The number of 3 × 2 matrices A, which can be formed using the elements of the set -2,-1,0,1,2 such that the sum of all the diagonal elements of A^ T…

For a 3×2 matrix A, the diagonal sum of $A^T A$ATA equals the sum of squares of all 6 elements of A. We need:

$a_{11}^2 + a_{21}^2 + a_{31}^2 + a_{12}^2 + a_{22}^2 + a_{32}^2 = 5$a112​+a212​+a312​+a122​+a222​+a322​=5 where each element is from $\{-2, -1, 0, 1, 2\}${−2,−1,0,1,2}.

Possible squares: $0^2=0$02=0, $(\pm 1)^2=1$(±1)2=1, $(\pm 2)^2=4$(±2)2=4.

Case 1:

Distribution $4+1+0+0+0+0$4+1+0+0+0+0 (one ±2, one ±1, four 0's).

Arrangements: $\binom{6}{1} \times \binom{5}{1} = 30$(16​)×(15​)=30.

Sign choices: $2 \times 2 = 4$2×2=4. Total: $30 \times 4 = 120$30×4=120.

Case 2:

Distribution $1+1+1+1+1+0$1+1+1+1+1+0 (five ±1, one 0).

Arrangements: $\binom{6}{1} = 6$(16​)=6 ways to place the 0.

Sign choices for five ±1: $2^5 = 32$25=32. Total: $6 \times 32 = 192$6×32=192.

Total matrices: $120 + 192 = 312$120+192=312