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Find $A^{50}$ if A= $ \left[ \begin{array}{ccc} 2 & 1 \\ 1 & 2 \end{array}\right] $

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

Marks : 6 M

Year : May 2015

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Characteristic equation is | A- λ I | =0

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

∴ $(2-λ)^2$ – 1 = 0

∴ $4 - 4 λ + λ^2 – 1 = 0$

∴ $λ^2-4 λ+3=0$

∴ λ=1 ,3

Eigen values λ=1,3

Since , A is 2 X 2 matrix

Let $ A^50= a_1 A+ a_2 I$-----(1)

We assume eigen value λ satisfies equation (1)

$λ^{50}= a_1 A+ a_2 I$-------(2)

∴ Let λ=1

∴ $1^{50}= a_1 (1)+ a_2 I$

∴$ 1 = a_1+ a_2$ -------(3)

Now let λ=3 in equation (2)

∴$ 3^{50}= a_1 (3)+ a_2$----(4)

Subtracting (3) and (4) we get

$ 3^{50}-1 =3a_1+ a_2- a_1-a_2$

∴$ 3^{50}-1=2a_1$

∴$ a_1= \dfrac{3^{50}-1}2$

Substituting value of $a_1= \dfrac{3^{50}-1}2$ in equation (3)we get ,

$1 = \dfrac{3^{50}-1}2 + a_2 $

∴ $\dfrac{3-3^{50}}2 = a_2$

∴ $a_2= \dfrac{3-3^{50}}2 $

Hence from equation (1)

$A^50$= $\dfrac{3^{50}-1}2 A+ \dfrac{3-3^{50}}2 I \\ \; \\ = \dfrac{1}{2} \{ 3^{50}-1 \left[ \begin{array}{ccc} 2 & 1 \\ 1 & 2 \end{array}\right] + 3 - 3^{50} \left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array}\right] \} \\ \; \\ = \dfrac{1}{2} \{ \left[ \begin{array}{ccc} 2* 3^{50}-2& 3^{50}-1 \\ 3^{50}-1 & 2* 3^{50}-2 \end{array}\right] + \left[ \begin{array}{ccc} 3- 3^{50} & 0 \\ 0 & 3- 3^{50} \end{array}\right] \} \\ \; \\ = \dfrac{1}{2} \left[ \begin{array}{ccc} 2* 3^{50}-2+3- 3^{50} & 3^{50}-1+0 \\ 3^{50}-1+0 & 2* 3^{50}-2+3-3^{50} \end{array}\right] $

$ A \;=\; \dfrac{1}{2} \left[ \begin{array}{ccc} 3^{50}+1& 3^{50}-1 \\ 3^{50}-1 & 3^{50}+1 \end{array}\right] $

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