Question

Let
${\Delta }_1=\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ac& bc& a^2+b^2\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|$
then

• A. ${\Delta }_1+{\Delta }_2=0$
• B. ${\Delta }_1={\Delta }_2$
• C. ${\Delta }_1^2={\Delta }_2$
• D. ${\Delta }_1={\Delta }_2^2$

Answer 1

Answer: (B)

${\Delta }_1={\Delta }_2$

${\Delta }_1=\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ac& bc& a^2+b^2\end{array}\right|$

Multiplying $C_1,C_2$ and $C_3$ by $a,b$ and $c$ respectively , we get

${\Delta }_1=\frac{1}{abc}\left|\begin{array}{ccc}a(b^2+c^2)& a{b}^2& a{c}^2\\ a^{2}b& b(c^2+a^2)& b{c}^2\\ a^{2}c& b^{2}c& c(a^2+b^2)\end{array}\right|$

Taking $a,b$ and $c$ common from $R_1,R_2$ and $R_3$, respectively, we get

${\Delta }_1=\frac{abc}{abc}\left|\begin{array}{ccc}(b^2+c^2)& b^2& c^2\\ a^2& (c^2+a^2)& c^2\\ a^2& b^2& (a^2+b^2)\end{array}\right|$

Applying $C_1\to C_1-C_2-C_3$

${\Delta }_1=\left|\begin{array}{ccc}0& b^2& c^2\\ -2c^2& (c^2+a^2)& c^2\\ -2b^2& b^2& (a^2+b^2)\end{array}\right|$

Again applying $C_2\to C_2-C_1,C_3\to C_3-C_1$, we get

${\Delta }_1=-2\left|\begin{array}{ccc}0& b^2& c^2\\ c^2& a^2& 0\\ b^2& 0& a^2\end{array}\right|=4a^2{b}^2{c}^2$
And for ${\Delta }_2=\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|$

Taking $a,b$ and $c$ common from $C_1,C_2$ and $C_3$

${\Delta }_2=abc\left|\begin{array}{ccc}-a& a& a\\ b& -b& b\\ c& c& -c\end{array}\right|$

Taking $a,b$ and $c$ common from $R_1,R_2$ and $R_3$, we get

${\Delta }_2=a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|=4a^2{b}^2{c}^2$

Hence ${\Delta }_1={\Delta }_2$