Compute the Jordan normal form of the product $$BA\operatorname{adj}(B),$$ where $$A = \begin{pmatrix} 1 & 3 & -1 \\ -1 & 4 & 0 \\ 0 & -1 & 4 \\ \end{pmatrix}, \qquad B = \begin{pmatrix} 2 & 1 & 3 \\ 4 & 0 & 2 \\ 1 & 1 & 3 \\ \end{pmatrix}$$
I tried to solve it head-on but it looks that it's a wrong way since $BA\operatorname{adj} B$ is too complicated to compute a JNF. Then I was thinking about computing eigenvalues and eigenvectors of $A$ and $B$, rewrote $BA\operatorname{adj}B$ as $detB*B*A*B^{-1}$ and tried to prove that since $A$ and $B$ have $n$ independent eigenvectors, JNF should $3$ $1x1$ blocks. However, $A$ doesn't have $n$ independent eigenvectors (see below) and I don't really believe it's a correct approach.
$p(\lambda) = - \lambda ^3 + 9 \lambda ^2- 27\lambda + 27 = -( \lambda -3)^3 = 0$
$(A- 3 I)x=0$ is satisfied only for $x=(1,1,1)$
Hint Computing gives $\det B \neq 0$, so $\operatorname{adj} B = (\det B) B^{-1},$ and thus $$B A \operatorname{adj} B = B ((\det B) A) B^{-1} \sim (\det B) A,$$ where you've essentially already found the equality. In particular, $B A \operatorname{adj} B$ and $(\det B) A$ have the same Jordan normal form.
Now, for a (nonzero) scalar $\mu$, how can we express the eigenvalues and corresponding eigenvectors of $\mu A$ in terms of those of $A$?