An n times n matrix is formed using 0, 1 and | vedclass

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(A) For a matrix $A = [a_{ij}]$ to be skew-symmetric, it must satisfy the condition $a_{ij} = -a_{ji}$ for all $i, j$ and $a_{ii} = 0$ for all $i$.$1$

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$3^{n(n-1)/2}$

Vedclass · Answer

(A) For a matrix $A = [a_{ij}]$ to be skew-symmetric, it must satisfy the condition $a_{ij} = -a_{ji}$ for all $i, j$ and $a_{ii} = 0$ for all $i$.$1$. The diagonal elements $a_{ii}$ must be $0$. There is only $1$ choice for each of the $n$ diagonal elements.$2$. For the off-diagonal elements, we only need to choose the elements $a_{ij}$ where $i $3$. The number of pairs $(i, j)$ such that $i $4$. Each of these $\frac{n(n-1)}{2}$ positions can be filled with any of the $3$ values: $\{0, 1, -1\}$.$5$. Therefore, the total number of such skew-symmetric matrices is $3^{\frac{n(n-1)}{2}}$.

An $n \times n$ matrix is formed using $0, 1$ and $-1$ as its elements. The number of such matrices which are skew-symmetric is

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