---
title: "Characteristic equation from state model [−5 −1; 3 −1]"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-char-eq-5-1-3"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-char-eq-5-1-3
site: BCT Engineering
---

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

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2081 Baisakh, 4 marks; 2070 Chaitra, 4 marks (asked 2 times)
- Video: [Characteristic Equation from State Equation (Lec 9)](https://www.youtube.com/watch?v=s6P11OWEobg) — Epsita Das
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-char-eq-5-1-3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

Obtain characteristic equation for the system having given state model. [ẋ1; ẋ2] = [−5 −1; 3 −1][x1; x2] + [2; 5]u and Y = [1 2][x1; x2]

## Short answer

Computing |sI - A| = 0 gives the characteristic equation s^2 + 6s + 8 = 0, with eigenvalues s = -2, s = -4.

## Answer

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 $-4$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the characteristic equation of a state model obtained?
   A. |sI - A - BC| = 0
   B. |sI - A| = 0
   C. |A| = 0
   D. C(sI - A)^-1 B = 0
   Answer: B. The characteristic equation is the determinant of sI - A set to zero; its roots are the eigenvalues of A. (Key point: Definition of characteristic equation)

2. For A = [-5 -1; 3 -1], what is |sI - A|?
   A. s^2 + 6s + 2
   B. s^2 - 6s + 8
   C. s^2 + 6s + 8
   D. s^2 + 4s + 8
   Answer: C. (s+5)(s+1) - (1)(-3) = s^2 + 6s + 5 + 3 = s^2 + 6s + 8. (Key point: Computing the determinant)

3. Which matrices affect the characteristic equation?
   A. A and B
   B. A and C
   C. A, B and C equally
   D. Only A
   Answer: D. The characteristic equation depends only on A; B and C do not change it. (Key point: Role of B and C)

4. The roots of s^2 + 6s + 8 = 0 are -2 and -4. What do they represent and imply?
   A. Eigenvalues of A (poles); the system is stable
   B. Zeros of the system; the system is stable
   C. Eigenvalues of A; the system is unstable
   D. Gains of B and C; stability cannot be decided
   Answer: A. Roots of the characteristic equation are the poles; both lie in the left half-plane, so the system is stable. (Key point: Roots and stability)

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