Linear Transformation: Solving Coefficient Matrix and Evaluating T(e1) and T(e2)

Clandry
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I attached the problem. I'm not sure if I'm misinterpreting the question, but this problem seems really easy, which is usually not the case with my class.

for part a) isn't that just the coefficient matrix of the right hand side?
This makes A:
1 -2
3 1
0 2



for part b) T(e1)=T[1 0]^t=[1 3 0]^t
T(e2)=T[0 1]^t=(-2 1 2)^t
Is this right?
 

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Clandry said:
I attached the problem. I'm not sure if I'm misinterpreting the question, but this problem seems really easy, which is usually not the case with my class.

for part a) isn't that just the coefficient matrix of the right hand side?
This makes A:
1 -2
3 1
0 2



for part b) T(e1)=T[1 0]^t=[1 3 0]^t
T(e2)=T[0 1]^t=(-2 1 2)^t
Is this right?

Looks good to me.
 
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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