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.

Corollary

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

From Determinant of Diagonal Matrix, it follows directly that:

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

$\blacksquare$