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If $A_{2x2}$ is a matrix such that $a_{iT}= \left ( \omega \right )^{i+T}$ where $\omega$ is the cube root of unity then $B=A+\overline{A}$ . Find $\overline{B}$ .

Option 1)
$\begin{bmatrix} \omega &1 \\ 1 & \omega \end{bmatrix}$

Option 2)
$\begin{bmatrix} 2\omega^{2} &2 \\ 2 & 2\omega \end{bmatrix}$

Option 3)
$\begin{bmatrix} -1 &2 \\ 2 & -1 \end{bmatrix}$

Option 4)
none of these

Answers (1)

best_answer

As we learnt

Property of Conjugate -

$\overline{A+B}= \overline{A}+\overline{B}$

- wherein

Conjugate of matrix $A$ is $\overline{A}$

$A=\begin{bmatrix} \omega ^{2} &\omega ^{3} \\ \omega ^{}3 & \omega ^{4} \end{bmatrix}$

$A=\begin{bmatrix} \omega ^{2} &1 \\ 1 & \omega \end{bmatrix}$ $\overline{A}=\begin{bmatrix} \omega &1 \\ 1 & \omega^{2} \end{bmatrix}$

Thus $A+\overline{A}=\begin{bmatrix} \omega^{2} +\omega &2 \\ 2 &\omega+ \omega^{2} \end{bmatrix}$

$=\begin{bmatrix} -1 &2 \\2 &-1 \end{bmatrix}$

Option 1)

$\begin{bmatrix} \omega &1 \\ 1 & \omega \end{bmatrix}$

Option 2)

$\begin{bmatrix} 2\omega^{2} &2 \\ 2 & 2\omega \end{bmatrix}$

Option 3)

$\begin{bmatrix} -1 &2 \\ 2 & -1 \end{bmatrix}$

Option 4)

none of these

Posted by

Himanshu

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