Determinant with Row Multiplied by Constant/Proof 2

Proof
Let:
 * $\mathbf A = \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}$


 * $\mathbf B = \begin{bmatrix}

b_{1 1} & b_{1 2} & \ldots & b_{1 n} \\ b_{2 1} & b_{2 2} & \ldots & b_{2 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} \\ 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}$

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: