1
Question
Which of the following value(s) 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 value(s) 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 D 9
∣∣
∣
∣∣(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+3α2∣∣
∣
∣∣=−648α
R2→R2−R1 and R3→R3−R2 gives
∣∣
∣
∣∣(1+α)2(1+2α)2(1+3α)2(3+2α)(3+4α)(3+6α)(5+2α)(5+4α)(5+6α)∣∣
∣
∣∣=−648α
R3→R3−R2 gives
∣∣
∣
∣∣(1+α)2(1+2α)2(1+3α)23+2α3+4α3+6α222∣∣
∣
∣∣=−648α
C2→C2−C1 and C3→C3−C2 gives
∣∣
∣
∣∣(1+α)2α(2+3α)α(2+5α)(3+2α)2α2α(2)00∣∣
∣
∣∣=−648α
⇒2(2α2)[(2+3α)−(2+5α)]=−648α
⇒−8α3=−648α
⇒α2=81 or α=0
⇒α=0 or ±9
∣∣ ∣ ∣∣(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+3α2∣∣ ∣ ∣∣=−648α
R2→R2−R1 and R3→R3−R2 gives
∣∣ ∣ ∣∣(1+α)2(1+2α)2(1+3α)2(3+2α)(3+4α)(3+6α)(5+2α)(5+4α)(5+6α)∣∣ ∣ ∣∣=−648α
R3→R3−R2 gives
∣∣ ∣ ∣∣(1+α)2(1+2α)2(1+3α)23+2α3+4α3+6α222∣∣ ∣ ∣∣=−648α
C2→C2−C1 and C3→C3−C2 gives
∣∣ ∣ ∣∣(1+α)2α(2+3α)α(2+5α)(3+2α)2α2α(2)00∣∣ ∣ ∣∣=−648α
⇒2(2α2)[(2+3α)−(2+5α)]=−648α
⇒−8α3=−648α
⇒α2=81 or α=0
⇒α=0 or ±9
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