Question

The value of the determinant is:

• A. $abc$
• B. $a+b+c$
• C. $0$
• D. $ab+bc+ca$

Answer 1

The correct option is C $0$

$\left[\begin{array}{ccc}{b}^2-ab& b-c& bc-ca\\ ab-a^2& a-b& b^2-ab\\ bc-ac& c-a& ab-a^2\end{array}\right]$

$\left[\begin{array}{ccc}b(b-a)& b-c& c(b-a)\\ a(b-a)& a-b& b(b-a)\\ c(b-a)& c-a& a(b-a)\end{array}\right]$

Taking out $(b-a)$ common from $c_1$ and $c_2$

${(b-a)}^2\left[\begin{array}{ccc}b& b-c& c\\ a& a-b& b\\ c& c-a& a\end{array}\right]$

$c_1\to c_1-c_3$

${(b-a)}^2\left[\begin{array}{ccc}b-c& b-c& c\\ a-b& a-b& b\\ c-a& c-a& a\end{array}\right]$

$c_1$ and $c_2$ are same.

$\Rightarrow$ The determinant is $0$.