Here & are state variables, µ(t) is a force vector & µ(t) being the system response.

The given state space system is

$\begin{pmatrix} x_1^{'}\\ x_2^{'}\\ \end{pmatrix}=\begin{bmatrix} -3&2\\ 1&1\\ \end{bmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}+\begin{bmatrix} 0\\ 2\\ \end{bmatrix}\mu(t)\\ T.F = \frac{Y(S)}{U(S)}\\ = C[SI - A]^{-1}B + D$

Here,

$A = \begin{bmatrix} -3&2\\ 1&1\\ \end{bmatrix} ; \\ B = \begin{bmatrix} 0\\ 2\\ \end{bmatrix} ;\\ C = \begin{bmatrix} 1&0\\ \end{bmatrix} \& \\ D = 0\\ [SI - A] = \begin{bmatrix} S&0\\ 0&S\\ \end{bmatrix} - \begin{bmatrix} -3&2\\ 1&1\\ \end{bmatrix}\\ = \begin{bmatrix} S+3&-2\\ -1&S-1\\ \end{bmatrix}$

The inverse of [SI - A] is

$[SI - A]^{-1} = \frac{adj[SI - A]}{det[SI - A]}\\ adj[SI - A] = \begin{bmatrix} S-1&2\\ 1&S+3\\ \end{bmatrix}\\ det[SI - A] = (S - 1)(S + 3) -2\\ \hspace{0.025cm} = s^2 + 3S - S - 3 - 2\\ \hspace{0.025cm} = S^2 + 2S - 5$