# Multiple of Row Added to Row of Determinant/Proof 2

## 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

$\map \det {\mathbf B} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} & a_{r2} & \cdots & a_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ k a_{s1} & k a_{s2} & \cdots & k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}$

$\map \det {\mathbf B} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} & a_{r2} & \cdots & a_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + k \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}$

$\begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = 0$

The result follows.

$\blacksquare$