Determinant with Row Multiplied by Constant

Theorem
Let $$\mathbf{A} = \left[{a}\right]_{n}$$ be a square matrix of order $n$.

Let $$\det \left({\mathbf{A}}\right)$$ 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 $$\det \left({\mathbf{B}}\right) = c \det \left({\mathbf{A}}\right)$$.

That is, multiplying one row (or column) of a square matrix by a constant multiplies its determinant by that constant.

Proof
Let $$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$$.

Let $$\mathbf{B} = \begin{bmatrix} b_{11} & b_{12} & \ldots & b_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{r1} & b_{r2} & \cdots & b_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \\ \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ ca_{r1} & ca_{r2} & \cdots & ca_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$$.

Then from the definition of the determinant:

$$\det \left({\mathbf{B}}\right) = \sum_{\lambda} \left({\sgn \left({\lambda}\right) \prod_{k=1}^n b_{k \lambda \left({k}\right)}}\right) = \sum_{\lambda} \sgn \left({\lambda}\right) a_{1 \lambda \left({1}\right)} a_{2 \lambda \left({2}\right)} \cdots ca_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)}$$

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

$$\det \left({\mathbf{B}}\right) = c \sum_{\lambda} \sgn \left({\lambda}\right) a_{1 \lambda \left({1}\right)} a_{2 \lambda \left({2}\right)} \cdots a_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)} = c \sum_{\lambda} \left({\sgn \left({\lambda}\right) \prod_{k=1}^n a_{k \lambda \left({k}\right)}}\right) = c \det \left({\mathbf{A}}\right)$$

The result for columns follows from Determinant of Transpose.