A linear time invariant (LTI) system is described by:
\[
\begin{bmatrix}\dot{x}_1\\ \dot{x}_2\end{bmatrix}=\begin{bmatrix}0 & 1\\ -10 & -7\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\begin{bmatrix}1\\ 1\end{bmatrix}u
\]
\[
y=\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}
\]
The poles of the system are ______.
Show Hint
The poles are the eigenvalues of the state matrix \(A\): solve \(\det(sI-A)=0\).
Step 1: Notice this is a companion-form matrix.
The matrix $A=\begin{bmatrix}0 & 1\\ -10 & -7\end{bmatrix}$ has the classic companion form $\begin{bmatrix}0 & 1\\ -a_0 & -a_1\end{bmatrix}$, whose characteristic polynomial can be read straight off as $s^2+a_1 s+a_0$. Here $a_1=7$ and $a_0=10$, so the polynomial is $s^2+7s+10$ without even expanding a determinant.
Step 2: Use the sum and product of the roots.
For $s^2+7s+10=0$, if the two poles are $p_1$ and $p_2$, then
\[ p_1+p_2=-7, \qquad p_1 p_2=10 \]
We need two numbers that multiply to $10$ and add to $-7$. Trying $-2$ and $-5$: their product is $10$ and their sum is $-7$, both match.
Step 3: Verify by direct substitution.
\[ (-2)^2+7(-2)+10=4-14+10=0 \]
\[ (-5)^2+7(-5)+10=25-35+10=0 \]
Both check out, so $s=-2$ and $s=-5$ are indeed the roots.
Step 4: Confirm these survive in the transfer function.
Computing $G(s)=C(sI-A)^{-1}B$ with $B=\begin{bmatrix}1\\1\end{bmatrix}$ and $C=\begin{bmatrix}1 & 1\end{bmatrix}$ gives $G(s)=\dfrac{2(s-1)}{(s+2)(s+5)}$, so neither factor cancels with the numerator's zero at $s=1$.
\[ \boxed{-2,\ -5} \]
