[THOUGHT] We need to count the number of 3×3 invertible matrices P with integer entries satisfying:

1. ( P^{-1} = P^T ) → ( P^T P = I ) → P is orthogonal over ℝ, but with integer entries, so P is an integer orthogonal matrix.
2. ( PM = MP ) where ( M = \text{diag}(2,2,3) ).

Since M is diagonal, the condition ( PM = MP ) means that P must commute with M. For a diagonal matrix with distinct diagonal entries, only diagonal matrices commute with it. But here M has two equal diagonal entries (2,2,3). So the commuting condition allows P to be block-diagonal with respect to the eigenspaces.

Let’s analyze: M = diag(2,2,3). The eigenspace for eigenvalue 2 is 2-dimensional (first two coordinates), and for eigenvalue 3 is 1-dimensional (third coordinate). Any matrix P that commutes with M must preserve these eigenspaces. That means P must be block-diagonal: [ P = \begin{pmatrix} A & 0 \ 0 & b \end{pmatrix} ] where A is 2×2, b is 1×1 (scalar), both integers.

Additionally, P must be orthogonal (over ℝ) with integer entries. Since P is integer orthogonal, each column must have integer entries and be orthonormal. For a 3×3 integer orthogonal matrix, each column has norm 1, so each column is a standard basis vector up to sign (since integer entries with squared sum 1 means entries are 0, ±1 with exactly one nonzero entry). So P must be a signed permutation matrix.

But wait: integer orthogonal matrices are exactly signed permutation matrices (since rows/columns are orthonormal and integer entries). Indeed, the only integer vectors with norm 1 are ±e_i. So P must be a signed permutation matrix.

Now, we also have the block structure from the commuting condition: P must be block-diagonal with respect to the partition {1,2} and {3}. That means P does not mix coordinates 1,2 with coordinate 3. So P is of the form: [ P = \begin{pmatrix} A & 0 \ 0 & b \end{pmatrix} ] where A is 2×2 signed permutation matrix (orthogonal with integer entries, so A is signed permutation of 2×2 identity), and b = ±1.

But careful: For a 2×2 integer orthogonal matrix, the possibilities are signed permutation matrices of size 2: they must be of the form ±I or ±(anti-diagonal). Actually, the 2×2 signed permutation matrices are: [ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}. ] But also we can have permutations with signs independently? Let's check: A signed permutation matrix means each row/column has exactly one nonzero entry which is ±1. For 2×2, the possible permutations are identity and swap. For each permutation, we can assign signs to each nonzero entry independently? Actually, if we assign signs independently, the matrix may not be orthogonal? Let's test: For identity, we can have diag(±1, ±1) → that's orthogonal. For swap, we can have matrix [[0, a],[b,0]] with a,b = ±1. Then rows: row1 norm = a^2 = 1, row2 norm = b^2 = 1, dot product = 0a? Wait, dot product of row1 and row2 = 0a + ab? Actually row1 = (0,a), row2 = (b,0), dot = 0b + a*0 = 0. So yes, any a,b = ±1 works. So total 2×2 signed permutation matrices: 2 permutations × 2 signs for each nonzero entry? But careful: For identity permutation, we have 2 signs (each diagonal entry ±1) → 2×2=4 possibilities. For swap permutation, we have two off-diagonal entries, each can be ±1 → 2×2=4 possibilities. But are they all orthogonal? Yes, as argued. However, note that the matrix [[0,1],[-1,0]] is not a signed permutation matrix? Wait, that is [[0,1],[-1,0]] has rows (0,1) and (-1,0), each has one nonzero entry ±1, so it is a signed permutation matrix (permutation swap with signs). So yes, 4 possibilities for swap as well. So total 8 possible 2×2 signed permutation matrices.

But we must also consider that the overall P is 3×3, block-diagonal with A (2×2) and b (scalar ±1). So b = ±1. So total possibilities: 8 (choices for A) × 2 (choices for b) = 16.

But wait: Are there any other integer orthogonal matrices that commute with M but are not block-diagonal? Since M has eigenspaces of dimensions 2 and 1, any matrix commuting with M must map each eigenspace to itself. That means P must map the first two coordinates among themselves and the third coordinate to itself (up to scalar). But for orthogonal integer matrices, the third coordinate must map to ±e_3 (since norm 1). So indeed, the structure is block-diagonal as described. However, could there be mixing within the 2-dimensional eigenspace? Yes, that's captured by A being any 2×2 signed permutation matrix. So 16 seems plausible.

But we need to verify that all such P are indeed invertible (they are orthogonal, so invertible). So number of such P is 16.

But the problem asks: "Find the square of the number of ... matrices P". So we need (number)^2 = 16^2 = 256.

However, we must double-check if there are any other integer orthogonal matrices that commute with M but are not of that block form? Let's analyze more systematically.

We require P^T P = I (orthogonal) and P M = M P.

Let P = [p_{ij}] integer. Since P is orthogonal, each column has norm 1: sum_i p_{ij}^2 = 1, so each column has exactly one nonzero entry which is ±1. So each column is a signed standard basis vector. Similarly for rows. So P is a signed permutation matrix.

Thus P corresponds to a permutation σ on {1,2,3} and signs ε_i = ±1 such that P e_i = ε_i e_{σ(i)}.

Now impose PM = MP.

Compute (PM){ij} = ∑k p{ik} M{kj} = p_{ij} M_{jj} because M is diagonal. So (PM){ij} = p{ij} M_{jj}.

Similarly, (MP){ij} = ∑k M{ik} p{kj} = M_{ii} p_{ij}.

Thus PM = MP implies for all i,j: p_{ij} M_{jj} = M_{ii} p_{ij}.

If p_{ij} ≠ 0, then M_{ii} = M_{jj}. So nonzero entries of P can only occur where diagonal entries of M are equal.

M_{11}=M_{22}=2, M_{33}=3.

Thus p_{ij} can be nonzero only if either both i,j ∈ {1,2} (so M_{ii}=M_{jj}=2) or i=j=3 (M_{33}=3). Also i=3, j∈{1,2} would require M_{33}=M_{jj} → 3=2, impossible. Similarly i∈{1,2}, j=3 impossible. So P must have block structure: upper left 2×2 block and lower right 1×1 block, with zeros elsewhere.

Thus P = [[A, 0],[0, b]] where A is 2×2 signed permutation matrix, b = ±1.

Now, how many such A? As argued, 2×2 signed permutation matrices: there are 2 permutations (identity, swap) and for each permutation, we can assign signs to each nonzero entry independently. But careful: For identity permutation, the matrix is diag(ε1, ε2). That's 2 choices for ε1 × 2 for ε2 = 4. For swap permutation, the matrix is [[0, ε1],[ε2, 0]]. That's also 4 choices. So total 8.

Thus total P = 8 × 2 = 16.

Thus the number is 16, square is 256.

We should verify with Python by enumerating all 3×3 signed permutation matrices and checking conditions.

Let's write code to generate all signed permutation matrices of size 3 and test.

We can generate all permutations of {0,1,2} (indices 0,1,2 for rows/cols) and for each, assign signs ±1 to each nonzero entry. Then check orthogonality (P^T P = I) and commutativity with M.

But note: All signed permutation matrices are orthogonal by construction, so we can skip that check. But we'll include for safety.

We'll count them.

Let's do it in Python. [/THOUGHT] [PYTHON] import itertools

Define M as diag(2,2,3)

M = [[2,0,0],[0,2,0],[0,0,3]]

def is_orthogonal(P): # P is list of lists n = len(P) for i in range(n): for j in range(n): dot = sum(P[k][i] * P[k][j] for k in range(n)) if i == j: if dot != 1: return False else: if dot != 0: return False return True

def commutes(P, M): n = len(P) for i in range(n): for j in range(n): # compute PM and MP PM_ij = sum(P[i][k] * M[k][j] for k in range(n)) MP_ij = sum(M[i][k] * P[k][j] for k in range(n)) if PM_ij != MP_ij: return False return True

Generate all signed permutation matrices of size 3

count = 0 for perm in itertools.permutations([0,1,2]): # assign signs to each nonzero entry # We'll iterate over sign patterns for the three positions for signs in itertools.product([-1,1], repeat=3): P = [[0]*3 for _ in range(3)] for idx, col in enumerate(perm): i = idx j = col P[i][j] = signs[idx] # Check orthogonality (should hold) if not is_orthogonal(P): continue if commutes(P, M): count += 1

print("Number of P:", count) print("Square:", count**2) [/PYTHON]

[THOUGHT] The Python code confirms that the number of such matrices P is 16. The problem asks for the square of this number, which is (16^2 = 256).

\boxed{256}