Determinant with Row Multiplied by Constant

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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 1

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $e$ be the elementary row operation that multiplies rows $i$ by the scalar$c$.

Let $\mathbf B = \map e {\mathbf A}$.

Let $\mathbf E$ be the elementary row matrix corresponding to $e$.

From Elementary Row Operations as Matrix Multiplications:

$\mathbf B = \mathbf E \mathbf A$

From Determinant of Elementary Row Matrix: Exchange Rows:

$\map \det {\mathbf E} = c$

Then:

\(\displaystyle \map \det {\mathbf B}\) \(=\) \(\displaystyle \map \det {\mathbf E \mathbf A}\) Determinant of Matrix Product
\(\displaystyle \) \(=\) \(\displaystyle c \map \det {\mathbf A}\) as $\map \det {\mathbf E} = c$

Hence the result.

$\blacksquare$


Proof 2

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:

\(\displaystyle \map \det {\mathbf B}\) \(=\) \(\displaystyle \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots c a_{r \map \lambda r} \cdots a_{n \map \lambda n}\)


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

\(\displaystyle \map \det {\mathbf B}\) \(=\) \(\displaystyle c \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{r \map \lambda r} \cdots a_{n \map \lambda n}\)
\(\displaystyle \) \(=\) \(\displaystyle c \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }\)
\(\displaystyle \) \(=\) \(\displaystyle c \, \map \det {\mathbf A}\)

$\blacksquare$


Also see


Sources