A-b amp- b-c amp- c-a- x-y amp- y-z amp- z-x- p-q amp- q-r amp- r-pa-x-y-z-b-p-q-r-c0abc-xyz-pqrNone of these

The correct option is B 0a b amp; b c amp; c a x y amp; y z amp; z x p q amp; q r amp; r p = 0 amp; b c amp; c a 0 amp; y z am ...

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Sarrus Rule

a-b amp; b...

Question

$∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a x - y & y - z & z - x p - q & q - r & r - p \end{matrix} ∣ ∣ ∣ ∣$

a(x+y+z)+b(p+q+r)+c

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0.0

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abc+xyz+pqr

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None of these

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Solution

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Question 5(a)
Add :p ( p - q ), q (q - r ) and r (r - p ).

|\begin{matrix} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{matrix}| = |\begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{matrix}| = |\begin{matrix} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{matrix}| = x y z (x - y) (y - z) (z - x) (x + y + z) .

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a x - y & y - z & z - x p - q & q - r & r - p \end{matrix} ∣ ∣ ∣ ∣

x + y + z = 0

\frac{(x + y) (y + z) (z + x)}{x y z} + 11

x + y + z = 0

\frac{x y z}{(x + y) (y + z) (z + x)} = - 1

x \neq - y, y \neq - z, z \neq - x

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