Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

For characteristic equation |A - $\lambda$ I| = 0

$$ \begin{vmatrix} 0 -\lambda&1 \\ -1&0-\lambda \end{vmatrix} = 0 \\ \begin{vmatrix} -\lambda&1 \\ -1&-\lambda \end{vmatrix} = 0 $$

$ \lambda^2 + 1 = 0 \implies \lambda = i,-i $

Let

$ \phi(A) = e^{At} \\ \phi(A) = aA + bI \\ \phi(\lambda) = e^{\lambda t} \\ \phi(\lambda) = a \lambda + b \\ \phi(i) = e^{it} \\ \phi(-i) = e^{-it} $

case(i) $\lambda$ = i

$ \phi(i) = a(i) + b \\ e^{it} = ai + b \\ ai + b = e^{it} \hspace{0.25cm} ...(1) $

case(ii) $\lambda$ = -i

$ \phi(-i) = a(-i) + b \\ e^{-it} = -ai + b \\ -ai + b = e^{-it} \hspace{0.25cm} ...(2) $

Solving (1) and (2), we get

b = cos t and a = sin t

$ \phi(A) = aA + bI \\ e^{At} = sin \,t \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} + cos \, t \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \\ = \begin{bmatrix} cos \,t & sin \, t \\ - sin \, t & cos \,t \end{bmatrix} $