Solveeit Logo

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+c33abc + a^{3} + b^{3} + c^{3}

B

3abca3b3c33abc - a^{3} - b^{3} - c^{3}

C

abca3+b3+c3abc - a^{3} + b^{3} + c^{3}

D

abc+a3b3c3abc + a^{3} - b^{3} - c^{3}

Answer

3abca3b3c33abc - a^{3} - b^{3} - c^{3}

Explanation

Solution

abcbcacab=a+b+ca+b+ca+b+cbcacab\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| = \left| \begin{matrix} a + b + c & a + b + c & a + b + c \\ b & c & a \\ c & a & b \end{matrix} \right|,

(R1R1+R2+R3)(R_{1} \rightarrow R_{1} + R_{2} + R_{3})

= (a+b+c)(a + b + c) k=1,1.k = 1, - 1.

= 3abca3b3c33abc - a^{3} - b^{3} - c^{3}, (After simplification).