Mathematics - Determinant Question with Solution | TestHub

Let $\Delta =\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ac& bc& a^2+b^2\end{array}\right|$ Multiply $C_1$ by $a,C_2$ by b and $C_3$ by $\mathrm{c}$ and hence divide by $abc$. $$=\frac{1}{abc}\left|\begin{array}{ccc}a(b^2+c^2& a{b}^2)& a{c}^2\\ a^{2}b& b(c^2+a^2& b{c}^2)\\ a^{2}c& b^{2}c& c(a^2+b^2\end{array}\right|$$ Take out $a,b,c$ common from $R_1,R_2$ and $R_3$ respectively. $$\begin{array}{rl}& \therefore \Delta =\frac{abc}{abc}\left|\begin{array}{ccc}{b}^2+c^2& b^2& c^2\\ a^2& c^2+a^2& c^2\\ a^2& b^2& a^2+b^2\end{array}\right|\\ & \text{ Apply }{C}_1\to C_1-C_2-C_3\\ & \Delta =\left|\begin{array}{ccc}0& b^2& c^2\\ -2c^2& c^2+a^2& c^2\\ -2b^2& b^2& a^2+b^2\end{array}\right|\\ & =-2\left|\begin{array}{ccc}0& b^2& c^2\\ c^2& c^2+a^2& c^2\\ b^2& b^2& a^2+b^2\end{array}\right|\\ & \end{array}$$ Apply $C_2-C_1$ and $C_3-C_1$ $$\begin{array}{rl}& =-2\left|\begin{array}{ccc}0& b^2& c^2\\ c^2& a^2& 0\\ b^2& 0& a^2\end{array}\right|\\ & =-2\left[-b^2\left(c^2{a}^2\right)+c^2\left(-a^2{b}^2\right)\right]\\ & =2a^2{b}^2{c}^2+2a^2{b}^2{c}^2=4a^2{b}^2{c}^2\\ & \text{ But }\Delta =k{a}^2{b}^2{c}^2\therefore k=4\end{array}$$