If matrix A=≤ft[{array}{cc} 1-i & i \\ -i & 1-i {array}] where i={-1}, then which of the following is correct? [NDA-I 2020]

Correct Answer is : (A‾)T+A({A})^{T}+A(A)T+A is hermitian

Correct Answer is : (A‾)T+A({A})^{T}+A(A)T+A is hermitian

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Question : 22 of 69

Marks: +1, -0

If matrix

A=\left[\begin{array}{cc} 1-i & i \\ -i & 1-i \end{array}\right]

i=\sqrt{-1}

Solution: 👈: Video Solution

A square

M

matrix is said to be Hermitian (or self-adjoint) if it is equal to its. Own Hermitian conjugate, i.e.

(\overline{M})^{T}=M

Given Matrix

A=\left[\begin{array}{cc} 1-i & i \\ -i & 1-i \end{array}\right]

(\overline{A})^{T}=\left[\begin{array}{cc} 1+i & i \\ -i & 1+i \end{array}\right]

Now,

A+(\overline{A})^{T}

=\left[\begin{array}{cc} 1-i & i \\ -i & 1-i \end{array}\right] +\left[\begin{array}{cc} 1+i & i \\ -i & 1+i \end{array}\right]

=\left[\begin{array}{cc} 2 & 2i \\ -2i & 2 \end{array}\right]

=2\left[\begin{array}{cc} 1 & i \\ -i & 1 \end{array}\right]

Conjugate transpose of

(A+(\overline{A})^{T})=2\left[\begin{array}{cc} 1 & i \\ -i & 1 \end{array}\right]

Hence,

(\overline{A})^{T}+A

is hermitian

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