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The number of skew symmetric matrices of order 3 can be formed by using all elements of set {0, -a, a, -b, b, -c, c} |
24 48 96 108 |
48 |
The correct answer is Option (2) → 48 $A=\begin{pmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{pmatrix}$ Number of skew-symmetric matrices = $3!×8$ $=6×8=48$ |
