Solving Canonical Question: Find Q, P, r for A={(1,2,1),(2,4,2),(3,6,3)}

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got stucked on this question but had a pritty good bash at it and might possibly be getting close to the answer

right so the question in full is let A={(1,2,1),(2,4,2),(3,6,3)}

find r and real invertible matrices Q and P such that Q-1AP={(Ir,0)(0,0)
where each zero denotes a matrix of zeros (not nessessarily the same size in each case)

Paying special attension to write down the bases of r3 with respect to which Q-1AP represents the mapping x->Ax

right now I've started off by row and column reducing A to get {(1,0,0)(0,0,0)(0,0,0)}
and then by applying the row and column opperations to the 3x3 and 3x3 respectivly Identity matrices i ended up with Q-1={(1,0,0)(-2,1,0)(-3,0,1)} Q={(1,0,0)(1/2,1,0)(1-3,0,1)} and finaly p={(1,-2,-1)(0,1,0)(0,0,1) which did indeed satisfy the equation Q-1AP=I1

now the next part of the question i didn't/don't really understand "Paying special attension to write down the bases of r3 with respect to which Q-1AP represents the mapping x->Ax"

however i looked at what i believe to be a similar question on my past homework questions (this question is from a previous exam paper so i don't have answers) and came to the conclusion (not sure if this is right or not this is my question to you really) that i was being asked to find a basis for my matrix A and then a basis for my canonical so this would just be {(1,2,1)} and {(1,0,0)} respectivly? I'm pritty sure this is wrong as it just seems too easy althought i believe i am on the right lines, could someone please elaberate on my findings. thanks
 
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posted on wrong post sorry
 
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thought i had a reply there, guessing no one has replied since terry posted and you all thought he had offered a solution, o well, anyone?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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