Let A and B be two invertible matrices of order $$3 \times 3$$. If $$\det(ABA^T) = 8$$ and $$\det(AB^{-1}) = 8$$, then $$\det(BA^{-1}B^T)$$ is equal to

We recall a basic property of determinants: for any two square matrices of the same order we have the formula $$\det(AB)=\det(A)\,\det(B)$$. We also know that $$\det(A^T)=\det(A)$$ and $$\det(A^{-1})=\dfrac{1}{\det(A)}$$. All matrices here are of order $$3 \times 3$$, so these facts apply directly.

First, we are told that

$$\det(ABA^T)=8.$$

Using the determinant product rule and the fact that a transpose does not change a determinant, we write

$$\det(ABA^T)=\det(A)\,\det(B)\,\det(A^T)=\det(A)\,\det(B)\,\det(A).$$

Simplifying, this becomes

$$\det(ABA^T)=\bigl(\det(A)\bigr)^2\,\det(B)=8.$$

Next, we are also given

$$\det(AB^{-1})=8.$$

Again applying the product rule and the inverse‐determinant relation, we have

$$\det(AB^{-1})=\det(A)\,\det(B^{-1})=\det(A)\,\dfrac{1}{\det(B)}=8.$$

To keep the algebra clear, let us denote

$$x=\det(A), \qquad y=\det(B).$$

The two pieces of information now translate to the pair of equations

$$x^{2}y=8 \qquad (1)$$

and

$$\dfrac{x}{y}=8 \qquad (2).$$

From equation (2) we immediately find

$$y=\dfrac{x}{8}.$$

Substituting this value of $$y$$ into equation (1) gives

$$x^{2}\left(\dfrac{x}{8}\right)=8.$$

This simplifies step by step:

$$\dfrac{x^{3}}{8}=8,$$

so

$$x^{3}=64,$$

and therefore

$$x=4.$$

Using $$x=4$$ in the relation $$y=\dfrac{x}{8}$$ we get

$$y=\dfrac{4}{8}=\dfrac{1}{2}.$$

Now we need the determinant

$$\det(BA^{-1}B^{T}).$$

Applying the same determinant properties, we write

$$\det(BA^{-1}B^{T})=\det(B)\,\det(A^{-1})\,\det(B^{T}).$$

Since $$\det(B^{T})=\det(B)$$ and $$\det(A^{-1})=\dfrac{1}{\det(A)}$$, this becomes

$$\det(BA^{-1}B^{T})=y \times \dfrac{1}{x} \times y=\dfrac{y^{2}}{x}.$$

Substituting $$x=4$$ and $$y=\dfrac{1}{2}$$, we calculate

$$\det(BA^{-1}B^{T})=\dfrac{\left(\dfrac{1}{2}\right)^{2}}{4}=\dfrac{\dfrac{1}{4}}{4}=\dfrac{1}{16}.$$

Hence, the correct answer is Option C.