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  • I need help with - Matrices and Determinants - JEE Main-2

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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