1
Question
Which of the following values of α satisfy the equation
∣∣
∣
∣∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣∣
∣
∣∣=−648α ?
Which of the following values of α satisfy the equation
∣∣
∣
∣∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣∣
∣
∣∣=−648α ?
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Solution
The correct option is C −9
Let Δ=∣∣
∣
∣∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣∣
∣
∣∣
C2→C2−C1,C3→C3−C1
Δ=2α2∣∣
∣
∣∣(1+α)23α+24α+2(2+α)23α+44α+4(3+α)23α+64α+6∣∣
∣
∣∣
R2→R2−R1,R3→R3−R1
Δ=8α2∣∣
∣
∣∣(1+α)23α+22α+12α+3212α+421∣∣
∣
∣∣
R3→R3−R2
Δ=8α2∣∣
∣
∣∣(1+α)23α+22α+12α+321100∣∣
∣
∣∣
Δ=−8α3
⇒−8α3=−648α
⇒α=±9,or α=0
Let Δ=∣∣ ∣ ∣∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣∣ ∣ ∣∣
C2→C2−C1,C3→C3−C1
Δ=2α2∣∣ ∣ ∣∣(1+α)23α+24α+2(2+α)23α+44α+4(3+α)23α+64α+6∣∣ ∣ ∣∣
R2→R2−R1,R3→R3−R1
Δ=8α2∣∣ ∣ ∣∣(1+α)23α+22α+12α+3212α+421∣∣ ∣ ∣∣
R3→R3−R2
Δ=8α2∣∣ ∣ ∣∣(1+α)23α+22α+12α+321100∣∣ ∣ ∣∣
Δ=−8α3
⇒−8α3=−648α
⇒α=±9,or α=0
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