Suppose there is matrix A.
I know that A2 = A $\cdot $A
But what if it is A3?
Is it A $\cdot $A $\cdot$A OR A2 $\cdot$ A OR A $\cdot$ A2?
So basically my question is what is An?
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Hint: $A(BC)=(AB)C$ usually on matrices, i.e. associativity holds. – Alexey Burdin Jun 05 '15 at 06:37
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Try it out with some random $2\times 2$ matrix, or if you're feeling up for it, a $3\times 3$ matrix (that's $108$ multiplications and $72$ additions to show that $A^2 A = AA^2$, though, lots of room to make a mistake). You should see that it doesn't matter which order you do it in. – Arthur Jun 05 '15 at 06:44
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@Arthur Oh I got it. I think example is best way. – Freddy Jun 05 '15 at 06:54
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Some related older posts: Do matrices have a “to the power of” operator? and Matrix P to the power of 4, i.e $P^4$, is this the same as $P^2 \cdot P^2$? – Martin Sleziak Jun 05 '15 at 11:16
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$A^3 = A(AA)=AA^2=(AA)A=A^2A$. This comes from the associativity of matrices. In the case of $A^n =A \cdot A\cdot ...\cdot A$, $n$-times
Joaquin Liniado
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Matrix multiplication is associative, so it is all those things.
Robert Israel
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nope. AB != BA. Consider $A^2$ as B. So we will get 3 different matrices. AAA, AB and BA – Freddy Jun 05 '15 at 06:38
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@Freddy It is true that matrices do not in general commute, but $A$ and $A^2$ do commute. – Arthur Jun 05 '15 at 06:42
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@Freddy $AB\neq BA$ has nothing to do with the fact that $A\cdot(A\cdot A) = (A\cdot A)\cdot A$ – 5xum Jun 05 '15 at 06:51
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1@5xum It might, if you look at it as a commutation rather than an association. However, Freddy, it is not the case that $AB \neq BA$. It's rather that "$AB = BA$ does not always hold" – Arthur Jun 05 '15 at 06:56
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You doubt is very common with people without a course in abstract algebra.
It doesn't matter if you calculate $(A\cdot A)\cdot A$ or $A\cdot (A\cdot A)$. This is true because the associativity holds for matrices multiplication. Note that you can write the multiplication without parentheses just if you have $(A\cdot A)\cdot A=A\cdot (A\cdot A)$.
Let's take some examples:
Products in the real numbers are associative, for example $(20\cdot 4)\cdot 2 =20\cdot (4\cdot 2)$, because $(20\cdot 4)\cdot 2=160$ and $20\cdot (4\cdot 2)=160$. Then the multiplication is well-defined in this example and you can write $20\cdot 4\cdot 2=160$.
Division are not associative, for example $(20\div 4)\div 2\neq 20\div(4\div2)$, because $(20\div 4)\div 2=2,5$ and $20\div(4\div2)=10$. So $20\div 4\div 2$ is not well-defined because it would have two results $10$ and $2,5$.
Any questions I would be glad to help.
user42912
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