If A = begin{bmatrix} 0 & 1+2i & i-2 -1-2i & | vedclass

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(C) For a matrix $A$ to be skew-symmetric,it must satisfy $A^T = -A$. Given the matrix $A = \begin{bmatrix} 0 & 1+2i & i-2 \ -1-2i & 0 & K \ 2-i & -

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$7$

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(C) For a matrix $A$ to be skew-symmetric,it must satisfy $A^T = -A$. Given the matrix $A = \begin{bmatrix} 0 & 1+2i & i-2 \ -1-2i & 0 & K \ 2-i & -7 & 0 \end{bmatrix}$. Comparing the elements $A_{ij} = -A_{ji}$,we observe: $A_{12} = 1+2i$ and $A_{21} = -1-2i = -(1+2i)$,which is consistent. $A_{13} = i-2$ and $A_{31} = 2-i = -(i-2)$,which is consistent. $A_{23} = K$ and $A_{32} = -7$. For skew-symmetry,$A_{23} = -A_{32}$,so $K = -(-7) = 7$. However,the problem states $A^{-1}$ does not exist,which implies $\det(A) = 0$. For a skew-symmetric matrix of odd order $(3 	imes 3)$,the determinant is always $0$. Thus,the condition $A^{-1}$ does not exist is satisfied for any $K$ that makes the matrix skew-symmetric. Therefore,$K = 7$.

If $A = \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & K \\ 2-i & -7 & 0 \end{bmatrix}$ and $A^{-1}$ does not exist,then $K = $ (where $i = \sqrt{-1}$)

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