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 the above

Answer 1

The correct option is B $4$

$△=\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|$
Applying $R_1\to a{R}_1,R_2\to b{R}_2,R_3\to c{R}_3,$ we get $△=\frac{1}{abc}\times \left|\begin{array}{ccc}a(b^2+c^2)& a^{2}b& a^{2}c\\ a{b}^2& b(c^2+a^2)& c{b}^2\\ a{c}^2& b{c}^2& c(a^2+b^2)\end{array}\right|$
Now, applying $C_1\to \frac{1}{a}{C}_1,C_2\to \frac{1}{b}{C}_2,C_3\to \frac{1}{c}{C}_3,$ we get $△=\frac{abc}{abc}\left|\begin{array}{ccc}{b}^2+c^2& a^2& a^2\\ b^2& c^2+a^2& b^2\\ c^2& c^2& a^2+b^2\end{array}\right|$
$=\left|\begin{array}{ccc}0& -2c^2& -2b^2\\ b^2& c^2+a^2& b^2\\ c^2& c^2& a^2+b^2\end{array}\right|[R_1\to R_1-R_2-R_3]$

Taking $2$ common from $R_1$ and applying $R_2\to R_2+R_1,R_3\to R_3+R_1$
$=2\left|\begin{array}{ccc}0& -c^2& -b^2\\ b^2& a^2& 0\\ c^2& 0& a^2\end{array}\right|$

Evaluating along $R_1,$ we get
$△=2\left[c^2\right(a^2{b}^2)-b^2(-a^2{c}^2\left)\right]$
$=4a^2{b}^2{c}^2$
Hence, $k=4.$