1 - 1 - 1 a - b - ca - 2 - b - 2- c - 2lend vmatrix -5 - then beginvmatrixb c- 2-b- 2 c - a- 2 c-a c- 2 - a b- 2-b a- 2 c-b - a-c - b-aend vmatrix is equal to -A- 25B- 35C- 5D- 15

The correct option is A 25|co factor matrix of A| =|A|^2 = 25 ...

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Finding Inverse Using Elementary Transformations

1 & 1 & 1 a &...

Question

$If ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ = 5$ , then

$∣ ∣ ∣ ∣ ∣ \begin{matrix} b c^{2} - b^{2} c & a^{2} c - a c^{2} & a b^{2} - b a^{2} b^{2} - c^{2} & c^{2} - a^{2} & a^{2} - b^{2} c - b & a - c & b - a \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to -

5

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15

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25

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35

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△ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + a^{4} & 1 + a b + a^{2} b^{2} & 1 + a c + a^{2} c^{2} 1 + a b + a^{2} b^{2} & 1 + b^{2} + b^{4} & 1 + b c + b^{2} c^{2} 1 + a c + a^{2} c^{2} & 1 + b c + b^{2} c^{2} & 1 + c^{2} + c^{4} \end{matrix} ∣ ∣ ∣ ∣ ∣

(a, b, c)

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} + 1 & a^{2} b & a^{2} c a b^{2} & b^{3} + 1 & b^{2} c a c^{2} & b c^{2} & c^{3} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 11

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

∣ ∣ ∣ ∣ ∣ \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

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

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