1. Interpret the condition

For a matrix $A=(a_{ij})$, the diagonal entries of $AA^T$ are

So the sum of all diagonal entries of $AA^T$ is

Thus the condition is:

where each $a_{ij}\in\{0,1,2,3\}$.

So we must count the number of $3\times 3$ matrices with 9 entries from $\{0,1,2,3\}$ such that the sum of squares of all 9 entries is 9.

2. Reduce to counting 9-tuples

Each entry contributes one of:

• $0^2=0$
• $1^2=1$
• $2^2=4$
• $3^2=9$

Let

• $x$ = number of entries equal to $1$
• $y$ = number of entries equal to $2$
• $z$ = number of entries equal to $3$
• remaining $9-x-y-z$ entries are $0$

Then

with $x,y,z\ge 0$ integers and $x+y+z\le 9$.

3. Find all possible cases

Solve

Case 1: $z=1$

Then

So we have exactly one entry equal to $3$, and the other 8 entries are $0$.

Number of matrices:

Case 2: $z=0$

Then

Possible values of $y$:

• $y=0 \Rightarrow x=9$
• $y=1 \Rightarrow x=5$
• $y=2 \Rightarrow x=1$

No larger $y$ works.

So the valid distributions are:

1. $(x,y,z)=(9,0,0)$
2. $(x,y,z)=(5,1,0)$
3. $(x,y,z)=(1,2,0)$

4. Count matrices in each distribution

(i) $(x,y,z)=(9,0,0)$

All 9 entries are $1$.

Number of matrices:

(ii) $(x,y,z)=(5,1,0)$

Choose 1 position for the entry $2$, then 5 of the remaining 8 positions for the entries $1$.

Number of matrices:

(iii) $(x,y,z)=(1,2,0)$

Choose 2 positions for the entries $2$, then 1 of the remaining 7 positions for the entry $1$.

Number of matrices:

(iv) $(x,y,z)=(0,0,1)$

Exactly one entry is $3$.

Number of matrices:

5. Add all cases

Total number of matrices:

6. Final answer

The required number of matrices is

This matches the stored correct answer.