# Determinant with Column 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 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 column of a square matrix by a constant multiplies its determinant by that constant.

## Proof

Let:

$\mathbf A = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 r} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$
$\mathbf B = \begin{bmatrix} b_{1 1} & b_{1 2} & \cdots & b_{1 r} & \cdots & b_{1 n} \\ b_{2 1} & b_{2 2} & \cdots & b_{2 r} & \cdots & b_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n r} & \cdots & b_{n n} \\ \end{bmatrix} = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & c a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & c a_{2 r} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$

We have that:

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

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

Similarly, we have that:

$\mathbf B^\intercal = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$
$\map \det {\mathbf B^\intercal} = c \map \det {\mathbf A^\intercal}$

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.

$\blacksquare$