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If A = $ \left[ \begin{array}{ccc} -1 & 4 \\ 2 & 1 \end{array}\right] $ then prove that 3tanA=Atan3.

Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks : 6 M

Year : May 2014

mumbai university maths4mech • 3.0k  views
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Let λ be the eigen value of matrix A

∴ $ \left| \begin{array}{ccc} -1-\lambda & 4 \\ 2 & 1-\lambda \end{array}\right| $ = 0

∴ (-1-λ) (1- λ)-8=0

∴ - $(1^2- λ^2)$-8=0

∴ $λ^2$-9=0

∴ λ= ±3

∴ Eigen values are 3 ,-3

Since matrix is of order 2X2 let $tan⁡A=a A+bI$-----(1)

Where a and b are constants.

We assume equation (1) is satisfied by eigen values λ

tan⁡λ=aλ+b-------------(2)

When λ=3

tan⁡3=a3+b-------------(3)

When λ=-3

-tan⁡(3)=-3a+b---------(4)

∴ Adding (3) and (4) we get ,

0 = 0 +b

∴ b = 0

From (3) tan (3) = 3a

∴ a = tan A / 3

Substituting a and b in (1)

tan A = $\dfrac{tan 3}{3}$ . A

∴ 3 tan A = A tan 3

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Please Can you explain how did you write tanA=aA+bI? when A is a 22 matrix and how can we extend it to nn?

how can u write tanA=aA+bI? when matrix is 2*2??

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