Mathematic Article - Matrix products
Certain mathematical problems exist where it is necessary to use mathematical tools to be solved quickly. This happens in problems where they request results to us that require an amount of quite great previous basic operations arithmetical (some excessive ones). For it it is important to know methods or theorems that will help us to solve them with simplicity. Today I am going to you to propose a quite peculiar problem, where I am sure that it will be generated quite interesting answers.
Be Matrix, to calculate and to explain the procedure:
a) M2 and M4
b) M2006
Note: For section b) it is not necessary to make 2005 matrix products.
not if he will be correct, but I made the algorithm for M^2 and it gave me that:
M: = (to a)
(- to a)
…
M^n: = (0 na^n)
(na^n 0)
I believe that I have solved it:
http://personal.telefonica.terra.es/web/tomas-m-s/matriz.gif
The method that Marc says is the one of induction, I do not know if it would be possible to be done for whole values of N, but the method that I have followed is for any value of real n.
In any case, Alfonso Jimenez, you have gone poquitín with the problem, would be necessary to know how mathematical of engineering level to remove the affluent solution.
Although or I say… I do not know if I have done it or or I have gone away by the hills of úbeda that did not have anything to do.
First powers of M:
M^2= (0 1)
(-1 0)
M^4= (- 1 0)
(0 -1)
M^8=M^16=M^32=… =M^512=M^1024= (1 0)
(0 1)
M^2006=M^1024*M^512*M^256^*M^128*M^64*M^16*M^4*M^2
M^2006= (1 0) *M^4*M^2= (0 -1)
(0 1) (1 0)
if I have not been mistaken.
What habria that to do is to calculate the matrix reduced of gaus to hayar its inverse one, soon to calculate the first diagonal and regulating, and simply soon is applied to an equation of the P^ type (−1) · M · P = D, M was sprightly,
M = P · D · P^ (- 1), and now unique that habria that to do is to elevate to 2005 or 2006, or the wished number the first diagonal, with which hariamos very many less calculations
I am in agreement, like M^8 = I, being I the first Unit, and since one is fulfilled that I*A=A*I =A, being To a matrix anyone, then:
M^2006= M^ (250*8) *M^6=I*M^6=M^6
Like M^6= (0 -1) | (1 0), then
M^2006= (0 -1) | (1 0)
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