Does this mean that the determinant of the inverse of $A$ is equal to the inverse of the determinate of $A$, and is this always true, for $n \times n$ matrices?
Suppose the $\det(A^{-1})=5$ Does that mean then that $(\det(A))^{-1} = \frac{1}{5}$ or am I missing something? Like do I have the wrong idea of the inverse?
On
Since $\det AB =\det A \det B,$ this is just saying $$ 1=\det I=\det (A)(A^{-1})=\det (A) \det(A^{-1}) $$
On
You will never hear me say "the inverse of (some number)". I learned years ago that it leads to exactly the confusion happening here.
The determinant of a matrix is a number and when you raise a number to the power $-1$, you obtain its reciprocal. The equation you have written says, "the determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix."
You have written $\det (A^{-1}) = (\det A)^{-1}$. It is perhaps more enlightening to write $$ \det (A^{-1}) = \frac{1}{\det A} \text{.} $$
You ask, if $\det (A^{-1}) = 5$, what is $(\det A)^{-1}$. Your equation says the calue of one of these is the same as the value of the other, so both of these are $5$. Here, however, is perhaps a clearer version of what you are thinking.
Suppose $\det A = 5$. Then $\frac{1}{\det A} = \frac{1}{5}$ and the equation says $\det(A^{-1}) = \frac{1}{5}$.
Suppose that $A$ is invertible. Then one has \begin{align*} 1 = \det I = \det(AA^{-1}) = \det(A)\det(A^{-1}) \Longrightarrow \det(A^{-1}) = [\det(A)]^{-1} \end{align*}