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  • Try this! - Complex numbers and quadratic equations - JEE Main-15

Let \alpha and \beta be the roots of the equation x^{2}+x+1=0. Then for y\neq 0 in R,\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1\\ \beta & 1& y+\alpha \end{vmatrix} is equal to :

  • Option 1)

    y\left ( y^{2}-1 \right )

  • Option 2)

    y\left ( y^{2}-3 \right )

  • Option 3)

    y^{3}

  • Option 4)

    y^{3}-1

Answers (1)

best_answer

\alpha ,\beta are roots of x^{2}+x+1=0

  \alpha +\beta =-1,\alpha \beta =1

\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1\\ \beta & 1& y+\alpha \end{vmatrix}

C_1\rightarrow C_1+C_{2}+C_{3}

\begin{vmatrix} y+1+\alpha +\beta & \alpha & \beta \\ \alpha +y+\beta +1& y+\beta & 1\\ \beta +1+y+\alpha & 1& y+\alpha \end{vmatrix}

\left ( y+1+\alpha +\beta \right )\begin{vmatrix} 1& \alpha & \beta \\ 1& y+\beta & 1\\ 1& 1& y+\alpha \end{vmatrix}

=\left ( y+1+\alpha +\beta \right )\left [ \left ( y+\beta \right )\left ( y+\alpha \right )-\alpha \left ( y+\alpha -1 \right )+\beta \left ( 1-y-\beta \right ) \right ]

\Rightarrow \left ( y+1+\alpha +\beta \right )\left [ y^{2}+y\left ( \alpha +\beta \right )+\alpha \beta -2y-2^{2} +2+\beta -y\beta -\beta ^{2}\right ]

\Rightarrow \left ( y+1-1 \right )\left [ y^{2}+y\left ( -1 \right )+1-y\left ( -1 \right )-\left ( \alpha ^{2} +\beta ^{2}\right ) -1\right ]

\Rightarrow y^{3}

 


Option 1)

y\left ( y^{2}-1 \right )

Option 2)

y\left ( y^{2}-3 \right )

Option 3)

y^{3}

Option 4)

y^{3}-1

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