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. $\alpha \beta \gamma$
• B. $1+\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$
• C. $0$
• D. $noneofthese$

Answer 1

Answer: (C)

$0$

$x^2(px+q)=r(x+1)$

$p{x}^3+q{x}^2-rx-r=0$

If $\alpha$, $\beta$ and $\gamma$ are the roots of above polynomial , then

$\alpha +\beta +\gamma =\frac{-q}{p}$

$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{-r}{p}$

$\alpha \beta \gamma =\frac{r}{p}$

Now solving the determinant,

$(1+\alpha )\left[\right(1+\beta \left)\right(1+\gamma )-1]+1[1-1-\gamma ]+1[1-1-\beta ]$

=$(1+\alpha )[1+\gamma +\beta +\beta \gamma -1]-\gamma -\beta$

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

=$\frac{-r}{p}$ + $\frac{r}{p}$ = $0$