Does $ABC=D\implies \det(ABC)=\det(D )$? $${\color{brown}{\text{Question I am trying to solve:}}}$$
Let $A,B$ and $X$ be 7 x 7 matrices such that $\det A=1$, $\det B=3$ and
$$A^{-1}XB^{t}=-I_7$$
where $I_7$ is the 7 x 7 identity matrix. Calculate $\det X$.
$$\color{brown}{------------------------------------}$$
The way I thought to solve this is if the following is true:
$$A^{-1}XB^{t}=-I_7 \implies \det(A^{-1}XB^{t})=\det(-I_7 )$$

$$\text{Then I could solve this by: }$$
$$\det (A^{-1}XB^{t})=\det(-I_7)$$

$$\text{Note (since A, X, B are of the same size):}$$$$\bbox[8pt, border: 1pt solid green]{\det(A^{-1}XB^{t})=\det(A^{-1})\det(X)\det(B^t)}$$
$$\text{Note:}$$
$$\det(B^t)=\det (B)$$

$$\det(A^{-1})\det(X)\det(B^t)=\det(-I_7)$$
$$\frac{1}{1} \det(X) \cdot 3 = -1 \implies \bbox[8pt,border: 2pt #06f solid]{\det X=-\frac{1}{3}}$$

So is it correct to do this :
$$A^{-1}XB^{t}=-I_7 \implies \det(A^{-1}XB^{t})=\det(-I_7 )$$
$$\color{gold}{\Large{?}}$$

I might be missing something here, but as far as I know:
$a = a' \Rightarrow f(a) = f(a')$
for all sets $A,B$; $a,a'\in A$ and functions $f : A\to B$.
