
Prove that det(A^2) = 1
If A is row equivalent to I and I can be obtained from A by using only row operations of type I (interchange rows) and III (adding a multiple of one row to another), prove $\displaystyle det(A^2) = 1$.
My attempt:
If A is row equivalent to I, then A is the product of some elementary matrices. As a result, det(EA) = det(E)det(A) where E is some elementary matrix.
*Assume det(A) is multiplied by the determinants of elementary matrices of type I. When this happens, that will change the value of det(A) by 1 since det(E) = 1 and det(EA) = det(E)det(A)
*Similarly, assume det(A) is multiplied by the determinants of elementary matrices of type III. When this happens, det(A) is unchanged because det(E) = 1 and det(EA)=det(E)det(A)
We can also consider the case with both row operations I and III, which again, will only change the det(A) by multiplying it by 1 or leaving it unchanged.
A is row equivalent to I, so
A = (E1)(E2) *** (Ek)(I)
This implies that
det(A) = det[(E1)(E2) *** (Ek)(I)]
det(A) = det(E1)det[(E2) *** (Ek)(I)]
...
det(A) = det(E1)det(E2) *** det(Ek)det(I)
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.
Proof done.
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.

This argument is correct; well done! At the end you can say $\displaystyle \text{det}(A^2)=[\text{det}(A)]^2$ and so the minus sign doesn't matter.

Thanks for the comment, I'll include that at the end of my proof.