How many $3\times 3$ matrices $M$ with entries from $\{0, 1, 2\}$ are there for which the sum of the diagonal entries of $M'M$ is 5.
Step-by-Step Solution
Key Concept: The trace of M'M equals the sum of squares of all entries of M. For a 3×3 matrix M with entries from {0,1,2}, we need sum of all 9 squared entries equal to 5, then count solutions using partitions of 5 into at most 9 parts with each part in {0,1,4}.
The matrix is symmetric with diagonal entries $(a, e, i)$ and off-diagonal pairs $(b,d), (c,g), (f,h)$. The sum of squares constraint is $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 = 5$. Using symmetry ($b=d, c=g, f=h$) and case analysis on the count of 1's and 2's, Case I gives $\binom{9}{5} = 126$ and Case II gives $\binom{9}{2} \cdot 2! = 72$, totaling 198 matrices.
Correct Answer: 1