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If $A = [a_{ij}]$ is skew symmetric matrix of order 'n', then |
$a_{ij}=\frac{1}{a_{ij}}$ for all $i, j$ $a_{ij}=0$ for $i= j$ $a_{ij}=0$ for all $i, j$ $a_{ij}≠0$ for $i= j$ |
$a_{ij}=0$ for $i= j$ |
The correct answer is Option (2) → $a_{ij}=0$ for $i= j$ For a skew-symmetric matrix $A = [a_{ij}]$, $A' = -A \Rightarrow a_{ji} = -a_{ij}$ When $i = j$, this gives $a_{ii} = -a_{ii} \Rightarrow 2a_{ii} = 0 \Rightarrow a_{ii} = 0$ $a_{ij} = 0$ for $i = j$ |
