but i think that i need to demostrate it. Could you please help me, please?
Thx
The last one really made me happy
I have a doubt on characteristic equation of a matrix..For a 3x3 matrix A , we know that sum of eigenvalues = trace of A(sum of diagonal elements of A), and product of eigenvalues= determinant of A..For a 3x3 matrix,is there any significance of sum of product of eigenvalues taken 2 at a time? (i.e. (coeff of A) )
I feel like a form of cryptographic key could be constructed with matrixes somehow.....maybe this will inspire me for the next week.
WHATTT the third way is actually witchcraft. I have been wasting my time doing the second-way smh.
3rd is nice
The 2nd way is familiar, and the third one is rather peculiarly interesting.
Can the 3rd method be extended for higher order matrices as well? That is, copy first 3 columns and rows for 4×4 instead of 2 which is for 3×3. And then take determinant for each 3×3 matrices formed inside
I think that will work for sizes of an odd number (but not even number) because when a column in a square matrix of size of an odd number is shifted to the opposite end the determinant doesn't change sign.
@@ShinichiKudou2008 ah that makes sense, thanks
It's not adjugate, it's just adjoint.
You should do topology or abstract algebra!
i like the 3rd way the most
Way 3 is better if you have the determinant if not then gotta go with way 1.
just happen to be taking linear right now so thanks for uploading! w00t w00t
Very good...👏👏👏👏👏👏from Brazil...
"6 - 2 is... Why is that so hard?"...FELT!!!!
I wish I watched this video yesterday. before my linear algebra final😂
If you have a prime determinant, you usually end up with that as the denominator of a fraction in at least one row.
If you have a compound determinant, you could have that as the denominator in one row, or you could choose to have fractions in different rows whose denominators multiply to that number.
There are occasional exceptions to both the above.
Good presentation !
11:00
"either you like it or you hate it"
Clearly hates it
It's the same matrix as the o e for the determinant trick. Is it special?
He : inverse of a matri-
Me : *adj(A) / |A|*
Adjugate ? I learned it as adjoint . Well both are same anyways so doesn't really matter
2020 raise to the power 2019 - 2020 divided by 2020 square + 2021=N
then find the sum of digits of n
bro plz solve this?? trying from last 5 weeks
Is that the same as N = 2021 + [ (2020^2019) - 2020 ] / 2020^2?
What if you have a 4 x 4 or 5 x 5 or anything like and n x n where n is greater than 3. Do we still add the two columns and rows to expand the matrix for the shortcut method ?
The sad part is i see this when i already completed my linear algebra course :'
Love your videos man❤️
can somebody clarify the rigorous name of the 3rd method in order to pre-quote the method before solving the exercise.
3rd method is great. To me, second method is way worse then the first one because it is a lot harder to compute and it is harder to understand why it gives you the correct result in the first place
Here's the ‘how to determinant‘ video: ruclips.net/video/TkNeDxoRikY/видео.html (Maybe put that link in the description, 曹?)
does the third method work for matrices with range > 3?
Does this work for all matrices for n x n matrices with n > 3 ?
dude thanks so much perfect timing
Needed this
Plz answer this question
Q- integral of 2x^2.dx divided by (x-1) (x-2) (x-3)
Try partial fractions.
@@davidgendron6955 yes I applied that formula but
Here 3 algebraic functions are present how can i solve?
"dididididida" (delete this, delete that)
Love ur vids, keep going on !!
21:00 MAH PROFESSOR IS GONNA MELT LOL
PLEASE DO LINEAR ALGEBRA PROOFS! I need some enlightening or else I’ll fail my class
I have lots of playlists of linear algebra proofs in case you’re interested
Data science is very popular now and data science requires linear algebra. So, it is good time to learn linear algebra.
Can you please show a way to turn an infinite product into an infinite sum......
Basically I wanna know a relation between f(x) and g(x) so that
(∑ [g(n)] as n=k to ∞)=(∏ [f(q)] as q=c to ∞)
Wouldn't that be g(n) = log(f(n)) ? There are quite a few caveats, but that's got to be the stratergy in the general case.
@@SimonClarkstone no that would not work .......you can try yourself ..
In order for g(n)=log(f(n)) to work
You have to take the log of both sides which isn't shown in the equation.
Ah, you're right; my idea would require an extra log outside the Π. Sorry about that.
There are an infinite variety of functions g(x) that could satisfy it for any f(x). For example:
g(n) = [ the correct product when n = k
[ 0 for all other n
@@SimonClarkstone at first I also thought that g(n)=log(f(n)) would be a solution to my question...then later I realized that it is not.
Actually I want an expression for g(x) in terms of f(x) ..so that if I have a product and I need to turn it into a sum,I can just substitute the expression.....and samely of I need to turn a sum into a product ..I can just substitute the expression for f(x) in terms of g(x).
Just when I need it ❤😭
I like the fact that you are bringing linear algebra to your channel but I hope you will focus more on theory, definitions and theorems (with proofs) and not just doing pretty standard exercises