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

Theorem
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
$\left| A \right| = \left| A^T \right|$
.