If \(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\), then the absolute value of the determinant of \(\left(A^{9}+A^{6}+A^{3}+A\right)\) is

The absolute determinant is $32$. Since $A^2=I$, the matrix is $I+3A$, with one diagonal entry $4$ outside the block $\begin{pmatrix}1&3\\3&1\end{pmatrix}$, giving determinant $4(1-9)=-32$. The value $64$ comes from replacing the odd powers of $A$ by $I$ as well.