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Question

Without expanding evaluate the determinant;
∣ ∣ ∣sinαcosαsin(α+δ)sinβcosβsin(β+δ)sinγcosγsin(γ+δ)∣ ∣ ∣

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Solution

∣ ∣ ∣sinαcosαsin(α+δ)sinβcosβsin(β+δ)sinγcosγsin(γ+δ)∣ ∣ ∣

=∣ ∣sinαcosαsinαcosδ+cosαsinδsinβcosβsinβcosδ+cosβ+sinδsinγcosγsinγcosδ+cosγsinδ∣ ∣

Using sin(A+B)=sinAcosB+cosAsinB

=∣ ∣sinαcosαsinαcosδsinβcosβsinβcosδsinγcosγsinγcosδ∣ ∣+∣ ∣sinαcosαcosαsinδsinβcosβcosβsinδsinγcosγcosγsinδ∣ ∣

Using ∣ ∣a+bgjc+dhke+fil∣ ∣=∣ ∣agjchkeil∣ ∣+∣ ∣bgjdhkfil∣ ∣

=cosδ∣ ∣sinαcosαsinαsinβcosβsinβsinγcosγsinγ∣ ∣+sinδ∣ ∣sinαcosαcosαsinβcosβcosβsinγcosγcosγ∣ ∣


taking cosδ from third column of 1st det
sinδ from third column of 2nd det

=cosδ.0+sinδ.0
=0
det having two same rows or columns have zero value.

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