1
Question
If ∣∣
∣
∣∣(β+γ−α−δ)4(β+γ−α−δ)21(γ+α−β−δ)4(γ+α−β−δ)21(α+β−γ−δ)4(α+β−γ−δ)21∣∣
∣
∣∣=−K(α−β)(α−γ)(β−γ)(β−δ)(γ−δ)
Then the value of K is:
If ∣∣
∣
∣∣(β+γ−α−δ)4(β+γ−α−δ)21(γ+α−β−δ)4(γ+α−β−δ)21(α+β−γ−δ)4(α+β−γ−δ)21∣∣
∣
∣∣=−K(α−β)(α−γ)(β−γ)(β−δ)(γ−δ)
Then the value of K is:
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Solution
The correct option is D 64
Let a=β+γ−δ−α,b=γ+α−β−δ,c=α+β−γ−δ
Thus determinant becomes ∣∣
∣
∣∣a4a21b2b21c4c21∣∣
∣
∣∣
R1→R1−R2,R2→R2−R3
=∣∣
∣
∣∣a4−b4a2−b20b4−c4b2−c20c4c21∣∣
∣
∣∣
=(a2−b2)(b2−c2)∣∣
∣
∣∣a2+b210b2+c210c4c21∣∣
∣
∣∣
=(a2−b2)(b2−c2)(a2−c2)
=(a−b)(a+b)(b−c)(b+c)(a+c)(a−c)
=−26(α−β)(α−γ)(β−γ)(α−δ)(β−δ)(γ−δ)
=−64(α−β)(α−γ)(β−γ)(α−δ)(β−δ)(γ−δ)
∴K=64
Let a=β+γ−δ−α,b=γ+α−β−δ,c=α+β−γ−δ
Thus determinant becomes ∣∣ ∣ ∣∣a4a21b2b21c4c21∣∣ ∣ ∣∣
R1→R1−R2,R2→R2−R3
=∣∣ ∣ ∣∣a4−b4a2−b20b4−c4b2−c20c4c21∣∣ ∣ ∣∣
=(a2−b2)(b2−c2)∣∣ ∣ ∣∣a2+b210b2+c210c4c21∣∣ ∣ ∣∣
=(a2−b2)(b2−c2)(a2−c2)
=(a−b)(a+b)(b−c)(b+c)(a+c)(a−c)
=−26(α−β)(α−γ)(β−γ)(α−δ)(β−δ)(γ−δ)
=−64(α−β)(α−γ)(β−γ)(α−δ)(β−δ)(γ−δ)
∴K=64
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