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_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end{bmatrix}$


 * $\mathbf B = \begin{bmatrix}

b_{1 1} & b_{1 2} & \ldots & b_{1 n} \\ \vdots & \vdots & \ddots &  \vdots \\ b_{r 1} & b_{r 2} & \cdots & b_{r n} \\ \vdots & \vdots & \ddots &  \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \\ \end{bmatrix} = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 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}$

Then from the definition of the determinant:

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

The result for columns follows from Determinant of Transpose.

Also see

 * Determinant with Column Multiplied by Constant