I have encountered a TF question which states that if $A$ is an $n$ by $n$ matrix with $\det(A-3I_n）= 0$， then there is a vector $v$ with $Av = 3v$.

I think about $A$ as a scaling vector, so $A$ should also be $kI_n$, and in this case it would be $3I_n$. Then $\det(A-3I_n)$ becomes $\det(0)$.

My question is that whether $\det(0)$ has a meaning and if so is it equal to zero?

Also, is there any logical fallacy in my approach of this question?