Question

If a,b and c are the roots of the equation
$x^2(px+q)=r(x+1)$
then the value of determinant of matrix is
$\left[\begin{array}{ccc}1+a&1&1\\1&1+b&1\\1&1&1+c\end{array}\right]$
is

Answer 1

$x^{2} (px+q)-rx-1=0$
$px^3+qx^2-rx-r=0$
as for cubic equation
$abc= \frac{-r}{p}$
$a+b+c=\frac{-q}{p}$
$ab+bc+ca= \frac{r}{p}$
now the value of determinant is
$=(1+a)(1+b)(1+c)-1-a-b-c=$
$=ab+bc+ca+abc$
as
$abc= \frac{-r}{p}$
$ab+bc+ca= \frac{r}{p}$
$ab+bc+ca+abc= \frac{r}{p}- \frac{r}{p}=0$
So the Value of determinant is zero

Answer 2

the value of determinat is abc(1+1/a+1/b+1/c)
found using elementary operations
and form the cubic equation u can say that it is 0