If you had to calculate 2^100 and 3^100 without a calculator, you would probably do exponentiation by squaring. That takes advantage of this exponentiation property: (a^b)^c = a^(bc). So by repeated squaring of your last result, you can compute 2^2 then 2^4 then 2^8 then 2^16 then 2^32 then 2^64. Once you have those, then you can do this operation to get 2^100: 2^64*2^32*2^4. That means that you can calculate 2^100 by multiplying 8 times! The first multiplication would be 2*2=2^2. The second would be 2^2*2^2 = 2^4. The third would be 2^4*2^4=2^8. The fourth would be 2^8*2^8=2^16. The fifth would be 2^16*2^16 = 2^32. The sixth would be 2^32*2^32=2^64. The seventh would be 2^64*2^32=2^96. The eighth and final would be 2^96*2^4=2^100.
Just came across your video, this is the first math video i've found on youtube that really helped me understand what I was learning! You are a great teacher thank you so much for your help and enthusiasm :)
Got this video in recommendation had a playback of first 30 seconds without clicking the video the great introduction made me open this in my other tab. Now I'm going To stalk your whole channel for Linear Algebra content 🤣
To raise anything to the 100th power, we multiply it by itself not 100 times, but 99 times. To square it, we multiply it by itself once. To cube it, we multiply it by itself twice... etc.
This channel is a gem! Great video quality, great lecture, great teacher with a great sense of fashion! One question though, shouldn't the inverse of the matrix be (1/det)[ d -b over -c a]?
So for any nxn matrix there are n! sets, of numbers, of cardinality n^2, that lead to its diagonalisation. I didn't know that before :-) (just messing ... good video)
I was taught that for these kind of questions First do A², ³, ⁴... until you see a pattern in these Analyse that pattern and then put for the exponent asked And it works, atleast the questions I solved were right
![ILLSLICK - KILLSHOT REMIX [FULL VERSION]](http://i.ytimg.com/vi/RIK8RrqIAzE/mqdefault.jpg)
when I think of maths, I think of a gentle man like him teaching. you really are the best
"Do not confuse yourself"
Great advice! 🙂👍
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Glad you liked it!
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If you had to calculate 2^100 and 3^100 without a calculator, you would probably do exponentiation by squaring. That takes advantage of this exponentiation property: (a^b)^c = a^(bc). So by repeated squaring of your last result, you can compute 2^2 then 2^4 then 2^8 then 2^16 then 2^32 then 2^64. Once you have those, then you can do this operation to get 2^100: 2^64*2^32*2^4. That means that you can calculate 2^100 by multiplying 8 times! The first multiplication would be 2*2=2^2. The second would be 2^2*2^2 = 2^4. The third would be 2^4*2^4=2^8. The fourth would be 2^8*2^8=2^16. The fifth would be 2^16*2^16 = 2^32. The sixth would be 2^32*2^32=2^64. The seventh would be 2^64*2^32=2^96. The eighth and final would be 2^96*2^4=2^100.
Thanks
you actually make me smile even though linear algebra is my least favorite course. thank you a lot
Just came across your video, this is the first math video i've found on youtube that really helped me understand what I was learning! You are a great teacher thank you so much for your help and enthusiasm :)
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Glad to hear that
Got this video in recommendation had a playback of first 30 seconds without clicking the video the great introduction made me open this in my other tab. Now I'm going To stalk your whole channel for Linear Algebra content 🤣
Didn't know kamaru usman was that good at matrices
I was thinking that I saw him somewhere else too
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Thanks, Great teaching!!
To raise anything to the 100th power, we multiply it by itself not 100 times, but 99 times.
To square it, we multiply it by itself once.
To cube it, we multiply it by itself twice... etc.
{{1,-1},{2,4}}^100={{2^101-3^100,2^100-3^100},{-2^101+2*3^100,-2^100+2*3^100}} It’s in my head.
This was SO SO helpful. Thank you!!!!
This channel is a gem! Great video quality, great lecture, great teacher with a great sense of fashion! One question though, shouldn't the inverse of the matrix be (1/det)[ d -b over -c a]?
You are really best man
Even without diagonalization you would only need to perform ceil(log2(100)) = 7 matrix multiplications.
So for any nxn matrix there are n! sets, of numbers, of cardinality n^2, that lead to its diagonalisation. I didn't know that before :-) (just messing ... good video)
Incroyable
@ 0:07 Great bit of theatre! LOL
We learn diagonalization but not aware of this type of application...
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I am now happy!
I was taught that for these kind of questions
First do A², ³, ⁴... until you see a pattern in these
Analyse that pattern and then put for the exponent asked
And it works, atleast the questions I solved were right
Smart!
@@PrimeNewtons yes Imagine without knowing this the question comes for a 3x3 Matrix to the power of 2023 (that was one question a remember)
What a king
Than you so much
Thank You
ty very much man!
thankyou sir
Just to make sure, you did the inverse P wrong or not because you didn’t take into consideration the sign of the cofactors
Correct
2^100 be casually having 30 digits lol
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nigerians can relate