Question

The value of $\Delta$ is

• A. $r^2/p^2$
• B. $p^2/r^2$
• C. $-s/p$
• D. none of these

Answer 1

Answer: (A)

$r^2/p^2$

$\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& b(b+c)& c(c+b)\\ a(a+c)& -ac& c(c+a)\\ a(a+b)& b(b+a)& -ab\end{array}\right|$
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}-abc& ab(b+c)& ac(c+b)\\ ab(a+c)& -abc& bc(c+a)\\ ac(a+b)& bc(b+a)& -abc\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& a(b+c)& a(c+b)\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$\Delta =\left|\begin{array}{ccc}ab+bc+ac& ab+bc+ac& ab+bc+ac\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}1& 1& 1\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}0& 0& 1\\ ab+bc+ac& -(ab+bc+ac)& b(c+a)\\ 0& ab+bc+ac& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& b(c+a)\\ 0& 1& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2$ ....(i)
Since, $a$,$b$,$c$ are the roots of the equation $p{x}^3+q{x}^2+rx+s=0$
$\Rightarrow ab+bc+ca=\frac{r}{p}$
So, $\Delta =\frac{r^2}{p^2}$ (by (i))
Hence, option 'A' is correct.