If both left( A - frac{I}{2} right) and left( | vedclass

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(B) Given that $\left( A - \frac{I}{2} \right)$ and $\left( A + \frac{I}{2} \right)$ are orthogonal matrices.For $\left( A - \frac{I}{2} \right)$ to b

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$A$ is a skew-symmetric matrix of even order

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(B) Given that $\left( A - \frac{I}{2} ight)$ and $\left( A + \frac{I}{2} ight)$ are orthogonal matrices.For $\left( A - \frac{I}{2} ight)$ to be orthogonal:$\left( A - \frac{I}{2} ight) \left( A - \frac{I}{2} ight)^T = I$$\left( A - \frac{I}{2} ight) \left( A^T - \frac{I}{2} ight) = I$$AA^T - \frac{A}{2} - \frac{A^T}{2} + \frac{I}{4} = I$$AA^T - \frac{1}{2}(A + A^T) = \frac{3}{4}I$  .......$(1)$For $\left( A + \frac{I}{2} ight)$ to be orthogonal:$\left( A + \frac{I}{2} ight) \left( A + \frac{I}{2} ight)^T = I$$\left( A + \frac{I}{2} ight) \left( A^T + \frac{I}{2} ight) = I$$AA^T + \frac{A}{2} + \frac{A^T}{2} + \frac{I}{4} = I$$AA^T + \frac{1}{2}(A + A^T) = \frac{3}{4}I$  .......$(2)$Subtracting $(1)$ from $(2)$:$AA^T + \frac{1}{2}(A + A^T) - (AA^T - \frac{1}{2}(A + A^T)) = \frac{3}{4}I - \frac{3}{4}I$$(A + A^T) = 0 \Rightarrow A^T = -A$This implies $A$ is a skew-symmetric matrix.Adding $(1)$ and $(2)$:$2AA^T = \frac{6}{4}I \Rightarrow AA^T = \frac{3}{4}I$Since $A^T = -A$,we have $A(-A) = \frac{3}{4}I \Rightarrow -A^2 = \frac{3}{4}I \Rightarrow A^2 = -\frac{3}{4}I$.For a skew-symmetric matrix $A$ of odd order,the determinant $|A| = 0$. However,from $AA^T = \frac{3}{4}I$,taking the determinant gives $|A||A^T| = |\frac{3}{4}I| = (\frac{3}{4})^n |I|$. Since $|A| = |A^T|$,we have $|A|^2 = (\frac{3}{4})^n$. This implies $|A| 
eq 0$,which is only possible if the order $n$ is even. Thus,$A$ is a skew-symmetric matrix of even order.

If both $\left( A - \frac{I}{2} \right)$ and $\left( A + \frac{I}{2} \right)$ are orthogonal matrices,then:

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