# Multiple of Column Added to Column of Determinant

## Theorem

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

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

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

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

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

## Proof

We have that:

$\mathbf A^\intercal = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{r 1} & a_{r 2} & \cdots & a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$

where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$.

Similarly, we have that:

$\mathbf B^\intercal = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{r 1} + \lambda a_{s 1} & a_{r 2} + \lambda a_{s 2} & \cdots & a_{r n} + \lambda a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\  a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$

$\map \det {\mathbf B^\intercal} = \map \det {\mathbf A^\intercal}$

From from Determinant of Transpose:

$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$

and the result follows.

$\blacksquare$