A) If n is odd, then A is an invertible matrix,
B) If n is even, then A is not an invertible matrix,
C) For all values of n, A is not invertible matrix,
D) A is a skew-symmetric matrix.
Correct Answer: D
Solution :
Let \[A=\left[ \begin{matrix} 0 & {{\sin }^{-1}}(-1) & {{\sin }^{-1}}(-2) \\ {{\sin }^{-1}}(1) & 0 & {{\sin }^{-1}}(-1) \\ {{\sin }^{-1}}(2) & {{\sin }^{-1}}(1) & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 0 & -{{\sin }^{-1}}(-1) & -{{\sin }^{-1}}(-2) \\ {{\sin }^{-1}}(1) & 0 & -{{\sin }^{-1}}(-1) \\ {{\sin }^{-1}}(2) & {{\sin }^{-1}}(1) & 0 \\ \end{matrix} \right]\] As the matrix is skew-symmetric hence, \[|A|\,=0,\] if n is odd. \[\Rightarrow \] A is not an invertible matrix, when n is odd.
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