Matrices and Determinants Part 4

Concept:The sum of diagonal elements of $A^T A$ equals the sum of squares of all entries of $A$.We need this sum to be 5 using entries from $\{-2,-1,0,1,2\}$.Explanation:Let $A$ be $3\times 2$ with entries $a,b,c,d,e,f$ from the set.Then $A^T A$ is $2\times 2$. Its diagonal sum is $a^2+b^2+c^2+d^2+e^2+f^2=5$.Possible squares are only $0,1,4$.We consider two cases for the multiset of squares:Case 1: One square is $4$, one square is $1$, and the remaining four are $0$.The square $4$ comes from $\pm 2$ (2 ways), $1$ from $\pm 1$ (2 ways), $0$ only from $0$ (1 way).Choose which position gets $4$: $\binom{6}{1}$ ways (6 choices).From remaining 5 positions, choose which gets $1$: $\binom{5}{1}$ ways.The other four automatically get $0$.Total for case 1: $\binom{6}{1}\times 2 \times \binom{5}{1}\times 2 = 12 \times 10 = 120$.Case 2: Five squares are $1$ and one square is $0$.Each $1$ comes from $\pm 1$ (2 ways each), $0$ from $0$ (1 way).Choose which five positions have $1$: $\binom{6}{5}=6$ ways.For those five positions, each has 2 choices, so $2^5=32$ ways.The remaining position is automatically $0$ (1 way).Total for case 2: $6 \times 32 = 192$.Total matrices = $120 + 192 = 312$.Answer:312 (Option A).