How many `3xx3` matrices `M` with entries from `{0,1,2}` are there, for which the sum of the diagonal entries of `M^T Mi s5?`

Options

A. 126
B. 198
C. 135
D. 162

Explanation

To solve the problem of finding how many \(3 \times 3\) matrices \(M\) with entries from \(\{0, 1, 2\}\) exist such that the sum of the diagonal entries of \(M^T M\) is 5, we can follow these steps:

### Step 1: Define the Matrix
Let \(M\) be a \(3 \times 3\) matrix represented as:
\[ M = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \]
where each entry \(a_{ij} \in \{0, 1, 2\}\).

### Step 2: Compute \(M^T M\)
The transpose \(M^T\) of matrix \(M\) is:
\[ M^T = \begin{pmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]
The product \(M^T M\) is calculated as follows:
\[ M^T M = \begin{pmatrix} a_{11}^2 + a_{21}^2 + a_{31}^2 & a_{11}a_{12} + a_{21}a_{22} + a_{31}a_{32} & a_{11}a_{13} + a_{21}a_{23} + a_{31}a_{33} \\ a_{12}a_{11} + a_{22}a_{21} + a_{32}a_{31} & a_{12}^2 + a_{22}^2 + a_{32}^2 & a_{12}a_{13} + a_{22}a_{23} + a_{32}a_{33} \\ a_{13}a_{11} + a_{23}a_{21} + a_{33}a_{31} & a_{13}a_{12} + a_{23}a_{22} + a_{33}a_{32} & a_{13}^2 + a_{23}^2 + a_{33}^2 \end{pmatrix} \]

### Step 3: Sum of Diagonal Entries
The sum of the diagonal entries of \(M^T M\) is:
\[ \text{Trace}(M^T M) = a_{11}^2 + a_{22}^2 + a_{33}^2 + a_{21}^2 + a_{12}^2 + a_{32}^2 + a_{31}^2 + a_{23}^2 + a_{13}^2 \]
This can be simplified to:
\[ S = a_{11}^2 + a_{21}^2 + a_{31}^2 + a_{12}^2 + a_{22}^2 + a_{32}^2 + a_{13}^2 + a_{23}^2 + a_{33}^2 \]
We need \(S = 5\).

### Step 4: Count the Combinations
We need to find all combinations of entries from \(\{0, 1, 2\}\) such that the sum of squares equals 5.

#### Case 1: Five 1's
- Choose 5 entries to be 1 and the rest to be 0.
- The number of ways to choose 5 positions from 9 is \(\binom{9}{5}\).

#### Case 2: One 2 and One 1
- Choose 1 entry to be 2, 1 entry to be 1, and the rest to be 0.
- The number of ways to choose 1 position for 2 from 9 is \(\binom{9}{1}\) and for 1 from the remaining 8 is \(\binom{8}{1}\).

### Step 5: Calculate Each Case
1. For Case 1:
\[ \binom{9}{5} = 126 \]

2. For Case 2:
\[ \binom{9}{1} \cdot \binom{8}{1} = 9 \cdot 8 = 72 \]

### Step 6: Total Combinations
The total number of \(3 \times 3\) matrices \(M\) such that the sum of the diagonal entries of \(M^T M\) is 5 is:
\[ 126 + 72 = 198 \]

### Final Answer
Thus, the total number of such matrices \(M\) is \(198\).

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