Here $x_1, x_2 \hspace{0.05cm}\& \hspace{0.05cm} x_3$ are atate variables, $\mu(t)$ is a false vaector & μ(t) being the sytem response. Obtain transfer function of the system.

Comparing the given equation with standard equation, we get

$A = \begin{bmatrix} 0&1&0\\ 0&0&1\\ -1&-3&-2\\ \end{bmatrix}\\ B = \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix}\\ C = \begin{bmatrix} 1&0&0\\ \end{bmatrix}\\ D = 0$

We know that

$T.F = \frac{Y(S)}{U(S)} = C[SI - A]^{-1}B + D$

Now,

$[SI - A] = \begin{bmatrix} S&0&0\\ 0&S&0\\ 0&0&S\\ \end{bmatrix} -\begin{bmatrix} 0&1&0\\ 0&0&1\\ -1&-3&-2\\ \end{bmatrix}\\ [SI -A] = \begin{bmatrix} S&-1&0\\ 0&S&1\\ -1&-3&S+2\\ \end{bmatrix} = P,say\\ \textit{Now, to find} [SI - A]^{-1} \textit{by adj method, we know}\\ P^{-1} = \frac{Adj P}{P}\\ [SI -A]^{-1} = \frac{1}{S^2 + 2S - 5}\begin{bmatrix} S-1&2\\ 1&S+3\\ \end{bmatrix}\\ \frac{Y(S)}{U(S)} = C[SI - A]^{-1}B\\ \hspace{0.25cm} = \begin{bmatrix} 1&0\\ \end{bmatrix}\frac{1}{S^2 + 2S - 5}\begin{bmatrix} S-1&2\\ 1&S+3\\ \end{bmatrix} \begin{bmatrix} 0\\ 2\\ \end{bmatrix}\\ \hspace{0.25cm} = \frac{\begin{bmatrix} S-1&0\\ \end{bmatrix}\begin{bmatrix} 0\\ 2\\ \end{bmatrix}}{S^2 + 2S - 5}\\ \frac{Y(S)}{U(S)} = \frac{4}{S^2 + 2S -5}\\ \textit{The cofactor matrix is,}\\ P(1, 1) = S(S + 2) + 3 = S^2 + 2S +3\\ P(1, 2) = 1\\ P(1, 3) = +S\\ P(2, 1) = -S-2\\ P(2,2) = S^2 + 2S\\ P(2,3) = 3S -1\\ P(3,1) = 1\\ P(3,2) = -S\\ P(3,3) = S^2$

Cofactors elements = $(-1)^{i + j}A_{ij}\\ = \begin{bmatrix} S^2+2S+3&1&S\\ S+2&S^2+2S&-3S+1\\ 1&S&S^2\\ \end{bmatrix}$

This is the matrix of cofactors

$Adj(P) = [\textit{Matrix of coefficient}]^T\\ \hspace{0.25cm} = \begin{bmatrix} S^2+2S+3&S+2&1\\ 1&S^2+2S&S\\ S&-3S+1&S^2\\ \end{bmatrix}\\ \textit{Now}\\ P = S[S^2 + 2S + 3] + 1[-1] +0\\ P = S^3 + 2S^2 + 3S - 1$

Substituting in equation,

$[SI - A]^{-1} = \frac{\begin{bmatrix} S^2+2S+3&S+2&1\\ 1&S^2+2S&S\\ S&-3S+1&S^2\\ \end{bmatrix}}{S^3+2S^2+3S-1}$

Substituting above value in equation of transfer function,

$T.F = \frac{Y(S)}{U(S)}\\ = \begin{bmatrix} 1&0&0\\ \end{bmatrix}\times \frac{1}{S^3+2S^2+3S-1}\begin{bmatrix} S^2+2S+3&S+2&1\\ 1&S^2+2S&S\\ S&-3+1&S^2\\ \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix}\\ T.F = \frac{1}{S^3+2S^2+3S-1}$