Answer
The determinant equals $k$.
Work Step by Step
An elementary scaling matrix looks like the identity matrix with a single $1$ replaced by a $k$. Since a square matrix with only entries along the main diagonal is both upper and lower triangular, we can apply Theorem 2, giving for the determinant $1\times1\times\cdots\times1\times k\times1\times\cdots\times1=k$.
