Matrices And Determinants Ques 69
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0$. Then, for $y \neq 0$ in $\mathbf{R}$, $\left|\begin{array}{ccc}y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha\end{array}\right|$ is equal to
(a) $y\left(y^{2}-1\right)$
(b) $y\left(y^{2}-3\right)$
(c) $y^{3}-1$
(d) $y^{3}$
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Answer:
Correct Answer: 69.(d)
Solution:
Formula:
- Given, quadratic equation is $x^{2}+x+1=0$ having roots $\alpha, \beta$.
Then, $\alpha+\beta=-1$ and $\alpha \beta=1$
Now, given determinant
$ \Delta=\left|\begin{array}{ccc} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{array}\right| $
On applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we get
$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} y+1+\alpha+\beta & y+1+\alpha+\beta & y+1+\alpha+\beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{array}\right| \\ & =\left|\begin{array}{ccc} y & y & y \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{array}\right| \end{aligned} $
On applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$, we get
$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} y & 0 & 0 \\ \alpha & y+\beta-\alpha & 1-\alpha \\ \beta & 1-\beta & y+\alpha-\beta \end{array}\right| \\ & =y[(y+(\beta-\alpha))(y-(\beta-\alpha))-(1-\alpha)(1-\beta)] \end{aligned} $
[expanding along $R_{1}$ ]
$=y\left[y^{2}-(\beta-\alpha)^{2}-(1-\alpha-\beta+\alpha \beta)\right]$
$=y\left[y^{2}-\beta^{2}-\alpha^{2}+2 \alpha \beta-1+(\alpha+\beta)-\alpha \beta\right]$
$=y\left[y^{2}-(\alpha+\beta)^{2}+2 \alpha \beta+2 \alpha \beta-1+(\alpha+\beta)-\alpha \beta\right]$
$=y\left[y^{2}-1+3-1-1\right]=y^{3} \quad[\because \alpha+\beta=-1$ and $\alpha \beta=1]$
BETA
