GO Classes DPP | GATE CS, DA | Linear Algebra | Determinant

Answer 1

$\alpha ,\beta ,\gamma$ are roots of equation.

$\alpha +\beta +\gamma=0$

$\alpha \beta + \beta \gamma+ \gamma\alpha =-p$

$\alpha \cdot\beta \cdot\gamma=q$

Now,

$\begin{vmatrix} \alpha &\beta &\gamma \\ \beta &\gamma &\alpha \\ \gamma &\alpha &\beta \end{vmatrix}$

Now, $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$

$\begin{vmatrix} \alpha+\beta+\gamma &\beta &\gamma \\ \alpha+\beta+\gamma &\gamma &\alpha \\ \alpha+\beta+\gamma &\alpha &\beta \end{vmatrix}$

$(\alpha+\beta+\gamma)$ $\begin {vmatrix} 1 &\beta &\gamma \\ 1 &\gamma &\alpha \\ 1 &\alpha &\beta \end{vmatrix}$

Since, $(\alpha+\beta+\gamma)$ $=0$ here

So, $0 \times $$\begin {vmatrix} 1 &\beta &\gamma \\ 1 &\gamma &\alpha \\ 1 &\alpha &\beta \end{vmatrix}$ $= 0$

Answer 2

Answer: C

Answer 3

A cubic equation with roots \( \alpha, \beta, \gamma \) can be written as:

\[
\begin{aligned}
 

& a(x - \alpha)(x - \beta)(x - \gamma) = 0 \\

\Longrightarrow& a[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - (\alpha\beta\gamma)] = 0 \\ \\

&\text{compare with standard cubic equation}: ax^3 + bx^2 + cx + d = 0 \\

& \alpha + \beta + \alpha = -\frac{b}{a} = 0 &\text{ (from the question)}\\

& \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -p \\

& \alpha\beta\gamma = d = q

\end{aligned}
\]

Rest of the problem is solved exactly like in the other answers.

\[
\begin{aligned}

&R_1 \rightarrow R_1 + R_2 + R_3 \\

&\begin{vmatrix}
\alpha & \beta & \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{vmatrix}

&\longrightarrow

&\begin{vmatrix}
\alpha + \beta + \gamma & \alpha + \beta + \gamma & \alpha + \beta + \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{vmatrix}

\\ \\

\longrightarrow&(\alpha + \beta + \gamma)\begin{vmatrix}
1 & 1 & 1 \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{vmatrix}

&={}

&0 \times \begin{vmatrix}
1 & 1 & 1 \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{vmatrix} \\ \\

={}& 0

\end{aligned}
\]

 

This question made me realise how dusty my algebra skills are.

Key search query/terms to google

> given alpha, beta, gamma are the roots of a cubic equation how is alpha + beta + gamma = 0
> Vieta's Formulas (look for cubic equation)