Question

If $a,b,c$ are the roots of $x^3+px+q=0$, then the determinant $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$ is equal to

• A. $p+q$
• B. $p-q$
• C. $pq$
• D. $2p+q$

Answer 1

The correct option is B $p-q$

The given equation $x^3+px+q=0$ can be written as $x^3+0.x^2+px+q=0$

Hence we can see that the coefficient of $x^3$ is $1$

the coefficient of the term $x^2$ is $0$,

coefficient of term $x$ is $p$ and coefficient of term $x$ is $q$.

If $a,b,c$ are the roots of the equation,

Then for a polynomial, we know that the sum of the roots of a eqaution,

,$\to a+b+c=-\frac{coefficientof{x}^2}{coefficientof{x}^3}$

$\to a+b+c=0$. $($coefficient of $x^2$ is zero.$)$
Also sum of the roots taken two as at time,

$\to ab+bc+ca=+\frac{coefficientofx}{coefficientof{x}^3}$
$\to ab+bc+ca=p$
The product of the roots of the polynomial,

$\to abc=-\frac{coefficientofconstantterm}{coefficientof{x}^3}$
$\to abc=q$
From above information we know that,

$\Rightarrow a+b+c=0$ ...$\left(1\right)$

$\Rightarrow ab+bc+ca=p$ ...$\left(2\right)$

$\Rightarrow abc=-q$...$\left(3\right)$

Now the value of the determinant $\left|\begin{array}{ccc}(1+a)& 1& 1\\ 1& (1+b)& 1\\ 1& 1& (1+c)\end{array}\right|$

$=(1+a)\left(\right(1+b\left)\right(1+c)-1)-1\left(\right(1+c\left)\right(1)-1))+1\left(\right(1\left)\right(1)-(1+b\left)\right)$
$\Rightarrow (1+a)(bc+b+c)-1-c+1+1-1-b$

$\Rightarrow bc+b+c+abc+ab+ac-1-c+1+1-1-b$

$\Rightarrow ab+bc+ac+abc$

Taking values from eq$\left(2\right)$ and eq.$\left(3\right)$, we get the Value of the determinant,

$\left|\begin{array}{ccc}(1+a)& 1& 1\\ 1& (1+b)& 1\\ 1& 1& (1+c)\end{array}\right|=p-q$

Hence correct option is $B$.