Modes observable and modes controllable

[Dustinsfl]

How do you find observable and controllable modes?
\[
\mathcal{L}\Big\{\big(\mathbf{A} - s\mathbf{I}\big)^{-1}\Big\}
=
\begin{bmatrix}
-e^{-t} - te^{-t} + \frac{1}{2}t^2e^{-t} & te^{-t}
& -te^{-t} + \frac{1}{2}t^2e^{-t}\\
-te^{-t} & -e^{-t} & te^{-t}\\
te^{-t} - \frac{1}{2}t^2e^{-t} & te^{-t}
& -e^{-t} + te^{-t} - \frac{1}{2}t^2e^{-t}
\end{bmatrix}\\
\]
and then
\[
X(t) = -e^{-t}\mathbb{I} - te^{-t}
\begin{bmatrix}
1 & -1 & -1\\
1 & 0 & -1\\
1 & -1 & -1
\end{bmatrix} - \frac{t^2}{2}e^{-t}
\begin{bmatrix}
-1 & 0 & 1\\
0 & 0 & 0\\
-1 & 0 & 1
\end{bmatrix} =
e^{-t}\mathbb{I} + te^{-t}
\begin{bmatrix}
1 & -1 & -1\\
1 & 0 & -1\\
1 & -1 & -1
\end{bmatrix} + \frac{t^2}{2}e^{-t}
\begin{bmatrix}
-1 & 0 & 1\\
0 & 0 & 0\\
-1 & 0 & 1
\end{bmatrix}.
\]
I read that the loss of observability or controllability then the residue is zero then the pole doesn't show up in the transfer function.

How can this be used or can it?

Also,
\begin{align}
\mathbf{A} &=
\begin{bmatrix}
0 & -1 & -1\\
1 & -1 & -1\\
1 & -1 & -2
\end{bmatrix}\\
\mathbf{B} &=
\begin{bmatrix}
1 & 0\\
0 & 1\\
0 & 0
\end{bmatrix}\\
\mathbf{U} &=
\begin{bmatrix}
u_1\\
u_2
\end{bmatrix}\\
\mathbf{T} &= (\mathbf{A} - s\mathbb{I})^{-1}\mathbf{B}\\
&=
-\frac{1}{(s+1)^3}
\begin{bmatrix}
s^2 + 3s + 1 & -(s + 1)\\
s+1 & (s+1)^2\\
s & -(s+1)
\end{bmatrix}\\
\mathbf{X}(s) & = \mathbf{T}\mathbf{U}
\end{align}

 

[Aaron Bex]

Using this, you can find the controllable and observable modes by computing the rank of $\mathbf{T}$ and $\mathbf{T}^T$. If the rank is n (the number of states), then all of the states are both controllable and observable. Otherwise, the states that have nonzero entries in the rows or columns with rank less than n are not controllable or observable.