## Proof That the Determinant of a Matrix Equals the Determinant of its Transpose

TheoremThe determinant of an

\[n \times n\]

matrix \[A\]

with \[n\]

rows and \[n\]

columns is equal to the determinant of its transpose so that \[\left| A \right| = \left| A^T \right| \]

.Proof

\[\left| A \right| = \sum_{\sigma} sign(\sigma ) a_{1k_1} a_{2k_2} ...a_{nk_n}\]

.where

\[sign(\sigma )\]

is the parity of the permutation \[1, \:2,..., \:n\]

onto \[k_1, \:k_2,..., \:k_n\]

.Each term of

\[\left| A \right|\]

is a product of \[n\]

elements of \[A\]

with one element taken from each row or column, and a further factor of 1 or -1 depending on whether the permutation given above is even or odd.The product can be written

\[a_{j_1 1} a_{j_2 2} ... a_{j_n n}\]

by rearranging the factors. Rearranging the factors in this way does not change the parity of the term.

The first term is a term in

\[\left| A \right|\]

and the second is a term in \[\left| A^T \right|\]

.Hence

\[\left| A \right| = \left| A^T \right| \]

.