The total number of 33 matrices A having entries from the set 0,1,2,3, such that the sum o…

Question

The total number of $3	imes3$ matrices $A$ having entries from the set $\{0,1,2,3\}$, such that the sum of all the diagonal entries of $AA^T$ is 9, is equal to

Answer: 766

Accepted Answer

$\operatorname{trace}(AA^T)$ equals the sum of squares of all 9 entries of $A$, each entry from $\{0,1,2,3\}$ (squares $0,1,4,9$). We need the 9 squares to sum to $9$.

Combinations of squares summing to 9 (each from $\{0,1,4,9\}$, 9 terms):
- Nine $1$'s: $1$ arrangement.
- One $9$, rest $0$: $\binom91=9$ arrangements.
- Two $4$'s + one $1$ + six $0$'s: $\dfrac{9!}{2!\,1!\,6!}=252$ arrangements.
- Four $1$'s + one $4$ + ... wait --- one $4$ + five $1$'s + three $0$'s ($4+5=9$): $\dfrac{9!}{1!\,5!\,3!}=504$ arrangements.

Total $=1+9+252+504=766$.

Solutions · 1