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

Marks : 8 M

Year : Dec 2015

The characteristic equation of A is :

$ \left| \begin{array}{ccc} 3/2-\lambda & 1/2 \\ 1/2 & 3/2-\lambda \end{array}\right| \;=\; 0 \\ \; \\ (\dfrac{3}{2}-λ)^2 – \dfrac{1}{4}=0 \\ \; \\ ∴ 9/4-3λ+λ^2-1/4=0 \\ \; \\ ∴λ^2-3λ+2=0 \\ \; \\ ∴ (λ-1)(λ-2)=0 \\ \; \\ ∴ λ=1,2 $

For λ=1 ,[A-λI]X=0----given

$ \left[ \begin{array}{ccc} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \end{array}\right] $

By $R_2+ R_1$ :

$ \left[ \begin{array}{ccc} -1/2 & 1/2 \\ 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ ∴ \dfrac{X_1}{2}+ \dfrac{X_2}{2}=0 \\ ∴ X_1= X_2 $

If $X_2$=1 ,$X_1$=1 , Hence eigen vector,

M=$ \left[ \begin{array}{ccc} 1 & 1 \\ -1 & 1 \end{array}\right] $

∴ M=2

Now D=$ \left[ \begin{array}{ccc} 1 & 0 \\ 0 & 2 \end{array}\right] $

If f(A)= $e^A$ , f(D)= $ e^D= \left[ \begin{array}{ccc} e^1 & 0 \\ 0 & e^2 \end{array}\right] $

If f(A)= $4^A$ , f(D)= $ 4^D= \left[ \begin{array}{ccc} 4^1 & 0 \\ 0 & 4^2 \end{array}\right] $

$ ∴ e^A=M f(D) M^{-1}= \left[ \begin{array}{ccc} 1 & 1 \\ -1 & 1 \end{array}\right] \left[ \begin{array}{ccc} e^1 & 0 \\ 0 & e^2 \end{array}\right] \dfrac{1}{2} \left[ \begin{array}{ccc} 1 & 1 \\ -1 & 1 \end{array}\right] \\ \; \\ = \left[ \begin{array}{ccc} e^1 & e^2 \\ -e & e^2 \end{array}\right] \left[ \begin{array}{ccc} 1 & 1 \\ -1 & 1 \end{array}\right] \\ \; \\ ∴ e^A= 1/2 \left[ \begin{array}{ccc} e+e^2 & -e+e^2 \\ -e+e^2 & e+e^2 \end{array}\right] \\ \; \\ $

∴ Replacing e by 4 , we get ,

$ 4^A= \dfrac{1}{2} \left[ \begin{array}{ccc} 20 & 12 \\ 12 & 20 \end{array}\right] \;=\; \dfrac{1}{2} \left[ \begin{array}{ccc} 10 & 6 \\ 6 & 10 \end{array}\right] $