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!
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!
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
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 !
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
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
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?
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
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
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
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 !! ;-)
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 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
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 ?
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.
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) )
Is it a coincidence that one of the rows of the inverse don't have the nasty 13 denominator, or is something deeper going on? It's suspicious to me that, in the cofactor matrix, the multiples of 13 happen to line up.
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!
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.
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 🇬🇷
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
please note i do not think the last method applies to matrices greater than 3 x 3
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!
I really appreciated the 3rd version! Many thanks for that!
Szerintem is ez a nyerő.
I have an Algebra exam next week, really appreciate these videos you are uploading.
Greetings from Chile!
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.
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. ??
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
Peyam - Funniest math teacher. Bprp - Coolest math teacher.
the last method is the life saver!!! :))
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
That last trick is so cool! I wish my lecturer taught me about it!
That 3rd method is actually very useful, thank you for showing that
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
As a HS math tutor, you are very entertaining!
Watching this while doing my linear algebra homework on inverse matrices
Ditto
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?
24:25: "It's not a new way"
Title: "Inverse of a 3 by 3 matrix (3 ways)"
The second method is best. You won't need to calculate the determinant separately if you don't have it.
Sir I found 3'rd method the best .Congratulations for it .DrRahul Rohtak Haryana India
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
Woohoo, I’m inverse ready 😇
do you play the sims ?
This video is so good, now I'm ready for tomorrow's exam, thx a lot
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.
One more reason to start watching bprp is that he is now making Linear Algebra videos
I am actually looping 9:40,13:42,15:40 those charming laughter
3nd way is very clever, thanks Steve!
Gracias por compartir sus conocimientos maestro redpen 💪🙌
matrices bless you man, thanks for this dead cool video. Much appreciated
I know all of the method.
But, i like your way of teaching ♥️
This is AMUUUSING! thank you! i love this trick
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
I think I like method #3 the best for manual longhand calculation, but #2 as the easiest to program..
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.
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
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.
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
Here's the ‘how to determinant‘ video: th-cam.com/video/TkNeDxoRikY/w-d-xo.html (Maybe put that link in the description, 曹?)
The method used in the thumbnail was already taught by my teacher last year
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 always used to use the second matrix. Thx for this
Thank you Very much!
You should cover pseudo-inverses!
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?
The third way is excellent. May I teach my students this method?
just happen to be taking linear right now so thanks for uploading! w00t w00t
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
21:00 MAH PROFESSOR IS GONNA MELT LOL
11:00
"either you like it or you hate it"
Clearly hates it
Can you do proof for second method? Thank you
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..
It's not adjugate, it's just adjoint.
Very good...👏👏👏👏👏👏from Brazil...
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 ?
"6 - 2 is... Why is that so hard?"...FELT!!!!
Superb video brother
Love your videos man❤️
"dididididida" (delete this, delete that)
Love ur vids, keep going on !!
I don't even know linear algebra but I am watching this because it seems smart.
Congratulations from Brazil.
WHATTT the third way is actually witchcraft. I have been wasting my time doing the second-way smh.
The last one really made me happy
thanks to this video I have coded an nxn inverse calculator in python. The determinant function was the trickiest, since it's recursive.
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 the same matrix as the o e for the determinant trick. Is it special?
The 1st method that you've done is Gauss-Jordan method
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
does the third method work for matrices with range > 3?
Does this work for all matrices for n x n matrices with n > 3 ?
can somebody clarify the rigorous name of the 3rd method in order to pre-quote the method before solving the exercise.
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) )
Good presentation !
Fun fact: Inverses can be found vertically using column operations.
You should do topology or abstract algebra!
I feel like a form of cryptographic key could be constructed with matrixes somehow.....maybe this will inspire me for the next week.
The 2nd way is familiar, and the third one is rather peculiarly interesting.
I wish I watched this video yesterday. before my linear algebra final😂
The sad part is i see this when i already completed my linear algebra course :'
很久沒看你影片了，怎麼突然留鬍子了XD
dude thanks so much perfect timing
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
Now that you have used a blue pen, are you going to rename your channel to blackpenredpenbluepen?
What is the name of 3rd method?
Is it a coincidence that one of the rows of the inverse don't have the nasty 13 denominator, or is something deeper going on? It's suspicious to me that, in the cofactor matrix, the multiples of 13 happen to line up.
Question: Do all of these methods work for matrices that are larger than 3x3?
The first method has to work for square matrices of any size (as long as their determinant is not zero of course)
3rd is nice
I've always loved method two.
Needed this
Cayley-Hamilton Theorem go brrrrr!
What is the application for finding the inverses of a matrix?
Well, if you know A,C and that C=AB then the inverse of A times C gives you B.
@@Apollorion I understand that. But what about an engineering application or a physics application?
@@michaeltimpanaro5622 Eh.. ever heard of CT-scan and MRI-scan?
@@Apollorion Of course. Ever hear of honest inquiry. Can you explain it?
i like the 3rd way the most
Just when I need it ❤😭
Way 3 is better if you have the determinant if not then gotta go with way 1.
The Ist and 2nd methods are too hard! The 3rd one is easy and super tricky! Where the hell you learnt the 3rd method???