If beginvmatrix a b cll b c a c a bll end vmatrix - k a - b - c a - 2- b - 2- c - 2- bc - ca - ab - then k -A- 1B- 2C- -1D- -2

The correct option is C 1a amp;b amp;c b amp;c amp;a c amp;a amp;b = a(bc a^2) b(b^2 ca)+c(ab c^2) = a^3 b^3 c^3 + 3abc = ...

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Adjoint of a Matrix

If beginvmatr...

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If $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = k (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b)$ , then k=

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∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} + c^{2} & a b & a c a b & c^{2} + a^{2} & b c c a & c b & a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = k a^{2} b^{2} c^{2},

k

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Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

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