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

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