Question

Let $\alpha ,\beta$ be the roots of the equation $x^2+x+1=0$. Then for $y\ne 0$ in R,
$$\left|\begin{array}{ccc}y+1& \alpha & \beta \\ \alpha & y+\beta & 1\\ \beta & 1& y+\alpha \end{array}\right|$$is equal to:

• A. $y(y^2-3)$
• B. $y^3-1$
• C. $y(y^2-1)$
• D. $y^3$

Answer 1

The correct option is D $y^3$

Roots of the equation $x^2+x+1=0$ will be $\alpha =w\text{and}\beta =w^2$, where $w$ and $w^2$ are the cube roots of unity.
$\Rightarrow \Delta =\left|\begin{array}{ccc}y+1& w& w^2\\ w& y+w^2& 1\\ w^2& 1& y+w\end{array}\right|$

$R_1\to R_1+R_2+R_3\Rightarrow \Delta =y\left|\begin{array}{ccc}1& 1& 1\\ w& y+w^2& 1\\ w^2& 1& y+w\end{array}\right|[\because 1+w+w^2=0]$

On expanding along $R_1$, we get
$\Delta =y\left(y^2\right)=y^3$