Can AB^2009 be calculated?Matrix Multiplication and Exponentiation Limitations

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It is possible to compute AB^2009 as both A and B are square matrices of the same size, resulting in an n x n matrix. The multiplication of two n x n matrices yields another n x n matrix, allowing for further multiplications. The exponentiation notation can be interpreted as either AB raised to the 2009th power or B raised to the 2009th power before multiplying by A, but both interpretations remain valid. The key point is that the dimensions remain consistent throughout the operations. Thus, calculating AB^2009 is feasible within the constraints of matrix multiplication.
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Homework Statement



Matrix A and B are both square matrices of the same size.

Is it possible to compute,

AB^2009

If not, why not?

The Attempt at a Solution



Just had this question in an exam, the exam did not allow calculators and I could not answer this question, I gather it is not possible but I cannot explain why.

Because A and B are square matrices it is possible to multiply them together but I am unsure about the exponent. Any ideas?
 
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Suppose that A and B are both n x n matrices. Then what is the size of AB? Then what about ABB = AB^2? And ABBB = AB^3? ...
 
Well if A and B are both n x n then they will always be n x n matrices. So does that mean that it is possible?
 
By the way, is that AB2009[/b] or (AB)2009?

It doesn' really matter. The only important point is that any two n by n matrices can be multiplied and give an n by n matrix as a result- which can be multiplied again by any other n by n matrix.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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