If a square matrix A is such that left(A^T-fr | vedclass

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(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$. Expanding bot

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skew-symmetric matrix

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(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$. Expanding both sides: $A^T A - \frac{1}{2} A^T - \frac{1}{2} A + \frac{1}{4} I = A^T A + \frac{1}{2} A^T + \frac{1}{2} A + \frac{1}{4} I$. Subtracting $A^T A + \frac{1}{4} I$ from both sides,we get: $-\frac{1}{2} A^T - \frac{1}{2} A = \frac{1}{2} A^T + \frac{1}{2} A$. Rearranging the terms: $0 = A^T + A$. Therefore,$A^T = -A$. This is the condition for a skew-symmetric matrix. Hence,$A$ is a skew-symmetric matrix.

If a square matrix $A$ is such that $\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$,where $I$ is a unit matrix,then $A$ is

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