
#1
Nov2412, 04:44 PM

P: 24

 x 1 0 
 0 x 1   0 0 x  I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this: Let A represent the matrix above, and let N represent the following matrix:  0 1 0   0 0 1   0 0 0  Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem. A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50k)N^k by the binomial theorem = (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3 Now if you take N^3 you will get 0, and if you take N^2, you get the following:  0 0 1   0 0 0   0 0 0  So we get: I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is,  X^50 50X^49 1225X^48   0 X^50 50X^49   0 0 X^50  I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks. 



#2
Nov2412, 04:48 PM

HW Helper
Thanks
P: 4,673

RGV 



#3
Nov2412, 05:03 PM

Sci Advisor
HW Helper
Thanks
P: 25,174

Your solution looks just fine to me.



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