Let S be the set containing all 3 × 3 matrices with entries from \...

Given that the sum of all the diagonal elements of ${A}^{T}A$ is $6$, means sum of all diagonal elements is equal to sum of square each element of the matrix.

i.e., $A =$\left[$\begin{cases}{a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{cases}$\right]$and${A}^{T} = \left[$\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}$\right]$</p><p>Then${t}_{r}\left(A \cdot {A}^{T}\right)$= {a}_{1}^{2} + {a}_{2}^{2} + {a}_{3}^{2} + {b}_{1}^{2} + {b}_{2}^{2} + {b}_{3}^{2} + {c}_{1}^{2} + {c}_{2}^{2} + {c}_{3}^{2}$

Given ${a}_{i}, {b}_{i}, {c}_{i} \in \left\{- 1, 0, 1\right\}$ for $i = 1, 2, 3$

Now sum to be $6$ exactly three of them should be zero and rest should be $1$ or $- 1$.

So, total number of possible matrices $C3 9 \times {2}^{6}$

$= \dfrac{9 \times 8 \times 7}{6} \times 64$

$= 5376$