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Find the Hermitian matrix of matrix A=\begin{bmatrix} 3 & 4-2i & 5+3i\\ 4+2i& 4 & 4+5i\\ 5-3i & 4-5i & 5 \end{bmatrix}

Option: 1

A^{\theta} = \begin{bmatrix} 3i & -2i & 5-3i\\ 4+2i& 4 & 4+5i\\ 5+3i & 4-5i & 5i \end{bmatrix}


Option: 2

A^{\theta} = \begin{bmatrix} 3 & 4-2i & 5+3i\\ 4+2i& 4 & 4+5i\\ 5-3i & 4-5i & 5 \end{bmatrix}


Option: 3

A^{\theta} = \begin{bmatrix} 3 & 4-2i & 5+3i\\ 4+2i& 4 & 4-5i\\ 5-3i & 4+5i & 5 \end{bmatrix}


Option: 4

A^{\theta} = \begin{bmatrix} 3 & 4-2i & 5-3i\\ 4+2i& 4 & 4+5i\\ 5+3i & 4-5i & 5 \end{bmatrix}


Answers (1)

best_answer

 

 

Hermitian matrix -

Hermitian matrix

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

i .e. \mathrm{A=A^\theta,\;\;[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 and -ve to +ve for imaginary part of all elements, So to satisfy the condition A?  = A diagonal elements must not change, ⇒ all diagonal element must be purely real, 

 

 

-

For matrix to be hermitian A?  = A

So we find A?  and verify that it is equal to A or not 

To find A? , we first take the transpose of A and then it's conjugate,

So taking the transpose of A, we have

\\\mathrm{A' = \begin{bmatrix} 3 & 4+2i & 5-3i\\ 4-2i& 4 & 4-5i\\ 5+3i & 4+5i & 5 \end{bmatrix}} \\ \\\mathrm{taking \; its\ conjugate\; now} \\\\\mathrm{\overline{A'} = A^{\theta} = \begin{bmatrix} 3 & 4-2i & 5+3i\\ 4+2i& 4 & 4+5i\\ 5-3i & 4-5i & 5 \end{bmatrix} = A}

 

Hence option (b) is correct

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Pankaj

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