Question

If none of $a,b,c$ is zero, Whether the given equation $\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|={(bc+ca+ab)}^3$ is ?

• A. True
• B. False

Answer 1

The correct option is A True

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

$\therefore D=\frac{1}{abc}\left|\begin{array}{ccc}-abc& a{b}^2+abc& a{c}^2+abc\\ a^{2}b+abc& -abc& b{c}^2+abc\\ c{a}^2+ab& c{b}^2+abc& -abc\end{array}\right|[R_1^{{}^{\prime }}=a{R}_1,R_2^{{}^{\prime }}=b{R}_2,R_3^{{}^{\prime }}=c{R}_3=\left|\begin{array}{ccc}-bc& ab+ac& ac+ab\\ ab+bc& -ac& bc+ab\\ ca+bc& bc+ac& -ab\end{array}\right|[\text{taking}a,b,c\text{common from}{c}_1,c_2\text{and}{c}_3\text{respectively}]=\left|\begin{array}{ccc}ab+bc+ca& ab+bc+ca& ab+bc+ca\\ ab+bc& -ca& bc+ab\\ ca+bc& bc+ca& -ab\end{array}\right|[R_1^{{}^{\prime }}=R_1+R_2+R_3]=(ab+bc+ca\left)\left|\begin{array}{ccc}1& 1& 1\\ ab+bc& -ca& bc+ab\\ ca+bc& bc+ca& -ab\end{array}\right|\right[\text{taking common from 1st row}]=(ab+bc+ca\left)\left|\begin{array}{ccc}0& 0& 0\\ ab+bc+ca& -(ab+bc+ca)& bc+ab\\ 0& ab+bc+ca& -ab\end{array}\right|\right[c_1^{{}^{\prime }}=c_1-c_2\text{and}{c}_2^{{}^{\prime }}=c_2-c_3]$

Now expanding w.r.t. $R_1$ we get,

$=(ab+bc+ca)\left[\right(ab+bc+ca{)}^2-0]=(ab+bc+ca{)}^3$