$A= $ $ \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 1 \\ \end{array} } \right] $

$e^{At}$=$ \left[ {\begin{array}{cc} cost & -sint \\ sint & cost \\ \end{array} } \right] $

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Medium

Characteristic equation of $\mid A-\lambda I \mid=0$

Therefore, $\lambda^2-(0)\lambda+\mid A \mid=0$

$\lambda^2+(0-(-1))=0$

$\lambda^2+1=0$

$\lambda=\pm i$

Let, $\phi(A)=e^At$

Consider, $\phi(A)=\alpha_1 A+\alpha_0 I$

$e^{At}=\alpha_1 A+\alpha_0 I $ (1)

$e^{\lambda t}=\alpha_1 \lambda+\alpha_0 $ (2)

Put $\lambda=i_1-1$ in (2), we get

$e^{it}=\alpha_1 i+\alpha_0$ (3)

$e^{-it}=-\alpha_1 i+\alpha_0$ (4)

adding (3) and (4) we get,

$e^{it}+e^{-it}=2\alpha_0$

$\alpha_0=\frac{e^{it}-e^{-it}}{2i}=sint$

Therefore, from (1) we get,

$e^{AT}=sintA+costI$

$e^{At}=sint\left[ {\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} } \right] $+$cost\left[ {\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array} } \right] $=$\left[ {\begin{array}{cc} 0 & -sint\\ sint & 0 \end{array} } \right] $+$\left[ {\begin{array}{cc} cost & 0\\ 0 & cost \end{array} } \right] $

=$\left[ {\begin{array}{cc} cost & -sint\\ sint & cost \end{array} } \right] $