Question

If $\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|=k{a}^2{b}^2{c}^2$, then the value of $k$ is

• A. 2
• B. 4
• C. 0
• D. none of these

Answer 1

Answer: (B)

4

Given, $\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|=k{a}^2{b}^2{c}^2$
$\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|$
Multiplying $R_1,R_2,R_3$ by $a,b,c$ respectively
$=\frac{1}{abc}\left|\begin{array}{ccc}{a(b}^2+c^2)& a^{2}b& a^{2}c\\ a{b}^2& {b(c}^2+a^2)& b^{2}c\\ c^{2}a& c^{2}b& {c(a}^2+b^2)\end{array}\right|$
Taking $a,b,c$ common from $C_1,C_2,C_3$ respectively
$=\left|\begin{array}{ccc}{(b}^2+c^2)& a^2& a^2\\ b^2& {(c}^2+a^2)& b^2\\ c^2& c^2& {(a}^2+b^2)\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$=\left|\begin{array}{ccc}2{(b}^2+c^2)& {2(c}^2+a^2)& 2{(a}^2+b^2)\\ b^2& {(c}^2+a^2)& b^2\\ c^2& c^2& {(a}^2+b^2)\end{array}\right|$
$=2\left|\begin{array}{ccc}{(b}^2+c^2)& {(c}^2+a^2)& {(a}^2+b^2)\\ b^2& {(c}^2+a^2)& b^2\\ c^2& c^2& {(a}^2+b^2)\end{array}\right|$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$=2\left|\begin{array}{ccc}{(b}^2+c^2)& {(c}^2+a^2)& {(a}^2+b^2)\\ {-c}^2& 0& {-a}^2\\ {-b}^2& {-a}^2& 0\end{array}\right|$
$=2[{a^{2}b}^2{c}^2+{a^{2}b}^2{c}^2]=4{a^{2}b}^2{c}^2$
Hence, $k=4$