Correct Answer: 312

Explanation

Let $$ A= \begin{pmatrix} a & b\\ c & d\\ e & f \end{pmatrix}, \qquad a,b,c,d,e,f \in \{-2,-1,0,1,2\}. $$

Then $$ A^T A= \begin{pmatrix} a^2+c^2+e^2 & ab+cd+ef\\ ab+cd+ef & b^2+d^2+f^2 \end{pmatrix}. $$

Sum of diagonal elements of $A^TA$ is $$ a^2+b^2+c^2+d^2+e^2+f^2. $$

Given $$ a^2+b^2+c^2+d^2+e^2+f^2=5. $$

Possible square values from the given set are $0,1,4$.

To get total sum $5$, the only possible combination is:

one entry with square $4$ and one entry with square $1$, remaining four entries $0$.

Choice of the position of $\pm2$ (square $4$): $6$ ways.

Choice of the position of $\pm1$ (square $1$) from remaining $5$ positions: $5$ ways.

Each $\pm2$ can be chosen in $2$ ways and each $\pm1$ in $2$ ways.

Remaining entries are forced to be $0$.

Hence total number of matrices: $$ 6 \times 5 \times 2 \times 2 = 120. $$

Additionally, cases where five entries have square $1$ and one entry has square $0$ also give sum $5$.

Choose the zero position: $6$ ways.

Remaining five positions each have $\pm1$: $2^5$ ways.

Total for this case: $$ 6 \times 32 = 192. $$

Therefore, total matrices: $$ 120+192=\boxed{312}. $$

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