Is an invertible matrix always lin. independent?

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Because according to my book,

If A's an invertible nxn matrix, then the eq. Ax=b is consistent for EACH b...

this consistency would imply lin. independency?
 
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Yes. Also, the determinant of an invertible matrix is non-zero. The determinant of a matrix of linearly dependent vectors is 0.
 
Strictly speaking, the term "independent" applies to sets of vectors, not matrices so it does not make sense to ask if a matrix is independent. What you are really asking is whether the rows or columns of the matrix, thought of as vectors are independent. The answer to that is yes. If a matrix is invertible, then its rows (or columns), thought of as a set of vectors, is indepependent.
 
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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