Given the equation $8CC^T=2I_4$, deduce $C^{-1}$

Question

The following was an exercise in my exam:

Let $C$ be a $4 \times 4$ matrix with nonzero determinant. If $$8CC^T=2I_4$$ then $C$ is invertible and $$C^{-1} = \frac{1}{16}C^T$$

I was wondering: how to I get to this conclusion?

Answer 1

We have $C\cdot 4C^T=I$. By uniqueness of the inverse in the group $GL_4(K)$ we know that $4C^T$ is the inverse of $C$.

Edit: Indeed, the scalar is not $\frac{1}{16}$, because otherwise the equation should have been $\frac{1}{8}CC^T=2I_4$.