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Question

If ∣ ∣ ∣xnxx+2xx+4ynyn+2yn+4znzn+2zn+4∣ ∣ ∣=(1y21x2)(1z21y2)(1x21z2) then n=


A

-4

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B

4

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C

2

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D

-2

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Solution

The correct option is A

-4


Δ=∣ ∣ ∣xnxx+2xx+4ynyn+2yn+4znzn+2zn+4∣ ∣ ∣=xnynzn∣ ∣ ∣1x2x41y2y41z2z4∣ ∣ ∣
Δ=xnynzn∣ ∣ ∣1x2x40y2z2y4z40z2x2z4x4∣ ∣ ∣(R2R2R3,R3R3R1)
Δ=xnynzn[(y2z2)(z2+x2)(z2x2) (z2x2)(y2+z2)(y2z2)]Δ=xnynzn(y2z2)(z2x2)(x2+z2y2z2)Δ=xnynzn(x2y2)(y2z2)(z2x2)Given that Δ=(1y21x2)(1z21y2)(1x21z2)
Δ=(x2y2x2y2)(y2z2y2z2)(z2x2x2z2)Δ=x4y4z4(x2y2)(y2z2)(z2x2)

Hence n =-4


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