Consider the matrix \(P=\left(\begin{matrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{matrix}\right)\) Let the transpose o…

Given

\(Q^{-1}=Q^T\)

So \(Q\) is an orthogonal matrix.

Since \(Q\) has integer entries, each row and each column must have exactly one non-zero entry equal to \(1\) or \(-1\).

Thus \(Q\) is a signed permutation matrix.

Now

\(PQ=QP\)

with

\(P=\operatorname{diag}(2,2,3)\).

Let

\(Q=(q_{ij})\).

Since \(PQ=QP\),

\((p_i-p_j)q_{ij}=0\)

for all \(i,j\),

where

\(p_1=p_2=2,\quad p_3=3\).

Hence

\(q_{13}=q_{31}=q_{23}=q_{32}=0\).

Therefore \(Q\) must have the form

\(\begin{pmatrix}
*&*&0\\
*&*&0\\
0&0&*
\end{pmatrix}\),

where \(Q\) is a signed permutation matrix.

The first two coordinates may be permuted in

\(2!\) ways.

Each non-zero entry can independently be assigned sign \(+1\) or \(-1\).

There are \(3\) non-zero entries, giving

\(2^3\) sign choices.

Therefore total number of such matrices

\(=2!\times2^3\)

\(=2\times8\)

\(=16\).