A) A is orthogonal
B) A is skew-symmetric of even order
C) \[{{A}^{2}}=\frac{3}{4}I\]
D) none of these
Correct Answer: B
Solution :
[b] \[\left( A'-\frac{1}{2}I \right)\left( A-\frac{1}{2}I \right)=I\]and \[\left( A'+\frac{1}{2}I \right)\left( A+\frac{1}{2}I \right)=I\] \[\Rightarrow A+A'=0\] (Subtracting the two results) or \[A'=-A\] \[\Rightarrow {{A}^{2}}=-\frac{3}{4}I\] Or \[{{\left( \frac{-3}{4} \right)}^{^{n}}}={{(det(A))}^{2}}\] Hence, n is even.
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