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A is a 3 × 3 matrix with entries from the set {–1, 0, 1}. Then the probability that A is neither symmetric nor skew-symmetric is:

My thoughts: There can be nine members on a $3*3$ matrix and there are three possibilities for each member. Therefore the total no. of matrices possible is $3^9$. Subtracting the no. of possibilities for skew symmetric and symmetric matrices from it and dividing by $3^9$ will give the required probability but, I'm not able to figure out the values to be subtracted. Is there a relationship between the total number of possibilities and the number of possibilities for symmetric or skew-symmetric matrix.

Please help and thanks in advance.

2 Answers2

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Skew-symmetric matrices have $0$s on the diagonal, and opposing entries are determined by each other. There are then $3$ entries we have to choose for a skew-symmetric matrix, with $3$ options for each entry, so $3^3$ skew-symmetric matrices.

For symmetric matrices, we can choose any entries on the diagonal, and again, opposing entries are determined by each other. Thus, there are $6$ entries we have to choose for a symmetric matrix, with $3$ options for each entry, so $3^6$ symmetric matrices.

The only remaining question is: which matrix or matrices are both symmetric and skew-symmetric? If the number of such matrices is $n$, then there are $3^3+3^6-n$ matrices that are skew-symmetric or symmetric.

Cameron Buie
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Symmetry is 3 independent relations, each realized with probability $\frac13$, hence symmetry has probability $\frac1{3^3}$. Skew-symmetry is 6 independent relations, each realized with probability $\frac13$, hence skew-symmetry has probability $\frac1{3^6}$. Symmetry and skew-symmetry means being the null matrix, this is 9 independent relations, each realized with probability $\frac13$, hence it has probability $\frac1{3^9}$. Thus, symmetry and skew-symmetry are independent and to be neither symmetric nor skew-symmetric hapens with probability $\left(1-\frac1{3^3}\right)\cdot\left(1-\frac1{3^6}\right)$.

Did
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