Why is the scale factor of a matrix the determinant squared?

Question

For a matrix e.g
$$\begin{bmatrix}2\sqrt{3} &-2\\2&2\sqrt{3} \end{bmatrix}$$ that is a rotation followed by an enlargement, why is the scale factor of the enlargement the root of the determinant of this matrix?
I would have thought that it would be the determinant of the enlargement matrix once separated out. Here the scale factor for the enlargement comes to $4$.

Answer 1

The determinant is the scaling factor of the area, so it makes sense that the square root of this is the scaling factor of the length of each vector, which is $4$ here.

Answer 2

You know that your matrix $M$ has the form $\lambda R$ where $\lambda \in \mathbb{R}$ and $R$ is a rotation. It follows that $\det R=1$ and $$ \det M = \det (\lambda R) = \lambda^2 \det R = \lambda^2 $$