] Sum of diagonal elements =a2+b2+c2+d2+e2+f2+g2+h2+i2=7 According to the question, the entries are {0,1,2}.‌‌[∵M⊤M=7] i.e. {a,b,c,...,h,i}={0,1,2} So, for Eq. (i) to be true, there are two cases. Case I When 7−1′ s are there and 2−0 's are there. ⇒‌‌‌9C7×‌2C2=36 ways of arrangements. Case II When 1−2 is there, 3−1′ s and 5−0′ s are there. ‌9C1×‌8C3×‌5C5=9×‌
8!
3!5!
×1 =504 ways of arrangements. ∴ Total possible arrangements =36+504=540