Prove that $\det(E^{\mathsf{T}})=\det(E)$, where $E$ is an $n \times n$ elementary matrix. 
Prove that $\det(E^{\mathsf{T}})=\det(E)$, where $E$ is an $n \times n$ elementary matrix.



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*I am not familiar with eigenvalues.

*I only know that the cofactor expansion can be done along any row of a square matrix, but not along any column.
 A: Thanks to all that gave comment on my question. After some thought, I think I know the answer to my question.
First, all elementary matrices are product of elementary matrices since all elementary matrices are invertible.
We know that if an elementary matrix is produced by 
1) times scalar $k$ to a row of an identity matrix
or 
2) adding scalar times row $i$ to row $r$ of an identity matrix
The output matrix $E$, must show the property $\det (E^t) = \det (E)$.
As the matrix would still be a lower triangular matrix or upper triangular matrix. Thus the $\det(E)$ and $\det(E^t)$ can both be calculated by multiplying the entries along the diagonal.
The only case left is an elementary matrix that is produced by an exchange in $2$ rows.
However, it is known that a row exchange operation can be expressed in terms of a series of the $2$ row operations mentioned above. Thus, if an elementary matrix is produced by such row exchange operation, it can be expressed as the product of elementary matrices that represent the other two row operations. By a simple matrix transpose rule and the results known from above and knowing that $\det (AB) = \det (A)\times \det (B)$, it can easily be shown that $\det (E^t) = \det (E)$ is also true for such E.
Thus it is proven that $\det (E^t) = \det (E)$
