Controllability and Observability by Rank Test for a 2D LTI System

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Sep 12, 2026

Consider the continuous-time LTI state-space model

\dot{x}=Ax+Bu,\qquad y=Cx+Du.

For the given system

A=\begin{bmatrix}0&1\\-1&-3\end{bmatrix},\quad B=\begin{bmatrix}0\\1\end{bmatrix},

we will check controllability using the rank of the controllability matrix. We will also discuss the observable vs. not observable possibilities—however, note that observability requires a measurement matrix $C$ (not provided in the prompt). So, the only fully determined part from the given data is controllability.

Key rank tests:

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We will compute the controllability matrix and its rank for the 2-state system ( $n=2$ ), then map the result to the multiple-choice statements.

Controllability and Observability (Rank Tests) - Intuition + Examples

Step 1: Form the controllability matrix

For a system of order $n=2$ , the controllability matrix is

\mathcal{C} = \begin{bmatrix} B & AB \end{bmatrix}.

Compute $AB$ :

AB = A\begin{bmatrix}0\\1\end{bmatrix} =\begin{bmatrix}0&1\\-1&-3\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix} =\begin{bmatrix}1\\-3\end{bmatrix}.

So,

\mathcal{C} =\begin{bmatrix}0&1\\ 1&-3\end{bmatrix}.

Step 2: Compute the rank

For a $2\times 2$ matrix, $\operatorname{rank}(\mathcal{C})=2$ iff $\det(\mathcal{C})\neq 0$ . Compute the determinant:

\det(\mathcal{C})= \det\begin{bmatrix}0&1\\1&-3\end{bmatrix} =0\cdot(-3)-1\cdot 1 =-1\neq 0.

Therefore,

\operatorname{rank}(\mathcal{C})=2=n.

Rank-based controllability check for $A,B$

1
Step 1
Here $A$ is $2\times 2$ , so $n=2$ .
2
Step 2
Multiply $A\,B$ : $AB=\begin{bmatrix}1\\-3\end{bmatrix}$ .
3
Step 3
$\mathcal{C}=[B\ AB]=\begin{bmatrix}0&1\\1&-3\end{bmatrix}$ .
4
Step 4
Since $\det(\mathcal{C})=-1\ eq 0$ , $\operatorname{rank}(\mathcal{C})=2$ .
5
Step 5
Because $\operatorname{rank}(\mathcal{C})=n=2$ , the system is controllable.

Pro Tip: Fast rank check for $2\times2$ controllability matrices

When $n=2$ , you can compute $\det([B\ AB])$ . If it’s nonzero, rank is $2$ immediately.

Observability cannot be concluded without $C$

The rank test for observability uses $\mathcal{O} = \begin{bmatrix} C\\ CA\end{bmatrix}$ for $n=2$ . Since $C$ (and $y=Cx$ ) is not given, options about 'observable/unobservable' cannot be uniquely determined.

Mapping the controllability result to the choices

We found:

\operatorname{rank}(\mathcal{C})=2=n.

So the correct controllability option is:

(i) Rank = 2, controllable

Why the other options are inconsistent with the given $A,B$

(ii) Rank = 1, uncontrollable contradicts $\\det([B\\ AB])=-1\\neq 0$ .
(iii) Rank = 2, not observable and (iv) Rank = 1, observable are observability statements and require $C$ . With only $A,B$ , we cannot determine observability rank at all.

How to decide controllability (rank test) for an $n$ -state system

Build matrices

Create $\mathcal{C}=[B\ AB\ \cdots\ A^{n-1}B]$ ."

Compute rank

Compute $\operatorname{rank}(\mathcal{C})$ ."

Conclude

If rank equals $n$ , system is controllable; otherwise uncontrollable."

Controllability matrix rank result

For this 2-state system, compare rank to $n=2$ .

Clarifications about controllability vs. observability rank tests

Quick self-check

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Question · Term

For $n=2$, what is the controllability matrix?

Click to reveal

Answer · Definition

It is $\mathcal{C}=[B\ AB]$ .

Knowledge Check

Question 1 of 4

Q1Single choice

For $A=\begin{bmatrix}0&1\\-1&-3\end{bmatrix}$ and $B=\begin{bmatrix}0\\1\end{bmatrix}$ , what is $AB$ ?

$\begin{bmatrix}1\\-3\end{bmatrix}$

$\begin{bmatrix}0\\1\end{bmatrix}$

$\begin{bmatrix}-1\\-3\end{bmatrix}$

$\begin{bmatrix}1\\3\end{bmatrix}$

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