# Multiple of Row Added to Row of Determinant/Proof 1

## Theorem

Let $\mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ be a square matrix of order $n$.

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

Let $\mathbf B = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} + k a_{s 1} & a_{r 2} + k a_{s 2} & \cdots & a_{r n} + k a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end{bmatrix}$.

Then $\map \det {\mathbf B} = \map \det {\mathbf A}$.

That is, the value of a determinant remains unchanged if a constant multiple of any row is added to any other row.

## Proof

Let $e$ be the elementary row operation that adds $k$ times row $r$ to row $s$.

Let $\mathbf B = \map e {\mathbf A}$.

Let $\mathbf E$ be the elementary row matrix corresponding to $e$.

$\mathbf B = \mathbf E \mathbf A$
$\map \det {\mathbf E} = 1$

Then:

 $\ds \map \det {\mathbf B}$ $=$ $\ds \map \det {\mathbf E \mathbf A}$ Determinant of Matrix Product $\ds$ $=$ $\ds \map \det {\mathbf A}$ as $\map \det {\mathbf E} = 1$

Hence the result.

$\blacksquare$