To solve the problem, we need to find the number of 3x3 matrices \( X \) with entries from the set \{0, 2, 3\} such that the sum of the principal diagonal elements of \( X \cdot X^T \) equals 28.
### Step-by-Step Solution:
1. **Understanding the Matrix Product**:
The matrix \( X \) is a 3x3 matrix, which we can denote as:
\[
X = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\]
The transpose of \( X \), denoted \( X^T \), is:
\[
X^T = \begin{pmatrix}
a & d & g \\
b & e & h \\
c & f & i
\end{pmatrix}
\]
2. **Calculating \( X \cdot X^T \)**:
The product \( X \cdot X^T \) results in a 3x3 matrix where the diagonal elements are given by:
\[
(X \cdot X^T)_{11} = a^2 + b^2 + c^2
\]
\[
(X \cdot X^T)_{22} = d^2 + e^2 + f^2
\]
\[
(X \cdot X^T)_{33} = g^2 + h^2 + i^2
\]
Therefore, the sum of the principal diagonal elements of \( X \cdot X^T \) is:
\[
S = a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2
\]
3. **Setting Up the Equation**:
We need this sum \( S \) to equal 28:
\[
S = a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 = 28
\]
4. **Identifying Possible Values**:
The entries of the matrix \( X \) can be 0, 2, or 3. Squaring these values gives:
- \( 0^2 = 0 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
5. **Finding Combinations**:
Let \( x_0 \), \( x_2 \), and \( x_3 \) be the number of entries that are 0, 2, and 3 respectively. The total number of entries is 9 (since it is a 3x3 matrix):
\[
x_0 + x_2 + x_3 = 9
\]
The equation for the sum of squares becomes:
\[
4x_2 + 9x_3 = 28
\]
6. **Solving the Equations**:
From \( 4x_2 + 9x_3 = 28 \), we can express \( x_2 \) in terms of \( x_3 \):
\[
x_2 = \frac{28 - 9x_3}{4}
\]
Since \( x_2 \) must be a non-negative integer, \( 28 - 9x_3 \) must be non-negative and divisible by 4.
Testing possible values for \( x_3 \):
- If \( x_3 = 0 \): \( 4x_2 = 28 \) → \( x_2 = 7 \) → \( x_0 = 2 \)
- If \( x_3 = 1 \): \( 4x_2 = 19 \) → Not possible (not divisible by 4)
- If \( x_3 = 2 \): \( 4x_2 = 10 \) → \( x_2 = 2 \) → \( x_0 = 5 \)
- If \( x_3 = 3 \): \( 4x_2 = 1 \) → Not possible (not divisible by 4)
The valid combinations are:
- \( (x_0, x_2, x_3) = (2, 7, 0) \)
- \( (x_0, x_2, x_3) = (5, 2, 2) \)
7. **Counting the Matrices**:
For \( (x_0, x_2, x_3) = (2, 7, 0) \):
The number of ways to choose 2 positions for 0s out of 9:
\[
\binom{9}{2} = 36
\]
For \( (x_0, x_2, x_3) = (5, 2, 2) \):
The number of ways to choose 5 positions for 0s, 2 positions for 2s, and the remaining 2 will be 3s:
\[
\frac{9!}{5! \cdot 2! \cdot 2!} = \frac{362880}{120 \cdot 2 \cdot 2} = 756
\]
8. **Total Count**:
The total number of matrices is:
\[
36 + 756 = 792
\]
### Final Answer:
The number of matrices \( X \) is **792**.
