Question

Let $\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$ and the equation $p{x}^2+q{x}^2+rx+s=0$ has roots, $a,b,c$ where $a,b,c\in R^{+}$

The value of $\Delta$ is

• A. $\frac{r^2}{p^2}$
• B. $\frac{r^3}{p^3}$
• C. $-\frac{s}{p}$
• D. None of these

Answer 1

The correct option is B $\frac{r^3}{p^3}$

Multiplying $R_1,R_2,R_3$ by $a,b,c$ respectively, and then taking $a,b,c$ common from $C_1,C_2,$ and $C_3,$ we get
$\Delta =\left|\begin{array}{ccc}-bc& ab+ac& ac+ab\\ ab+bc& -ac& bc+ab\\ ac+bc& bc+ac& -ab\end{array}\right|$

Now, using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1,$ and then taking $(ab+bc+ca)$ common from $C_2$ and $C_3,$ we get

$\Delta =\left|\begin{array}{ccc}-bc& 1& 1\\ ab+bc& -1& 0\\ ac+bc& 0& -1\end{array}\right|\times (ab+bc+ca{)}^2$

Now, applying $R_2\to R_2+R_1,$ we get
$\Delta =\left|\begin{array}{ccc}-bc& 1& 1\\ ab& 0& 1\\ ac+bc& 0& -1\end{array}\right|(ab+bc+ca{)}^2$

Expanding along $C_2,$ we get
$\Delta =(ab+bc+ca{)}^2[ac+bc+ab]$
$=(ab+bc+ca{)}^3$
$=(r/p{)}^3=r^3/p^3$