Let $B$ be a matrix of order $3$ with $\det(B) = -5$
Does there exist an invertible matrix $M^{-1}$ with real entries where $M = 4 B M^{-1}$?
I am very confused because I thought I had to verify that $\det(M) \not= 0$. So I computed $\det(M) = det(4 B M^{-1}) = 4^3 \det(B) \frac{1}{\det(M)}$. I got $\det(M) = \sqrt{-320}$. So the answer might be simply NO because $\det(M)$ is not a real value, so we might find complex value entries, right?
The equation $M = 4 B M^{-1}$ is equivalent to $M^2 = 4B$. As you noticed, it is not possible to find a real matrix $M$ satisfying such equation as it implies
$$(\det M)^2 = -320.$$
However, the equation has at least one solution in term of complex matrix using Schur decomposition. See this wikipedia entry for more details.