Proof That Multiplying a Row or Column of a Matrix By a Constant Multiplies the Determinant by that Constant

Theorem
Multiplying a row or column of a square matrix by a constant multiplies the determinant of the matrix by that constant.
Proof
For a matrix  
\[A\]
  with elements  
\[(a_{ij})\]
  the determinant is
\[\left| A \right| = \sum_{\sigma} sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... a_{nk_n}\]
.
where  
\[sgn (\sigma )\]
  is the parity of the  
\[k_1 , \: k_2 ,..., \: k_n \]
ie the number of transitions required to bring  
\[k_1 , \: k_2 ,..., \: k_n \]
  to the order 
\[1, \: 2, \: ..., \: n \]
.
If an even number of transitions is required then  
\[sgn( \sigma )=1 \]
  and if an odd number of transitions is required then  
\[sgn( \sigma )=-1 \]
.
Suppose a row of  
\[A\]
  is multiplied by  
\[\alpha\]
  to give a matrix  
\[A.\]
.
Every term in the determinant now includes a factor  
\[c\]
  and looks like  
\[sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... ca_{jk_j} ... a_{nk_n}= c sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... a_{nk_n}\]
.
\[\left| A' \right| =c \sum_{\sigma} sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... a_{nk_n}= c \left| A \right|\]
.
Since  
\[\left| A \right| = \left| A^T \right| \]
  from this result, multiplying a column by a constant also multiplies a determinant by the same constant.