61 Conclusion of the proof

Note that $A$ is a companion matrix, so by Proposition 20,

$$det(sI-A)=s^n+{\alpha }_{n-1}{s}^{n-1}+\mathrm{\dots }+{\alpha }_{1}s+{\alpha }_0.$$

Hence

$$T(s)=\frac{C\text{adj}(sI-A)B}{det(sI-A)}$$

$$=\frac{{\gamma }_0+{\gamma }_{1}s+\mathrm{\dots }+{\gamma }_{n-1}{s}^{n-1}}{s^n+{\alpha }_{n-1}{s}^{n-1}+\mathrm{\dots }+{\alpha }_0}.$$