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(D) To determine the nature of the matrix $A = \begin{bmatrix} i & 1 - 2i \ -1 - 2i & 0 \end{bmatrix}$,we calculate the conjugate transpose $(\bar{A}

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Skew-hermitian

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(D) To determine the nature of the matrix $A = \begin{bmatrix} i & 1 - 2i \ -1 - 2i & 0 \end{bmatrix}$,we calculate the conjugate transpose $(\bar{A})^T$.First,find the conjugate matrix $\bar{A}$ by changing the sign of the imaginary part of each element:$\bar{A} = \begin{bmatrix} -i & 1 + 2i \ -1 + 2i & 0 \end{bmatrix}$.Next,find the transpose of the conjugate matrix $(\bar{A})^T$:$(\bar{A})^T = \begin{bmatrix} -i & -1 + 2i \ 1 + 2i & 0 \end{bmatrix}$.Now,compare this with $-A$:$-A = -\begin{bmatrix} i & 1 - 2i \ -1 - 2i & 0 \end{bmatrix} = \begin{bmatrix} -i & -1 + 2i \ 1 + 2i & 0 \end{bmatrix}$.Since $(\bar{A})^T = -A$,the matrix $A$ is skew-hermitian.

The matrix $A = \begin{bmatrix} i & 1 - 2i \\ -1 - 2i & 0 \end{bmatrix}$ is which of the following?

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