Easy Determinants??

Algebra Level 4

The value of

c o s ( θ + ϕ ) s i n ( θ + ϕ ) c o s 2 ϕ s i n θ c o s θ s i n ϕ c o s θ s i n θ λ c o s ϕ \huge{\left| \begin{matrix} cos(\theta +\phi ) & -sin(\theta +\phi ) & cos2\phi \\ sin\theta & cos\theta & sin\phi \\ -cos\theta & sin\theta & \lambda cos\phi \end{matrix} \right| }

i s is

Do l i k e like and s h a r e share

dependent on θ \theta , ϕ \phi and λ \lambda independent of ϕ \phi and λ \lambda independent of λ \lambda independent of θ \theta None independent of all variables

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2 solutions

Tanishq Varshney
Mar 15, 2015

its simple

R 1 R 2 s i n ( ϕ ) + R 1 R_{1} \rightarrow R_{2}sin(\phi)+R_{1}

then

R 1 R 3 c o s ( ϕ ) + R 1 R_{1} \rightarrow R_{3}cos(\phi)+R_{1}

u will get final answer as ( λ + 1 ) c o s 2 ( ϕ ) (\lambda+1) cos^{2}(\phi)

yep or simply differentiate it with respect to theta and get that zero!

aryan goyat - 4 years, 6 months ago
Sourabh Jangid
Dec 29, 2016

*JUST SIMPLY OPEN IT WITH RESPECT TO THIRD COLUMN *

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