SATHEE: Matrices and Determinants 2 Question 5

Matrices and Determinants 2 Question 5

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5. Let $α$ and $β$ be the roots of the equation $x^{2} + x + 1 = 0$ . Then, for $y \neq 0$ in $R$ , $| \begin{array}{ccc} y + 1 & α & β α & y + β & 1 β & 1 & y + α \end{array} |$ is equal to

======= ####5. Let $α$ and $β$ be the roots of the equation $x^{2} + x + 1 = 0$ . Then, for $y \neq 0$ in $R$ , $| \begin{array}{ccc} y + 1 & α & β \\ α & y + β & 1 \\ β & 1 & y + α \end{array} |$ is equal to

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(a) $y (y^{2} - 1)$

(b) $y (y^{2} - 3)$

(d) $y^{3}$

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Answer:

Correct Answer: 5. (d)

Solution:

Given, quadratic equation is $x^{2} + x + 1 = 0$ having roots $α, β$ .

Then, $α + β = - 1$ and $α β = 1$

Now, given determinant

$Δ = | \begin{array}{ccc} y + 1 & α & β \\ α & y + β & 1 \\ β & 1 & y + α \end{array} |$

On applying $R_{1} \to R_{1} + R_{2} + R_{3}$ , we get

$\begin{aligned} Δ & = | \begin{array}{ccc} y + 1 + α + β & y + 1 + α + β & y + 1 + α + β \\ α & y + β & 1 \\ β & 1 & y + α \end{array} | \\ = | \begin{array}{ccc} y & y & y \\ α & y + β & 1 \\ β & 1 & y + α \end{array} | \end{aligned}$

On applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ , we get

$\begin{aligned} Δ & = | \begin{array}{ccc} y & 0 & 0 \\ α & y + β - α & 1 - α \\ β & 1 - β & y + α - β \end{array} | \\ = y [(y + (β - α)) (y - (β - α)) - (1 - α) (1 - β)] \end{aligned}$