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Step-by-Step Solution
Step 1: Understand the Problem
We are given a 3×3 matrix M whose entries come from the set {0, 1, 2}. We need the sum of the diagonal elements of the product MTM to be exactly 7. Recall that if
M = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix},
then
M^T M = \begin{pmatrix}
a & d & g \\
b & e & h \\
c & f & i
\end{pmatrix}
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}.
The diagonal elements of M^T M are
a^2 + b^2 + c^2, \quad d^2 + e^2 + f^2, \quad g^2 + h^2 + i^2
when summed across all three diagonal positions, they simplify to
a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2.
Therefore, the condition “sum of diagonal elements of M^T M is 7” translates to
a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 = 7
with each variable taking values from {0, 1, 2}.
Step 2: Possible Squares for Entries in {0,1,2}
Since each entry can be 0, 1, or 2, their squares can be:
0^2 = 0
1^2 = 1
2^2 = 4
We want to form a total of 7 by summing nine terms, each of which can be 0, 1, or 4.
Step 3: Identify Valid Combinations of Squares Summing to 7
We look for all possible ways to sum up to 7 using only 0, 1, and 4:
Case I: Using only 1's and 0's. We need seven 1's and two 0's, since 7 \times 1 + 2 \times 0 = 7.
Case II: Using one 4 and three 1's (with the rest 0's). We need 4 + 3 \times 1 = 7, and thus one 2 (whose square is 4), three 1's, and five 0's in total.
Step 4: Count the Number of Such Matrices for Each Case
Since each matrix entry can be placed independently in one of the nine positions:
Case I: Seven 1's and two 0's.
We choose 7 positions (out of 9) to place 1's, and the remaining 2 positions are automatically 0's. The number of ways is
\binom{9}{7} = \binom{9}{2} = 36.
Case II: One 2, three 1's, and five 0's.
First choose 1 position (out of 9) for the entry 2. Then from the remaining 8 positions, choose 3 for the entry 1; the remaining 5 are 0's. Hence the number of ways is
\binom{9}{1} \times \binom{8}{3} = 9 \times 56 = 504.
Step 5: Summation of the Two Cases
Adding the numbers from the two cases:
36 + 504 = 540.
Thus, the maximum number of 3×3 matrices with entries from {0,1,2} that satisfy the given condition is 540.
Final Answer
540
