# Determinant with Row Multiplied by Constant

## Theorem

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $\mathbf B$ be the matrix resulting from one row (or column) of $\mathbf A$ having been multiplied by a constant $c$.

Then:

$\map \det {\mathbf B} = c \, \map \det {\mathbf A}$

That is, multiplying one row (or column) of a square matrix by a constant multiplies its determinant by that constant.

## Proof

Let:

$\mathbf A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$
$\mathbf B = \begin{bmatrix} b_{11} & b_{12} & \ldots & b_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{r1} & b_{r2} & \cdots & b_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \\ \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ ca_{r1} & ca_{r2} & \cdots & ca_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$

Then from the definition of the determinant:

 $\displaystyle \map \det {\mathbf B}$ $=$ $\displaystyle \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }$ $\displaystyle$ $=$ $\displaystyle \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots ca_{r \map \lambda r} \cdots a_{n \map \lambda n}$

The constant $c$ is a factor of all the terms in the $\sum_\lambda$ expression and can be taken outside the summation:

 $\displaystyle \map \det {\mathbf B}$ $=$ $\displaystyle c \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{r \map \lambda r} \cdots a_{n \map \lambda n}$ $\displaystyle$ $=$ $\displaystyle c \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$ $\displaystyle$ $=$ $\displaystyle c \, \map \det {\mathbf A}$

The result for columns follows from Determinant of Transpose.

$\blacksquare$