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anonymous
 5 years ago
For these matrices:
S=[u,v,v,u] T=[x,y,y,x]
NOTE: these matrices are 2x2 matrices, the first two on the top row and the last two on the bottom row. (sorry I can't express them properly here).
a) Find the determinants of S, T and ST
b) Use ordinary algebra to verify that det(S)det(T)=det(ST)
c) Express (7^2 + 2^2)(4^2 + 1^2) as a sum of squares of two integers.
anonymous
 5 years ago
For these matrices: S=[u,v,v,u] T=[x,y,y,x] NOTE: these matrices are 2x2 matrices, the first two on the top row and the last two on the bottom row. (sorry I can't express them properly here). a) Find the determinants of S, T and ST b) Use ordinary algebra to verify that det(S)det(T)=det(ST) c) Express (7^2 + 2^2)(4^2 + 1^2) as a sum of squares of two integers.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for any 2X2 matrix a b c d det = ad  bc

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0give me just a sec k?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i will scan what i have so far... it isn't complete

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks everyone  especially myininaya :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i was fixing to type this i think i will just scan it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good idea, much easier :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you know a^2+2ab+b^2=(a+b)^2 and a^22ab+b^2=(ab)^2 right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you see i add in a zero so I could write (uxvy)^2+(vx+yu)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes indeed :) thank you so much!!!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0k cool are there any questions? please let me know if you don't understand something

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i understand it all :) thanks :)
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