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:

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

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

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

The result for columns follows from Determinant of Transpose.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\text{A}.2$: Theorem $\text{A}.10(3)$