The correct answer is: A. Is stable.
The characteristic equation of a linear discrete-time system is a polynomial equation in the variable $z$. The roots of the characteristic equation determine the stability of the system. If all of the roots of the characteristic equation lie inside the unit circle, then the system is stable. If any of the roots of the characteristic equation lie outside the unit circle, then the system is unstable.
In this case, the characteristic equation is $z^3 – 0.81 z = 0$. This equation has one real root and two complex conjugate roots. The real root is $z = 0.81$, which lies inside the unit circle. The complex conjugate roots are $z = \pm 0.54955 i$, which also lie inside the unit circle. Therefore, all of the roots of the characteristic equation lie inside the unit circle, and the system is stable.
Here is a brief explanation of each option:
- Option A: The system is stable. This is the correct answer, as explained above.
- Option B: The system is marginally stable. A marginally stable system is a system that is on the verge of being unstable. In other words, if the system is perturbed slightly, it will become unstable. However, if the system is not perturbed, it will remain stable.
- Option C: The system is unstable. An unstable system is a system that will diverge if it is perturbed. In other words, if the system is perturbed, it will continue to move away from its equilibrium point.
- Option D: Stability cannot be assessed from the given information. This option is incorrect, as the given information is sufficient to determine that the system is stable.