Question

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

• A. ${(ab+bc+ca)}^2$
• B. ${(ab+bc+ca)}^3$
• C. $(ab+bc+ca)$
• D. ${(a+b+c)}^3$

Answer 1

The correct option is B ${(ab+bc+ca)}^3$

$\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$
Multiplying $R_1,R_2,R_3$ by a,b,c respectively
$=\frac{1}{abc}\left|\begin{array}{ccc}-abc& a{b}^2+abc& a{c}^2+abc\\ b{a}^2+abc& -abc& b{c}^2+abc\\ c{a}^2+abc& c{b}^2+abc& -abc\end{array}\right|$
$=\frac{1}{abc}\left|\begin{array}{ccc}-abc& b(ab+ac)& c(ac+ab\\ a(ab+bc)& -abc& c(bc+ab)\\ a(ca+bc)& b(cb+ac)& -abc\end{array}\right|$
Taking a,b,c common from$C_1,C_2,C_3$ respectively $=\left|\begin{array}{ccc}-bc& (ab+ac)& (ac+ab)\\ (ab+bc)& -ac& (bc+ab)\\ (ca+bc)& (cb+ac)& -ab\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$=\left|\begin{array}{ccc}ab+bc+ca& ab+bc+ac)& ab+bc+ca)\\ (ab+bc)& -ac& (bc+ab)\\ (ca+bc)& (cb+ac)& -ab\end{array}\right|=(ab+bc+ac)\left|\begin{array}{ccc}1& 1& 1\\ (ab+bc)& -ac& (bc+ab)\\ (ca+bc)& (cb+ac)& -ab\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=(ab+bc+ac)\left|\begin{array}{ccc}0& 0& 1\\ ab+bc+ac& -(ab+bc+ac)& (bc+ab)\\ 0& (ab+bc+ac)& -ab\end{array}\right|$
$={(ab+bc+ac)}^3$