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Step-by-Step Solution
Step 1: Express the product AAT
Let
A = \begin{bmatrix}
x & y & z\\
a & b & c\\
d & e & f
\end{bmatrix}
where each entry is from the set {0, 1, 2, 3}. Then
A A^T is given by
A A^T = \begin{bmatrix}
x & y & z\\
a & b & c\\
d & e & f
\end{bmatrix}
\begin{bmatrix}
x & a & d\\
y & b & e\\
z & c & f
\end{bmatrix}
=
\begin{bmatrix}
x^2 + y^2 + z^2 & ax + by + cz & dx + ey + fz\\
ax + by + cz & a^2 + b^2 + c^2 & ad + be + cf\\
dx + ey + fz & ad + be + cf & d^2 + e^2 + f^2
\end{bmatrix}.
Step 2: Identify the trace condition
The trace of A A^T is the sum of its diagonal elements. Hence,
\text{Tr}(A A^T) = (x^2 + y^2 + z^2) + (a^2 + b^2 + c^2) + (d^2 + e^2 + f^2).
We are given that the trace equals 9:
x^2 + y^2 + z^2 + a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 9,
where each of x, y, z, a, b, c, d, e, f is in {0, 1, 2, 3}.
Step 3: List the possible ways to sum squares to 9
Each variable can only take the values 0, 1, 2, or 3. Their squares are 0, 1, 4, and 9 respectively. We must count the number of ways to distribute these values among the nine entries so that their squares sum to 9.
Case I: Nine 1's
If all nine entries are 1, then
1^2 + 1^2 + \cdots + 1^2 = 9.
This configuration is unique:
All entries are 1, so there is exactly 1 way.
Case II: One 3 and the rest 0's
If exactly one entry is 3 (square is 9) and the remaining eight entries are 0 (square is 0), then the sum of the squares is 9. We just need to choose which of the 9 positions gets the value 3:
Number of ways = 9.
Case III: Two 2's and one 1, remaining six 0's
Squares: 2^2 = 4, so two 2's contribute 4 + 4 = 8, and one 1 contributes 1, making 9 in total. We choose places for the two 2's and one 1:
Choose 2 positions out of 9 for the two 2's: \binom{9}{2} = 36.
Choose 1 position out of the remaining 7 for the 1: \binom{7}{1} = 7.
Total ways = 36 \times 7 = 252.
Case IV: One 2 and five 1's, remaining three 0's
Squares: 2^2 = 4 and 1^2 = 1. Thus one 2 and five 1's sum to 4 + 5 = 9. The remaining three entries are 0. We count the ways:
Choose 1 position out of 9 for the single 2: \binom{9}{1} = 9.
Choose 5 positions out of the remaining 8 for the ones: \binom{8}{5} = 56.
Total ways = 9 \times 56 = 504.
Step 4: Sum all possible cases
Adding the results from all four cases gives the total number of valid matrices:
\text{Total} = 1 + 9 + 252 + 504 = 766.
Final Answer
The total number of 3×3 matrices A (with entries in {0,1,2,3}) such that the trace of A A^T is 9, is
766.
