Characteristic equation from state model [−5 −1; 3 −1]

The characteristic equation of a state model is $|sI - A| = 0$; its roots are the eigenvalues of $A$, which are the poles of the system.

$$A = \begin{bmatrix} -5 & -1 \\ 3 & -1 \end{bmatrix},\quad B = \begin{bmatrix} 2 \\ 5 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 2 \end{bmatrix}$$

Step 1: Form $sI - A$

$$sI - A = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -5 & -1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} s+5 & 1 \\ -3 & s+1 \end{bmatrix}$$

Step 2: Determinant

$$\begin{aligned} |sI - A| &= (s+5)(s+1) - (1)(-3) \\ &= s^2 + 6s + 5 + 3 \\ &= s^2 + 6s + 8 \end{aligned}$$

Characteristic equation:

$$s^2 + 6s + 8 = 0 \quad\Rightarrow\quad (s+2)(s+4) = 0$$

Roots (eigenvalues): $s = -2$ and $s = -4$. Both are in the left half of the s-plane, so the system is stable. Note that $B$ and $C$ do not affect the characteristic equation.

Check through the transfer function

$\operatorname{adj}(sI-A) = \begin{bmatrix} s+1 & -1 \\ 3 & s+5 \end{bmatrix}$, so

$$\begin{aligned} \operatorname{adj}(sI-A)B &= \begin{bmatrix} 2(s+1) - 5 \\ 6 + 5(s+5) \end{bmatrix} = \begin{bmatrix} 2s - 3 \\ 5s + 31 \end{bmatrix} \\ \frac{Y(s)}{U(s)} &= \frac{(2s-3) + 2(5s+31)}{s^2+6s+8} = \frac{12s + 59}{s^2 + 6s + 8} \end{aligned}$$

The denominator of the transfer function is the same characteristic polynomial, $s^2 + 6s + 8$.

Answer: $s^2 + 6s + 8 = 0$, roots $-2$ and $-4$.