Identity Matrix: Is Inverse Always True for n>=2?

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Homework Statement



let I_n be as an identity matrix where a_ij = 1 when i=j
I just want to ask that is it true that all identity matrix has an inverse (determinant is not 0) for n>=2?



The Attempt at a Solution

 
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The determinant of In is 1, as you can easily compute (expanding along any row or column, you find that det(In) = 1 . det(In - 1) and clearly det(I1) = 1).

Also it is invertible, and it is its own inverse. You can check this directly from the definition.
 
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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