Determinant with Row Multiplied by Constant/Proof 2
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 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 of a square matrix by a constant multiplies its determinant by that constant.
Proof
Let:
- $\mathbf A = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$
- $\mathbf B = \begin{bmatrix}
b_{1 1} & b_{1 2} & \ldots & b_{1 n} \\ b_{2 1} & b_{2 2} & \ldots & b_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
b_{r 1} & b_{r 2} & \cdots & b_{r n} \\
\vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n} \\ \end{bmatrix} = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end{bmatrix}$
Then from the definition of the determinant:
| \(\ds \map \det {\mathbf B}\) | \(=\) | \(\ds \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots c a_{r \map \lambda r} \cdots a_{n \map \lambda n}\) |
The constant $c$ is a factor of all the terms in the $\sum_\lambda$ expression and can be taken outside the summation:
| \(\ds \map \det {\mathbf B}\) | \(=\) | \(\ds c \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{r \map \lambda r} \cdots a_{n \map \lambda n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c \, \map \det {\mathbf A}\) |
$\blacksquare$