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Find the skew-hermitian matrix of matrix \begin{bmatrix} i & 1-i & 2\\ -1-i & 3i & i\\ -2 & i & 0 \end{bmatrix}.

Option: 1

\begin{bmatrix} -i & -1+i & -2\\ 1+i & -3i & -i\\ 2 & -i & 0 \end{bmatrix}


Option: 2

\begin{bmatrix} i & 1-i & 2\\ -1-i & 3i & i\\ -2 & i & 0 \end{bmatrix}


Option: 3

\begin{bmatrix} i & -1+i & -2\\ 1+i & 3i & -i\\ 2 & -i & 0 \end{bmatrix}


Option: 4

\begin{bmatrix} -i & -1+i & 2\\ 1+i & 3i & -i\\ -2 & -i & 0 \end{bmatrix}


Answers (1)

best_answer

 

 

 

Skew-hermitian matrix -

Skew-hermitian matrix

A square matrix \mathrm{A=[a_{ij}]_{n\times n}} is said to be Skew-Hermitian matrix if \mathrm{\mathit{a_{ij}=-\overline{a_{ij}}}} ∀ i, j,  

i .e.\mathrm{A^\theta=-A,\;\;[where\;A^\theta\;is\;conjugate\;transpose \;of\;matrix \;A]}

 

We know that when we take the transpose of a matrix, its diagonal elements remain the same, and while taking conjugate we just change sign from +ve to -ve OR -ve to +ve in imaginary part of all elements, So to satisfy the condition A?  = - A, all diagonal element must be purely imaginary. As A?  = - A so 

 

\\\mathrm{a_{ij}=-\overline{a_{ij}} \;\; \forall \;i,j} \\\mathrm{if \; we \; put \; i=j, \; we\; have} \\\mathrm{a_{ii}= -\overline{a_{ii}} \Rightarrow a_{ii}+\overline{a_{ii}}=0} \\\mathrm{\Rightarrow a_{ii}=0}

Hence all diagonal element should be purely imaginary

 

-

 

  first, we take the transpose and then it's conjugate and equate it to -A.

 

\\\mathrm{A' = \begin{bmatrix} i & -1-i & -2\\ 1-i & 3i & i\\ 2 & i & 0 \end{bmatrix}} \\\\\mathrm{now\; taking\; conjugate \; of \; the \; transpose} \\ \\\mathrm{\overline{A'} = \begin{bmatrix} -i & -1+i & -2\\ 1+i & -3i & -i\\ 2 & -i & 0 \end{bmatrix} = -A}

hence option (a) is correct

Posted by

Ajit Kumar Dubey

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