Transfer function from third-order state model

For $\dot{x} = Ax + Bu$, $y = Cx + Du$:

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3 \end{bmatrix},\quad B = \begin{bmatrix} 10 \\ 0 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix},\quad D = 0$$

Step 1: $sI - A$ and its determinant

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 0 & s & -1 \\ 1 & 2 & s+3 \end{bmatrix}$$ $$\begin{aligned} |sI - A| &= s\left[s(s+3) + 2\right] - (-1)\left[0\cdot(s+3) - (-1)(1)\right] + 0 \\ &= s^3 + 3s^2 + 2s + 1 \end{aligned}$$

Step 2: Required part of the adjoint

Because $B$ has only its first element non-zero and $C$ picks only the first row, only the (1,1) element of the adjoint is needed:

$$\operatorname{adj}(sI-A)_{11} = \begin{vmatrix} s & -1 \\ 2 & s+3 \end{vmatrix} = s^2 + 3s + 2$$

Full first column of the adjoint: $\begin{bmatrix} s^2+3s+2 \\ -1 \\ -s \end{bmatrix}$, so

$$\operatorname{adj}(sI-A)B = 10\begin{bmatrix} s^2+3s+2 \\ -1 \\ -s \end{bmatrix}$$

Step 3: Transfer function

$$C\,\operatorname{adj}(sI-A)B = 10(s^2 + 3s + 2)$$ $$\frac{Y(s)}{U(s)} = \frac{10(s^2+3s+2)}{s^3+3s^2+2s+1} = \frac{10(s+1)(s+2)}{s^3+3s^2+2s+1}$$

Stability check

Routh array for $s^3 + 3s^2 + 2s + 1$:

No sign change, so the system is stable. The poles are $-2.325$ and $-0.338 \pm j0.562$; the zeros are $-1$ and $-2$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{10(s^2+3s+2)}{s^3+3s^2+2s+1}$.