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