Count 3x3 Matrices with Trace of M Transpose M Equal to 5

To find the total number of $3 \times 3$ matrices $M$ whose elements belong to the set $\{0, 1, 2\}$ such that $\text{Tr}(M^T M) = 5$, we proceed as follows:

Let $M = (a_{ij})_{3 \times 3}$ where $a_{ij} \in \{0, 1, 2\}$ for $1 \le i, j \le 3$.

The trace of the matrix $M^T M$ is given by the sum of the squares of all the elements of $M$: $\text{Tr}(M^T M) = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^2$

Let:

• $x_0$ be the number of entries in $M$ equal to $0$,
• $x_1$ be the number of entries in $M$ equal to $1$,
• $x_2$ be the number of entries in $M$ equal to $2$.

Since $M$ has $3 \times 3 = 9$ entries, we have: $x_0 + x_1 + x_2 = 9$

The condition $\text{Tr}(M^T M) = 5$ gives: $x_0 \cdot (0^2) + x_1 \cdot (1^2) + x_2 \cdot (2^2) = 5$ $x_1 + 4x_2 = 5$

Since $x_1$ and $x_2$ are non-negative integers, we consider the possible values for $x_2$:

1. Case 1: $x_2 = 1$ $x_1 + 4(1) = 5 \implies x_1 = 1$ $x_0 = 9 - 1 - 1 = 7$ In this case, the matrix contains one $2$, one $1$, and seven $0$'s. The number of such matrices is the number of ways to arrange these entries in 9 positions: $N_1 = \frac{9!}{1! \cdot 1! \cdot 7!} = 9 \times 8 = 72$

2. Case 2: $x_2 = 0$ $x_1 + 4(0) = 5 \implies x_1 = 5$ $x_0 = 9 - 5 - 0 = 4$ In this case, the matrix contains five $1$'s and four $0$'s. The number of such matrices is the number of ways to arrange these entries in 9 positions: $N_2 = \frac{9!}{5! \cdot 4!} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

Therefore, the total number of such matrices $M$ is: $\text{Total} = N_1 + N_2 = 72 + 126 = 198$

Hence, the correct option is B.