Q

# 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)

$y^{3}-1$

Views

$\alpha ,\beta$ 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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