Hi, I'm trying to solve such a problem that says: A= [a,b;0,1] "2x2 matrix" Find A^n I've tried A,A^2,A^3 to get the formula but I couldn't
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Hi ahmedzoro10! What did you get for A^2 and A^3?
Originally Posted by ILikeSerena Hi ahmedzoro10! What did you get for A^2 and A^3? I got a new matrix for each
Originally Posted by ahmedzoro10 I got a new matrix for each Which new matrices?
A^2= [a^2,ab+b;0,1] A^3= [a^3,b(a^2+a+1);0,1] A^4= [a^4,b(a^3+a^2+a+1);0,1]
Right! I see a clear pattern. Do you? Hint: multiply b(a^3+a^2+a+1) with (a-1)/(a-1) to find a simpler expression that does not have an infinite number of terms (if n tends to infinity).
Originally Posted by ILikeSerena Right! I see a clear pattern. Do you? Hint: multiply b(a^3+a^2+a+1) with (a-1)/(a-1) to find a simpler expression that does not have an infinite number of terms (if n tends to infinity). Woow! I got it,Thanks!! it's [a^n,b((a^n -1)/a-1),0,1],right?
Right!
Originally Posted by ILikeSerena Right! Thanks man!
another question,please I'm trying to prove that: A(A+B)^-1 B=B((A+B)^-1 A)= (A^-1 + B^-1)^-1 A,B and A+B are non singular matrices
Originally Posted by ahmedzoro10 another question,please I'm trying to prove that: A(A+B)^-1 B=B((A+B)^-1 A)= (A^-1 + B^-1)^-1 A,B and A+B are non singular matrices Hmm, since matrix multiplication is associative, is not generally true...
Originally Posted by ILikeSerena Hmm, since matrix multiplication is associative, is not generally true... Oh ok
Originally Posted by ahmedzoro10 Oh ok It seems your problem statement is a little off. Is it perhaps incomplete? Or is there a typo in it?
Originally Posted by ILikeSerena It seems your problem statement is a little off. Is it perhaps incomplete? Or is there a typo in it? I've already revised it and it seems cool The second term could be like that B(A+B)^{-1} A (i.e. with no parenthesis)
Originally Posted by ahmedzoro10 I've already revised it and it seems cool The second term could be like that B(A+B)^{-1} A (i.e. with no parenthesis) Hmm, as yet I'm not sure of it. Either way, as you can see here, your equality with the 3rd part does not hold: WolframAlpha - counter example
Last edited by ILikeSerena; Dec 30th 2011 at 02:27 PM.