The value of the determinant ∣∣
∣
∣∣1eiπ/3eiπ/4e−iπ/31e2iπ/3e−iπ/4e−2iπ/31∣∣
∣
∣∣ is ..................
A
(2+√2)
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B
(2−√2)
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C
−(2+√2)
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D
−(2−√2)
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Solution
The correct option is C−(2+√2) Let Δ=∣∣
∣
∣∣1eiπ/3eiπ/4e−iπ/31ei2π/3e−iπ/4e−i2π/31∣∣
∣
∣∣ Expanding along first column we get, Δ=(1)∣∣∣1ei2π/3e−i2π/31∣∣∣−e−iπ/3∣∣∣eiπ/3eiπ/4e−i2π/31∣∣∣+e−iπ/4∣∣∣eiπ/3eiπ/41ei2π/3∣∣∣ ⇒Δ=1(1−1)−e−iπ/3(eiπ/3−e−5iπ/12)+e−iπ/4(eπ−eiπ/4) ⇒Δ=−2+e3iπ/4+e−3iπ/4=−2+2cos(3π/4)=−2−√2=−(2+√2)