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  • I have a doubt, kindly clarify. - Matrices and Determinants - JEE Main-3

            Let  \omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}.    Then the value of the determinant \begin{vmatrix} 1 &1 & 1\\ 1 & -1-\omega ^2&\omega ^2 \\ 1 & \omega ^2 & \omega ^4 \end{vmatrix} is

 

 

      

  • Option 1)

    3\omega

  • Option 2)

    3\omega (\omega-1)

  • Option 3)

    3\omega ^2

  • Option 4)

    3\omega (1-\omega)

 

Answers (1)

best_answer

As we have learned

Property of determinant -

If to each element of a line ( row or column ) of a determinant be added the equimultiples of the corresponding elements of one or more parallel lines , the determinant remains unaltered

- wherein

 

 

Given determinant = \begin{vmatrix} 1 &1 &1 \\ 1 &-1 -\omega ^2& \omega ^2\\ 1 & \omega ^2 & \omega \end{vmatrix}

                        Applying,          R1\rightarrowR1 + R2 + R3, we get \begin{vmatrix} 3 &0 &0 \\ 1 &-1 -\omega ^2& \omega ^2\\ 1 & \omega ^2 & \omega \end{vmatrix}

                                    = 3(-\omega - \omega3- \omega4) = -3( 2\omega + 1) = 3\omega( \omega- 1 )

 


Option 1)

3\omega

Option 2)

3\omega (\omega-1)

Option 3)

3\omega ^2

Option 4)

3\omega (1-\omega)

Posted by

gaurav

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