Consider the matrix $P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$. Let the transpose of a matrix $X$ be denoted by $X^{\text{T}}$. Then the number of $3 \times 3$ invertible matrices $Q$, with integer entries, such that $Q^{-1} = Q^{\text{T}}$ and $PQ = QP$, is

P = diag(2,2,3) = 2 0 0 0 2 0 0 0 3

$Q \in M_3(\mathbb{Z})\ \text{invertible};\quad Q^{-1} = Q^{\text{T}};\quad PQ = QP$

$\text{To find: number of such } Q$

$Q^{-1} = Q^{\text{T}} \text{ with integer entries} \Rightarrow Q \text{ is a signed permutation matrix}$

$Q \text{ orthogonal}, \text{ integer entries} \Rightarrow \text{each column has one } \pm 1$

$\Rightarrow\ Q \text{ is a signed permutation matrix}$

$\text{Total signed permutation matrices} = 3! \times 2^3 = 48$

$PQ = QP \Rightarrow Q \text{ preserves eigenspaces of } P$

$\text{span}(e_1,e_2) \to \text{itself},\qquad \text{span}(e_3) \to \text{itself}$

Q = R 0 0 ± 1 , R ∈ 2u00D72 signed permutation

$\#R = 2! \times 2^2 = 8,\qquad \text{third entry: } 2 \text{ choices}$

$\text{Number of such } Q = 8 \times 2 = 16$

$\boxed{16 \text{ — option } (3)}$