## 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.