Let A=[adgbehcfi],AT=[abcdefghi]
Diagonal elements of AATisa2+b2+c2,d2+e2+f2,g2+b2+c2.
Sum =a2+b2+c2+d2+e2+f2+g2+h2+i2=9
a,b,c,d,e,f,g,h,i∈0,1,2,3
| Case | No. of Matrices |
| (1) | All – 1′s | 9!9!=1 |
| (2) | One →3′s, remaining-0′s | 1!×8!9!=9 |
| (3) | One-2′s Five-1′s Three-0′s | 1!×5!×3!9!=8×63 |
| (4) | Two -2′s One-1′s Six-0′s | 2!×6!9!=63×4 |
Then, the total no. of ways
=1+9+8×63+63×4
=766