The total number of 3 × 3 matrices A having enteries from the set 0 , 1 , 2 , 3 such that the sum of all the diagonal entries of A A T is 9 , is equal to

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Let A = a b c d e f g h i , A T = a d g b e h c f i Diagonal elements of A A T is a 2 + b 2 + c 2 , d 2 + e 2 + f 2 , g 2 + b 2 + c 2 . Sum = a 2 + b 2 + c 2 + d 2 + e 2 + f 2 + g 2 + h 2 + i 2 = 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 9 ! 1 ! × 8 ! = 9 (3) One- 2 ' s Five- 1 ' s Three- 0 ' s 9 ! 1 ! × 5 ! × 3 ! = 8 × 63 (4) Two - 2 ' s One- 1 ' s Six- 0 ' s 9 ! 2 ! × 6 ! = 63 × 4 Then, the total no. of ways = 1 + 9 + 8 × 63 + 63 × 4 = 766

	Case	No. of Matrices
(1)	All – $1^{'} s$	$\frac{9!}{9!} = 1$
(2)	One $\to 3^{'} s$ , remaining- $0^{'} s$	$\frac{9!}{1! \times 8!} = 9$
(3)	One- $2^{'} s$ Five- $1^{'} s$ Three- $0^{'} s$	$\frac{9!}{1! \times 5! \times 3!} = 8 \times 63$
(4)	Two - $2' s$ One- $1^{'} s$ Six- $0' s$	$\frac{9!}{2! \times 6!} = 63 \times 4$

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