Let \alpha and \beta be the roots of the equation x2 + x + 1 = 0. Then for y e 0 in R, | \beginmatrix y + 1 & \alpha & \beta \alpha & y + \beta & 1 \beta & 1 & y + \alpha \endmatrix | is equal to

Matrices and Determinants PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q37 with Solution

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Correct answer: C

We are given that $\alpha$ and $\beta$ are roots of $x^2+x+1=0.$ Hence, $\alpha+\beta=-1, \qquad \alpha\beta=1.$ Also, since $\alpha,\beta$ are cube roots of unity other than $1$ , $\alpha^2+\alpha+1=0, \quad \beta^2+\beta+1=0.$
Consider the determinant

y+1 & \alpha & \beta\\ \alpha & y+\beta & 1\\ \beta & 1 & y+\alpha \end{vmatrix}.$$ We expand along the first row: $$D=(y+1)\begin{vmatrix}y+\beta & 1\\ 1 & y+\alpha\end{vmatrix}-\alpha\begin{vmatrix}\alpha & 1\\ \beta & y+\alpha\end{vmatrix}+\beta\begin{vmatrix}\alpha & y+\beta\\ \beta & 1\end{vmatrix}.$$ 3. Compute each minor: First minor: $$\begin{vmatrix}y+\beta & 1\\ 1 & y+\alpha\end{vmatrix}=(y+\beta)(y+\alpha)-1 = y^2+y(\alpha+\beta)+\alpha\beta-1.$$ Using $\alpha+\beta=-1$ and $\alpha\beta=1$, $$=y^2-y+1-1=y^2-y.$$ So first term is $$(y+1)(y^2-y)=y(y^2-1).$$ Second minor: $$\begin{vmatrix}\alpha & 1\\ \beta & y+\alpha\end{vmatrix}=\alpha(y+\alpha)-\beta.$$ Thus second term is $$-\alpha[\alpha(y+\alpha)-\beta] =-\alpha^2 y-\alpha^3+\alpha\beta.$$ Since $\alpha\beta=1$ and $\alpha^3=1$, $$=-\alpha^2 y-1+1=-\alpha^2 y.$$ Third minor: $$\begin{vmatrix}\alpha & y+\beta\\ \beta & 1\end{vmatrix}=\alpha-\beta(y+\beta).

Thus third term is

=\alpha\beta-\beta^2 y-\beta^3.

Using $\alpha\beta=1$ and $\beta^3=1$ , $=1-\beta^2 y-1=-\beta^2 y.$

Factor out $y$ : $D=y\big[(y^2-1)- (\alpha^2+\beta^2)\big].$

Now, $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=1-2=-1.$ So, $D=y\big[(y^2-1)-(-1)\big]=y(y^2)=y^3.$

So the correct option is C.