A) abc
B) 2abc
C) 4abc
D) 0
Correct Answer: D
Solution :
\[\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \\ \end{matrix} \right|\] \[[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}]\] \[=0\]
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