The number of all 3 times 3 matrices A,with e | vedclass

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(B) Let $A = [a_{ij}]$ be a $3 	imes 3$ matrix where $a_{ij} \in \{-1, 0, 1\}$.The sum of the diagonal elements of $AA^{T}$ is given by $\operatornam

Vedclass · Accepted Answer

$672$

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(B) Let $A = [a_{ij}]$ be a $3 	imes 3$ matrix where $a_{ij} \in \{-1, 0, 1\}$.The sum of the diagonal elements of $AA^{T}$ is given by $\operatorname{trace}(AA^{T})$.We know that $\operatorname{trace}(AA^{T}) = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2}$.Given that $\operatorname{trace}(AA^{T}) = 3$,we have $\sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2} = 3$.Since $a_{ij} \in \{-1, 0, 1\}$,$a_{ij}^{2}$ can only be $0$ or $1$.For the sum of nine such squares to be $3$,exactly three of the entries $a_{ij}$ must be $\pm 1$,and the remaining six entries must be $0$.First,we choose $3$ positions out of $9$ for the non-zero entries in $\binom{9}{3}$ ways.Then,for each of these $3$ chosen positions,the entry can be either $1$ or $-1$,which gives $2^{3}$ possibilities.Therefore,the total number of such matrices is $\binom{9}{3} 	imes 2^{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes 8 = 84 	imes 8 = 672$.

The number of all $3 \times 3$ matrices $A$,with entries from the set $\{-1, 0, 1\}$ such that the sum of the diagonal elements of $AA^{T}$ is $3$,is

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