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(D) Let $A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}$.The diagonal entries of $AA^{T}$ are given by the sum of squares of th

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$766$

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(D) Let $A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}$.The diagonal entries of $AA^{T}$ are given by the sum of squares of the elements in each row of $A$.Specifically,the diagonal entries are $(a^{2}+b^{2}+c^{2})$,$(d^{2}+e^{2}+f^{2})$,and $(g^{2}+h^{2}+i^{2})$.The sum of all diagonal entries is $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+i^{2} = 9$,where $a, b, c, d, e, f, g, h, i \in \{0, 1, 2, 3\}$.We need to find the number of ways to choose these $9$ elements such that their squares sum to $9$.CaseDescriptionNumber of Matrices$1$. All $1$s$9$ ones,$0$ zeros$\frac{9!}{9!} = 1$$2$. One $3$One $3$,$8$ zeros$\frac{9!}{1! 	imes 8!} = 9$$3$. One $2$,one $1$One $2$,one $1$,$7$ zeros$\frac{9!}{1! 	imes 1! 	imes 7!} = 72$$4$. Two $2$s,one $1$Two $2$s,one $1$,$6$ zeros$\frac{9!}{2! 	imes 1! 	imes 6!} = 252$$5$. Three $1$s,two $2$sWait,let us re-evaluate combinations: $x_1^2 + ... + x_9^2 = 9$. Possible sets of squares: $\{1,1,1,1,1,1,1,1,1\}$,$\{9,0,0,0,0,0,0,0,0\}$,$\{4,4,1,0,0,0,0,0,0\}$,$\{4,1,1,1,1,1,0,0,0\}$.Recalculating based on partitions of $9$ into $9$ squares from $\{0, 1, 4, 9\}$:$1$. Nine $1$s: $\frac{9!}{9!} = 1$$2$. One $3$ $(3^2=9)$: $\frac{9!}{1!8!} = 9$$3$. One $2$ $(2^2=4)$ and five $1$s $(1^2=1)$: $\frac{9!}{1!5!3!} = 504$$4$. Two $2$s $(2^2=4)$ and one $1$ $(1^2=1)$: $\frac{9!}{2!1!6!} = 252$Total $= 1 + 9 + 504 + 252 = 766$.

The total number of $3 \times 3$ 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........

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