Prove identity matrix cannot be product of an odd number of row exchanges

Ara macao
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Problem: #29 in Strang Linear Algebra

Prove that the identity matrix cannot be the product of 3 row exchanges (or five). It can be the product of 2 exchanges (or 4).

Now, to start, we try to count the number of rows that are different from the identity matrix. For the first row exchange, it's 2. Second, it becomes either 0, 3, or 4. Third, it can go to 1,2,3,4,6. Four, 0,1,2,3,4,5,6,8...

Is this proof sufficient enough?
 
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Is this proof sufficient enough?
No, for two reasons.

(1) You had to prove something about 5 swaps, but have said nothing about that.

(2) You haven't offered any proof of why, for example, you can have 0, 3, or 4 rows different after two swaps.


(Incidentally, you are wrong. For example, I can easily get 8 rows different after 4 swaps, if my matrix is large enough)
 
Oh, sorry, title post was wrong, actually, identity matrix cannot be a product of 3 or 5 row exchanges - but did not say odd number of row exchanges.

hm. After 1 swap, we have 2 rows diferent. After 2 swaps, both rows can swap back to identity, one dif row can swap with one same row, or two same rows swap to different. So we have 0,2,4. And then the thing continues...?
 
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Can you take the row reduction rules for determinants as given?
 
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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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