The characteristic equation of matrix A is

$$|A - \lambda I| = 0 \\ \lambda^2 - \bigg(\frac{3\pi}{2} \bigg) \lambda - \pi^2 = 0$$

Characteristic roots are

$$\therefore \lambda = 2\pi, -\frac{\pi}{2}$$

Considering $\phi(A) = sinA$ as a rational function of A, and since A is a matrix of order 2, it can be expressed as

$$sinA = \alpha_{1}A + \alpha_{0}I..........................(1)$$

$\therefore sin\lambda$ is the eigen value of sinA

$\therefore$ we have $sin\lambda = \alpha_{1}\lambda + \alpha_{0}........................(2)$

Put $\lambda = 2\pi$ and $-\frac{\pi}{2}$ in equation (2)

$$sin2\pi = 2\pi \alpha_{1} + \alpha_{0} \\ i.e. 2\pi \alpha_{1} + \alpha_{0} = 0....................(3) \\ sin\frac{-\pi}{2} = \frac{-\pi}{2} \alpha_{1} + \alpha_{0} \\ \frac{-\pi}{2} \alpha_{1} + \alpha_{0} = -1 ........................(4)$$

Solving equation (3) and (4) we get,

$$\alpha_{1} = \frac{2}{5\pi}, \alpha_{0} = - \frac{4}{5}$$

Substituting the above values in equation (1) we get

sinA $$= \frac{2}{5\pi}A - \frac{4}{5}I \\ = \frac{2}{5\pi} \begin{bmatrix} \frac{\pi}{2} & \frac{3\pi}{2} \\ \pi & \pi \end{bmatrix} - \frac{4}{5} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \frac{2}{5} \begin{bmatrix} \frac{1}{2} & \frac{3}{2} \\ 1 & 1 \end{bmatrix} - \frac{4}{5} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} \frac{-3}{5} & \frac{3}{5} \\ \frac{2}{5} & \frac{-2}{5} \end{bmatrix}$$

Hence sinA = $\begin{bmatrix} \frac{-3}{5} & \frac{3}{5} \\ \frac{2}{5} & \frac{-2}{5} \end{bmatrix}$