Question

The number of all $3\times 3$ matrices $A$, with entries from the set $\{-1,0,1\}$ such that the sum of the diagonal elements of $\left(A{A}^T\right)$ is $3$, is

Answer 1

$\text{tr}\left(A{A}^T\right)=3$
Let $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Then $A^T=\left[\begin{array}{ccc}{a}_{11}& a_{21}& a_{31}\\ a_{12}& a_{22}& a_{32}\\ a_{13}& a_{23}& a_{33}\end{array}\right]$

$\text{tr}\left(A{A}^T\right)=a_{11}^2+a_{12}^2+a_{13}^2+a_{21}^2+a_{22}^2+a_{23}^2+a_{31}^2+a_{32}^2+a_{33}^2=3$
So, out of nine elements $a_{ij}^{\prime }s$, three elements must be equal to $1$ or $-1$ and rest elements must be $0$.
So, the possible cases will be:
For six $0^{\prime }s$ and three $1^{\prime }s$, total possibilities is ${}^9{C}_6$
For six $0^{\prime }s$ , two $1^{\prime }s$ and one $-1^{\prime }s$, total possibilities is ${}^9{C}_6\times 3$
For six $0^{\prime }s$, one $1^{\prime }s$ and two $-1^{\prime }s$, total possibilities is ${}^9{C}_6\times 3$
For six $0^{\prime }s$ and three $-1^{\prime }s$, total possibilities is ${}^9{C}_6$
$\therefore$ Total number of cases $={}^9{C}_6\times 8=672$