The determinant of the identity matrix I is 1 since the identity matrix is triangular, and the product of its diagonal elements is 1. Therefore, multiplying it by the determinants of elementary matrices I and III mean only two possible values of det(A) exist: 1 and -1. Squaring that in det(A^2) mean that det(A^2) = 1.
Note: The proofs for why det(E) = 1 or -1 where E is type I or III, respectively, are already in my textbook, so I can simply cite the proofs there. I inserted an asterik (*) at the start of those 2 statements just to make things clear for you as viewers.
Mar 30th 2011, 12:31 AM
This argument is correct; well done! At the end you can say and so the minus sign doesn't matter.
Mar 30th 2011, 06:38 PM
Thanks for the comment, I'll include that at the end of my proof.