What is the rank of an nxn matrix and how is it determined?

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Homework Statement


http://img94.imageshack.us/img94/5227/nxnmatrix.png

Homework Equations



Rank(A) = the number of pivots in Matrix A.

The Attempt at a Solution



I've spent some time rewriting the matrix and other operations. I really just feel like I'm banging my head against the wall. Not making any progress. Just need to be pointed in the right direction. Oh and the answer is "2." No idea how they get that. Thanks for any help in advance.

Edit: For reference this problem is from Applied Linear Algebra by Olver. Problem 1.8.17.
 
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JonathanT said:

Homework Statement


http://img94.imageshack.us/img94/5227/nxnmatrix.png


Homework Equations



Rank(A) = the number of pivots in Matrix A.

The Attempt at a Solution



I've spent some time rewriting the matrix and other operations. I really just feel like I'm banging my head against the wall. Not making any progress. Just need to be pointed in the right direction. Oh and the answer is "2." No idea how they get that. Thanks for any help in advance.

Stop and ask yourself how many linearly independent columns compose your matrix A.

If that notion confuses you, ask yourself how many zero ROWS you have after row reduction...
 
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I think I get it. I was reducing the matrix wrong. I think I have the correct rref of the matrix now. Take a look and let me know what you think.

Solution
 
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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