Consider the matrix

P=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)

\left(\begin{array}{c}\mathrm{<br>}\end{array}\right)

Let the transpose of a matrix X be denoted by X^T. Then the number of 3 × 3 invertible matrices Q with integer entries, such that

Q^{-1}=Q^{T}and\mathrm{PQ}=\mathrm{QP}

is equal to

The correct option is 16.

Let's analyze the given conditions step-by-step.

We are given the diagonal matrix $P$ defined as:
$P=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)$

We need to find the number of $3\times 3$ invertible matrices $Q$ with integer entries that satisfy two conditions:
1. $Q^{-1}=Q^T$, which means $Q$ is an orthogonal matrix: $Q{Q}^T=Q^{T}Q=I$.
2. $PQ=QP$, which means $Q$ commutes with $P$.

Let the matrix $Q$ be represented as:
$Q=\left(\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\end{array}\right)$

First, let's find the constraint imposed by the commutation relation $PQ=QP$.

Computing $PQ$:
$PQ=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)\left(\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\end{array}\right)=\left(\begin{array}{ccc}2q_{11}& 2q_{12}& 2q_{13}\\ 2q_{21}& 2q_{22}& 2q_{23}\\ 3q_{31}& 3q_{32}& 3q_{33}\end{array}\right)$

Computing $QP$:
$QP=\left(\begin{array}{ccc}{q}_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\end{array}\right)\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)=\left(\begin{array}{ccc}2q_{11}& 2q_{12}& 3q_{13}\\ 2q_{21}& 2q_{22}& 3q_{23}\\ 2q_{31}& 2q_{32}& 3q_{33}\end{array}\right)$

Equating $PQ=QP$ element-by-element gives:
- From $(1,3)$ entry: $2q_{13}=3q_{13}\Rightarrow q_{13}=0$.
- From $(2,3)$ entry: $2q_{23}=3q_{23}\Rightarrow q_{23}=0$.
- From $(3,1)$ entry: $3q_{31}=2q_{31}\Rightarrow q_{31}=0$.
- From $(3,2)$ entry: $3q_{32}=2q_{32}\Rightarrow q_{32}=0$.

Therefore, the matrix $Q$ must be of the block-diagonal form:
$Q=\left(\begin{array}{ccc}{q}_{11}& q_{12}& 0\\ q_{21}& q_{22}& 0\\ 0& 0& q_{33}\end{array}\right)$

Now, let's apply the orthogonality condition $Q{Q}^T=I$:
$\left(\begin{array}{ccc}{q}_{11}& q_{12}& 0\\ q_{21}& q_{22}& 0\\ 0& 0& q_{33}\end{array}\right)\left(\begin{array}{ccc}{q}_{11}& q_{21}& 0\\ q_{12}& q_{22}& 0\\ 0& 0& q_{33}\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

This gives the following system of equations:
1) $q_{11}^2+q_{12}^2=1$
2) $q_{21}^2+q_{22}^2=1$
3) $q_{11}{q}_{21}+q_{12}{q}_{22}=0$
4) $q_{33}^2=1$

Since the entries of $Q$ must be integers:
- From equation (4), $q_{33}$ can be either $1$ or $-1$ (2 choices).
- From equation (1), since $q_{11},q_{12}\in \mathbb{Z}$, the only possible pairs for $(q_{11},q_{12})$ are:
$(1,0),(-1,0),(0,1),(0,-1)$ (4 choices).
- Similarly, from equation (2), the only possible pairs for $(q_{21},q_{22})$ are:
$(1,0),(-1,0),(0,1),(0,-1)$ (4 choices).

Let's find the valid combinations for the $2\times 2$ submatrix under the orthogonality condition (3) $q_{11}{q}_{21}+q_{12}{q}_{22}=0$:
- **Case A**: If $(q_{11},q_{12})=(\pm 1,0)$ (2 choices):
Then $q_{11}{q}_{21}+q_{12}{q}_{22}=0\Rightarrow (\pm 1)q_{21}+0=0\Rightarrow q_{21}=0$.
This forces $q_{22}=\pm 1$ (2 choices).
So we have $2\times 2=4$ combinations here.
- **Case B**: If $(q_{11},q_{12})=(0,\pm 1)$ (2 choices):
Then $q_{11}{q}_{21}+q_{12}{q}_{22}=0\Rightarrow 0+(\pm 1)q_{22}=0\Rightarrow q_{22}=0$.
This forces $q_{21}=\pm 1$ (2 choices).
So we have another $2\times 2=4$ combinations here.

Thus, the number of choices for the $2\times 2$ upper-left block is:
$4+4=8$ choices.

Since the choice of $q_{33}$ is independent of the $2\times 2$ block, the total number of such matrices $Q$ is:
$\text{Total number of matrices}Q=(\text{Choices for}2\times 2\text{block})\times \left(\text{Choices for}{q}_{33}\right)=8\times 2=16$

The total number of invertible matrices $Q$ satisfying the given conditions is 16.