If b+c c+a a+ba+b b+c c+ac+a a+b b+c=ka b cc a bb c a, then the value of k is

We haveb+c c+a a+ba+b b+c c+ac+a a+b b+c=ka b cc a bb c aor 2(a+b+c) c+a a+b2(a+b+c) b+c c+a2(a+b+c) a+b b+c=kabccabbca [Applying C1→C1+C2+C3 on L.H.S.]or 2a+b+c −b −ca+b+c −a −ba+b+c −c −a=kabccabbca [Applying C2→C2−C1,C3→C3−C1 on L.H.S]or a −b −cc −a −bb −c −a=ka b cc a bb c a [Applying C1→C1+C2+C3 on L.H.S]or 2a b cc a bb c a=ka b cc a bb c a∴ k=2

If b+c c+a a+ba+b b+c c+ac+a a+b b+c=ka b cc a bb c a, then the value of k is

We haveb+c c+a a+ba+b b+c c+ac+a a+b b+c=ka b cc a bb c aor 2(a+b+c) c+a a+b2(a+b+c) b+c c+a2(a+b+c) a+b b+c=kabccabbca [Applying C1→C1+C2+C3 on L.H.S.]or 2a+b+c −b −ca+b+c −a −ba+b+c −c −a=kabccabbca [Applying C2→C2−C1,C3→C3−C1 on L.H.S]or a −b −cc −a −bb −c −a=ka b cc a bb c a [Applying C1→C1+C2+C3 on L.H.S]or 2a b cc a bb c a=ka b cc a bb c a∴ k=2

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If $|\begin{array}{lll} b + c c + a a + b \\ a + b b + c c + a \\ c + a a + b b + c \end{array}| = k |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}|$ , then the value of k is

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answer is B.

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Detailed Solution

We have

$|\begin{array}{lll} b + c c + a a + b \\ a + b b + c c + a \\ c + a a + b b + c \end{array}| = k |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}|$

or $|\begin{array}{lll} 2 (a + b + c) c + a a + b \\ 2 (a + b + c) b + c c + a \\ 2 (a + b + c) a + b b + c \end{array}| = k |\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}| [Applying C_{1} \to C_{1} + (C_{2} + C_{3}) on L . H . S .]$

or $2 |\begin{array}{lll} a + b + c - b - c \\ a + b + c - a - b \\ a + b + c - c - a \end{array}| = k |\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}| [Applying C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} on L . H . S]$

or $|\begin{array}{lll} a - b - c \\ c - a - b \\ b - c - a \end{array}| = k |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}| [Applying C_{1} \to C_{1} + C_{2} + C_{3} on L . H . S]$