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

TheoremMultiplying 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 \[sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... ca_{jk_j} ... a_{nk_n}= c sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... a_{nk_n}\]

.\[\left| A' \right| =c \sum_{\sigma} sgn ( \sigma ) a_{1k_1} a_{2 k_2} ... a_{nk_n}= c \left| A \right|\]