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.

Proof
From Determinant of Diagonal Matrix, it follows directly that:


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