Determinants and Matrices — Maths STD 11 Science — Question

Taking bc, ca, ab common from $R_1, R_2, R_3$

respectively, we get

L.H.S. $=(b c)(c a)(a b)\left|\begin{array}{ccc}\frac{b+c}{b c} & 1 & b c \\ \frac{c+a}{c a} & 1 & c a \\ \frac{a+b}{a b} & 1 & a b\end{array}\right|$

Taking abc common from $\mathrm{C}_3$, we get

L.H.S. $=\left(\mathrm{a}^2 \mathrm{~b}^2 \mathrm{c}^2\right)(a b c)\left|\begin{array}{lll}\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{~b}} & 1 & \frac{1}{a} \\ \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}} & 1 & \frac{1}{b} \\ \frac{1}{b}+\frac{1}{\mathrm{a}} & 1 & \frac{1}{c}\end{array}\right|$

Applying $C_1 \rightarrow C_1+C_3$, we get

L.H.S. $=a^3 b^3 c^3\left|\begin{array}{lll}\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{a} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{b} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{c}\end{array}\right|$

Taking $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ common from $C_1$, we get

$=a^3 b^3 c^3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(0)$

$\ldots\left[\because C_1\right.$ and $C_2$ are identical $]$

$=0=$ R.H.S.