Determinant of $ A^{-1}B (A^{-1})^{T} $ is equal to the determinant of B.

Question

In the solution to one of the questions I am attempting it says, $$\det{(A^{-1}B (A^{-1})^{T})} = \det B $$

where A is a lower triangular matrix and B is a diagonal matrix.

Could someone explain why this hold?

Answer 1

This statement is ONLY true if $\det(B)=0$ or $\det(A)=\pm 1$.

Observe that since the determinant is multiplicative $(\det(CD)=\det(C)\det(D))$, $$ \det(A^{-1}B(A^{-1})^T)=\det(A^{-1})\det(B)\det((A^{-1})^T). $$

Since $\det(C^T)=\det(C)$, this can be simplified to $$ \det(A^{-1})\det(B)\det((A^{-1})^T)=\det(A^{-1})\det(B)\det(A^{-1}). $$

Finally, since $\det(C^{-1})=\frac{1}{\det(C)}$, we can further simplify to get $$ \det(A^{-1})\det(B)\det(A^{-1})=\frac{\det(B)}{\det(A)^2}. $$

For this to equal $\det(B)$, either $\det(B)=0$ or $\det(A)^2=1$.