A be the set of 3 × 3 matrices formed by entries 0, –1 and 1 only. There are three (1), three (–1) and three (0). The number of symmetric matrices with trace (A) = 0 is K then $\frac{K}{6}$ is

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answer is 6.

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Detailed Solution

The elements 0, –1, 1 can be arranged in the principal
diagnol in 3! ways (and remaining places can be filled
3! ways corresponding to each case $(∵ a_{i j} = a_{j i})$
No of matrices $= 3! 3! = 36 \Rightarrow \frac{K}{6} = 6$

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