Let's solve each part step-by-step. --- ### **a) Test controllability** Given: \[ A = \begin{bmatrix} 0 & 1 \\ -5 & -6 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] **Controllability Test:** A system \((A,B)\) is controllable if the controllability matrix \[ \mathcal{C} = [B \quad AB] \] has full rank (for an \(n\)-state system, rank \(n\)). Let's compute: **Step 1: Compute \(AB\):** \[ AB = A \cdot B = \begin{bmatrix} 0 & 1 \\ -5 & -6 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -6 \end{bmatrix} \] **Step 2: Form the controllability matrix \(\mathcal{C}\):** \[ \mathcal{C} = [B \quad AB] = \begin{bmatrix} 0 & 1 \\ 1 & -6 \end{bmatrix} \] **Step 3: Compute the rank of \(\mathcal{C}\):** The determinant for a \(2 \times 2\) matrix is: \[ \det(\mathcal{C}) = (0)(-6) - (1)(1) = -1 \neq 0 \] So, \(\mathcal{C}\) is full rank (rank 2). **Conclusion:** **The system is controllable.** --- ### **b) Is the system stable?** Stability is determined by the eigenvalues of \(A\): \[ A = \begin{bmatrix} 0 & 1 \\ -5 & -6 \end{bmatrix} \] Find the characteristic equation: \[ \det(A - \lambda I) = 0 \] \[ \left| \begin{array}{cc} 0 - \lambda & 1 \\ -5 & -6 - \lambda \end{array} \right| = 0 \] \[ (0-\lambda)(-6-\lambda) - (1)(-5) = 0 \] \[ (-\lambda)(-6-\lambda) + 5 = 0 \] \[ (\lambda)(6+\lambda) + 5 = 0 \] \[ \lambda^2 + 6\lambda + 5 = 0 \] Solve for \(\lambda\): \[ \lambda^2 + 6\lambda + 5 = 0 \] \[ \lambda = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2} \] So, \[ \lambda_1 = \frac{-6+4}{2} = -1 \] \[ \lambda_2 = \frac{-6-4}{2} = -5 \] Both eigenvalues are negative. **Conclusion:** **The system is stable.** --- ### **c) Is a stable uncontrollable system reliable in practical design?** **Explanation:** - **Stable uncontrollable system:** Such a system has at least one mode (state) that cannot be controlled (i.e., reached or changed by the input), but all modes are stable (i.e., eigenvalues have negative real parts). - **Reliability in design:** Even if the system is stable, uncontrollable modes mean some states cannot be influenced by the control input. If an external disturbance or modeling error excites these modes, you have no way to correct or control them. Therefore, in practical design, you **want all modes to be controllable** (and observable) so you can fully regulate the system's behavior. **Conclusion:** **A stable uncontrollable system is NOT considered reliable in practical design, because you cannot control all states, which may cause issues if uncontrollable states are disturbed.** --- ## **Final Answers** **a)** The system is **controllable**. **b)** The system is **stable**. **c)** **No**, a stable uncontrollable system is **not reliable** in practical design. All modes should be controllable to ensure full control over the system.