$A^TJA = J$ $ \rightarrow \det(A) = 1$ Why is this true? I cant seem to understand why the $\det(A) = 1$ for this to hold?
It is not indeed the case. We might have $$\det (A)=-1$$ since $$\det(A^TJA)=\det(J)$$yields to $$\det(A^T)\det(J)\det(A)=\det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$\det(A)^2=1$$if $J$ is singuler ($\det(J)=0$) we cannot determine $\det (A)$
It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $A\cdot A\cdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $\det (A^T)=\det(A)$. This gives $\det(A)^2 \times \det(J) = \det(J)$, so if $\det(J)$ is not zero, then $\det(A)=1$ or $\det(A)=-1$.