Let S be the set containing all 3× 3 matrices with entries from -1,0,1. The total number of matrices A∈ S such that the sum of all the diagonal…

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

i.e., $A=[\begin{matrix}{a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{matrix}]$A=[a1​b1​c1​​a2​b2​c2​​a3​b3​c3​​] and ${A}^{T}=[\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}]$AT=[a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​]

Then ${t}_{r}(A\cdot {A}^{T})$tr​(A⋅AT)$={a}_{1}^{2}+{a}_{2}^{2}+{a}_{3}^{2}+{b}_{1}^{2}+{b}_{2}^{2}+{b}_{3}^{2}+{c}_{1}^{2}+{c}_{2}^{2}+{c}_{3}^{2}$=a12​+a22​+a32​+b12​+b22​+b32​+c12​+c22​+c32​

Given ${a}_{i},{b}_{i},{c}_{i}\in {-1,0,1}$ai​,bi​,ci​∈−1,0,1 for $i=1,2,3$i=1,2,3

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

So, total number of possible matrices $C39\times {2}^{6}$C39×26

$$=\frac{9\times 8\times 7}{6}\times 64$$=69×8×7​×64

$$=5376$$=5376