# Determinant of Rescaling Matrix

## Theorem

Let $R$ be a commutative ring.

Let $r \in R$.

Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by:

$\sqbrk {r \, \mathbf I_n}_{i j} = \begin{cases} r & : i = j \\ 0 & : i \ne j \end{cases}$

Then:

$\map \det {r \, \mathbf I_n} = r^n$

where $\det$ denotes determinant.

## Theorem

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

Let $\lambda$ be a scalar.

Let $\lambda \mathbf A$ denote the scalar product of $\mathbf A$ by $\lambda$.

Then:

$\map \det {\lambda \mathbf A} = \lambda^n \map \det {\mathbf A}$

where $\det$ denotes determinant.

## Proof

For $1 \le k \le n$, let $e_k$ be the elementary row operation that multiplies row $k$ of $\mathbf A$ by $\lambda$.

By definition of the scalar product, $\lambda \mathbf A$ is obtained by multiplying every row of $\mathbf A$ by $\lambda$.

That is the same as applying $e_k$ to $\mathbf A$ for each of $k \in \set {1, 2, \ldots, n}$.

Let $\mathbf E_k$ denote the elementary row matrix corresponding to $e_k$.

$\map \det {\mathbf E_k} = \lambda$

Then we have:

 $\ds \lambda \mathbf A$ $=$ $\ds \prod_{k \mathop = 1}^n \mathbf E_k \mathbf A$ $\ds \leadsto \ \$ $\ds \map \det {\lambda \mathbf A}$ $=$ $\ds \map \det {\prod_{k \mathop = 1}^n \mathbf E_k \mathbf A}$ $\ds$ $=$ $\ds \paren {\prod_{k \mathop = 1}^n \map \det {\mathbf E_k} } \map \det {\mathbf A}$ Determinant of Matrix Product $\ds$ $=$ $\ds \paren {\prod_{k \mathop = 1}^n \lambda} \map \det {\mathbf A}$ Determinant of Elementary Row Matrix: Scale Row $\ds$ $=$ $\ds \lambda^n \map \det {\mathbf A}$

$\blacksquare$

## Proof

From Determinant of Diagonal Matrix, it follows directly that:

$\map \det {r \, \mathbf I_n} = \displaystyle \prod_{i \mathop = 1}^n r = r^n$

$\blacksquare$