# Use row reduction to prove that $\det(\mathbf{A})=\det(\mathbf{A}^{T})$

I need to prove that the determinant of a matrix is equal to the determinant of its transpose. This fact is obviously easy to prove using the definition of the determinant, but the question stipulates that the proof must be by row reduction.

It would never occur to me to do it this way. How do you even get to the transpose using row reduction? I've always just swapped the rows and columns. Any advice would be appreciated.

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Row-reducing $A$ to $R$ is done by multiplying $A$ by a sequence of elementary matrices $E_i$: $$R = E_n E_{n-1}\ldots E_1 A$$ where each $E_i$ is one of three types of elementary matrices: row switching, row multiplication, or row addition (see http://en.wikipedia.org/wiki/Elementary_row_operations).
Then $R^T = A^T E_1^T \ldots E_n^T$ and \begin{align*} \det R &= \det E_1 \ldots \det E_n \det A\\ \det R^T = \det A^T \det E_1^T \ldots \det E_n^T &= \det E_1^T \ldots \det E_n^T \det A^T. \end{align*} Can you deduce the relationship between $\det R$ and $\det R^T$? What about between $\det E_i$ and $\det E_i^T$, for each of the three types of elementary matrices?
An exercise of the kind posted in this Question might be especially motivated if the text has already outlined a proof that $\det(AB) = \det(A) \det(B)$ using this same technique, e.g. expressing an invertible matrix as a product of elementary matrices. – hardmath Jan 21 '13 at 19:02
There was no hint explicitly here about it, but of course the last thing to point out is what $det(A)$ is for an upper triangular matrix (and also for a lower triangular matrix). – rschwieb Jan 21 '13 at 19:16
@hardmath, rschwieb Yes, good points, the above assumes you know that $\det$ is multiplicative, and how to calculate the determinate of a triangular matrix. Joe, let me know if you want a more complete solution, I'm assuming for now the hint is sufficient. – user7530 Jan 21 '13 at 19:18