Question

$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right|=$

• A. $0$
• B. $1$
• C. $abc$
• D. $(a-b),(b-c),(c-a)$

Answer 1

Answer: (A)

$0$

$=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right|$

$=\left|\begin{array}{ccc}a& 1& 0\\ a-b& b& (c-b)\\ (a-b)(a+b)+c(a-b)& b^2-ca& (c-b)(c+b)+a(c-b)\end{array}\right|$ .......... $C_1\to C_1-C_2$ and $C_3\to C_3-C_2$

$=\left|\begin{array}{ccc}0& 1& 0\\ (a-b)& b& (c-b)\\ (a-b)(a+b+c)& b^2-ac& (c-b)(a+b+c)\end{array}\right|$

$=(a-b)(c-b)\left|\begin{array}{ccc}0& 1& 0\\ 1& b& 1\\ a+b+c& b^2-ac& a+b+c\end{array}\right|$

$=(a-b)(c-b)\left|\begin{array}{ccc}0& 1& 0\\ 0& b& 1\\ 0& b^2-ac& a+b+c\end{array}\right|$

$=0$