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Question

The value of x which satisfies ∣ ∣x22x33x4x42x93x16x82x273x64∣ ∣=0 is:

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Solution

Δ=∣ ∣x22x33x4x42x93x16x82x273x64∣ ∣=0
Applying R2R2R1 & R3R3R1, then we get

Δ=∣ ∣x22x33x4261262460∣ ∣=0

Δ=∣ ∣x22x33x41361410∣ ∣=0

Expanding along 1st row, we have:
(x2).6(2x3).4+(3x4).1=0
x=4

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