Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

$$ \begin{vmatrix} \pi/2 - \lambda &\pi \\ 0& 3\pi/2 - \lambda \end{vmatrix} = 0 $$

$ (\frac{\pi}{2} - \lambda)(\frac{3 \pi}{2} - \lambda) = 0 \\ \lambda = \frac{\pi}{2} \hspace{1cm} \lambda = \frac{3 \pi}{2}$

$ \phi(A) = sin\,A \\ \phi(A) = aA + bI \\ \phi(\lambda) = sin \, \lambda \\ \phi(\lambda) = a \lambda + b \\ \phi(\frac{\pi}{2}) = sin(\frac{\pi}{2}) = 1 \\ \phi(\frac{3 \pi}{2}) = sin(\frac{3 \pi}{2}) = -1 $

case(i): $\lambda = \frac{pi}{2}$

$ \phi(\frac{\pi}{2} = a(\frac{ \pi}{2}) + b \\ 1 = \frac{\pi}{2}a + b \\ \frac{\pi}{2}a + b = 1 $

case(ii): $\lambda = \frac{3 \pi}{2}$

$ \phi(\frac{3 \pi}{2}) = a(\frac{3 \pi}{2}) + b \\ \frac{3 \pi}{2}a + b = 1 $

Solving cases (i) and (ii), a = $ \frac{-2}{\pi} $, b = 2

$ \phi(A) = aA + bI \\ = \frac{-2}{\pi} \begin{bmatrix} \frac{\pi}{2} & \pi \\ 0 & \frac{3 \pi}{2} \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} -1 & -2 \\ 0 & -3 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \\ = \begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix} \\ \therefore sin \,A = \begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix} $