Question
Question: \[\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| =\]...
a & b & c \\
b & c & a \\
c & a & b
\end{matrix} \right| =$$
A
3abc+a3+b3+c3
B
3abc−a3−b3−c3
C
abc−a3+b3+c3
D
abc+a3−b3−c3
Answer
3abc−a3−b3−c3
Explanation
Solution
abcbcacab=a+b+cbca+b+ccaa+b+cab,
(R1→R1+R2+R3)
= (a+b+c) k=1,−1.
= 3abc−a3−b3−c3, (After simplification).