Question

If $\alpha ,\beta ,\gamma$ are roots of the equation $x^2(px+q)=r(x+1)$. Then the value of determinant $\left|\begin{array}{ccc}1+\alpha & 1& 1\\ 1& 1+\beta & 1\\ 1& 1& 1+\gamma \end{array}\right|$ is

• A. $\frac{r}{p}$
• B. $-\left(\frac{q+r}{p}\right)$
• C. 0.0
• D. None of these

Answer 1

The correct option is C 0.0

We have,
$x^2(px+q)=r(x+1)p{x}^3+q{x}^2-rx-r=0$
$\alpha +\beta +\gamma =\frac{-q}{p},\alpha \beta +\beta \gamma +\gamma \alpha =\frac{-r}{p},\alpha \beta \gamma =\frac{r}{p}$
$D=\alpha \beta \gamma \left|\begin{array}{ccc}\frac{1}{\alpha }+1& \frac{1}{\alpha }& \frac{1}{\alpha }\\ \frac{1}{\beta }& \frac{1}{\beta }+1& \frac{1}{\beta }\\ \frac{1}{\gamma }& \frac{1}{\gamma }& \frac{1}{\gamma }+1\end{array}\right|$
$R_1\to R_1+R_2+R_{3}D=\left(\alpha \beta \gamma \right)(1+\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })$ $\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{\beta }& \frac{1}{\beta }+1& \frac{1}{\beta }\\ \frac{1}{\gamma }& \frac{1}{\gamma }& \frac{1}{\gamma }+1\end{array}\right|$
$C_1\to C_1-C_2$
$C_2\to C_2-C_3$
$D=\left(\alpha \beta \gamma \right)(1+\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })\left|\begin{array}{ccc}0& 0& 1\\ -1& 1& \frac{1}{\beta }\\ 0& -1& \frac{1}{\gamma }+1\end{array}\right|$
$D=\left(\alpha \beta \gamma \right)\left(\frac{\alpha \beta \gamma +\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)$
$D=(\frac{r}{p}-\frac{r}{p})=0$