Computing the determinant using $\det (A^2 C+I) = 1$

Question

I'm trying to obtain the determinant $$\det \left((A^2 B^{-1})^{-1}+BC \right)$$ where $\det(A) = 2$, $\det(B) = -8$, $\det(A^2 C+I)=1$ and $A,B,C\in\mathbb{R}^{n\times n}$.

What I have done: Assuming $C$ regular, taking common factor of $B$ and $C$ and using $\det(SR+I)=\det(RS+I)$ we have

\begin{align} \det \left((A^2 B^{-1})^{-1}+BC \right) &= \det(B)\det(C)\det \left(B^{-1}(A^2 B^{-1})^{-1}C^{-1}+I \right)\\ &= \det(B)\det(C)\det \left((A^2 )^{-1}C^{-1}+I \right)\\ &=\det(B)\det(C)\det \left(C^{-1}(A^2)^{-1}+I \right)\\ &=\det(B)\det(C)\det \left((A^2C)^{-1}+I \right) \end{align}

But now I don't know how to continue.

Answer 1

Just factorise by $\frac{\det B}{\det A^2}$, you will see that the term in your determinant is exactly what you have as an info. And I believe you formula of $\det(RS + I) = \det(SR + I)$ is false, you only have $\det(RS) = \det(SR)$

Edit: your formula is valid, but not useful here

Answer 2

\begin{align} \det \left((A^2 B^{-1})^{-1}+BC \right) &= \det \left( BA^{-2} + BC \right) \\ &= \det(B)\det(A^{-2})\det \left(I + A^2C \right)\\ \end{align}

Answer 3

You do not have to solve the problem in the way in which you attempted to. It is not known whether $C$ is regular or not.

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Well, I will play along, assuming that the determinant of $C$ is nonzero. Note that $$ \det {(X^{-1} + I)} = \det {(IX^{-1} + XX^{-1})} = \det {(I + X)} \det {(X^{-1})} = \frac{\det {(I + X)}}{\det {(X)}}. $$ Hence $$ \begin{aligned} & \det {(B)} \det {(C)} \det {((A^2 C)^{-1} + I)} \\ = {} & {\det {(B)} \det {(C)} \det {(I + A^2 C)} \over \det {(A^2 C)}} \\ = {} & {\det {(B)} \det {(C)} \det {(I + A^2 C)} \over (\det {(A)})^2 \det {(C)}} \\ = {} & {\det {(B)} \det {(I + A^2 C)} \over (\det {(A)})^2} \\ = {} & {-8 \cdot 1 \over 2^2} \\ = {} & {-2}. \end{aligned} $$

Remember that you will still have to handle the case in which the determinant of $C$ is zero.

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A better solution has been suggested by Offlaw:

$$ \begin{aligned} & \det {((A^2 B^{-1})^{-1}+BC )} \\ = {} & \det {( BA^{-2} + BC )} \\ = {} & \det {(B)} \det {(A^{-2})} \det {(I + A^2C)} \\ = {} & \det {(B)} (\det {(A)})^{-2} \det {(I + A^2C)} \\ = {} & (-8) \cdot 2^{-2} \cdot 1 \\ = {} & {-2}. \end{aligned} $$