Math Is Fun Forum

Are you sure its not this:

because when im trying to solve it im getting negative of the expression

i don't like playing with alpha, beta and gammas so im using
alpha = x   beta = y      gamma = z
Here it is:
you take the right hand side(RHS) of the expression and then expand it

RHS= (y-x)(z-x)(y-z)
      = (yz-yx-xz+x²)(y-z)
      = y²z-xy²-xyz+x²y-yz²+xyz+xz²-x²z
(now rearrange them and combine like terms)
      = y²z-yz²+xz²-x²z-xy²+x²y
      = -1(-y²z+yz²-xz²+x²z+xy²-x²y)
      = -1[(yz²-y²z)-(xz²-x²z)+(xy²-x²y)]
      =

this is what im getting...so check your question again

i think its more like this

x²+(1/x)²

his  ²  looks like its only on 1 not on the whole 1/x

anyways...

x²+(1/x)² = 9²

yea try to tell us what are the variables and what does the question look like.....im then we can have a look at it

okay now i don't understand this part:
x + y is mapped to T(x) + T(y),

i know we have to prove: T(x+y) = T(x) + T(y) and then  T(cx)=cT(x) but i don't understand how you got there.....can you explain please...

thanks polylog....i mean i really appretiate it....but its too little too late...i did not check late at night since i wanted to get it done....but your answer is the correct one.....if only you could finish on time.....or i checked early in the morning but i did not think anyone would get it, so i just turned it in....all not so very correct....hopefully i'll get some credit on it....but i really want to thank you....there is a new one that is due day after tommorrow that i will post now.....for this one you would have a lot more time....i will try to do it my self and then check my work with what ever responses i get on here.....

let A be nxn

We need to use these facts:

(1) (A^-1)^-1 = A
(2) (k A)^-1 = (1/k) A^-1
(3) det(adj A) = (det(A))^(n-1) where n is the dimension
(4) A^-1 = 1/det(A) adj(A)

begin with (4):

A^-1 = 1/det(A) adj(A)

(A^-1)^-1 = (1/det(A) adj(A))^-1

A = (det A)(adj A)^-1

A = (det A) [ 1/(det (adj A)) adj(adj A) ]

A = (det A)/(det (adj A)) adj(adj A)

A = (det A)/(det A)^(n-1) adj(adj A)

A = (det A)^(2-n) adj(adj A)

adj(adj A) = (det A)^(-(2-n)) A

adj(adj A) = (det A)^(n-2) A

and clearly as a special case for 3x3 matrices:

det(A) = 1 -> adj(adj A) = A

that is the correct answer!!!!....again thx for the effort:D