## 3 Application of the Drazin Inverse  
Let us assume that the matrix pencil of the pair \((E,A)\) of the system (1) is regular. As a consequence, we have \(\text{det}[Ec-A]\
eq 0\) for some \(c\in\mathbb{C}\). Premultiplication of (1) by \([Ec-A]^{-1}\) yields  
\[\bar{E}\Delta^{\alpha}x_{i+1}=\bar{A}x_{i}+\bar{B}u_{i}, \tag{4}\]  
where  
\[\bar{E}=[Ec-A]^{-1}E,\quad\bar{A}=[Ec-A]^{-1}A,\quad\bar{B}=[Ec-A]^{-1}B. \tag{5}\]It is well known that premultiplication of a matrix equation by the nonsingular matrix does not change its solution. Therefore, both Equations (1) and (4) have the same solution \(x_{i}\). The substitution of (2) into (4) yields  
\[\bar{E}x_{i+1}=\bar{A}_{\alpha}x_{i}+\sum_{j=1}^{i}\bar{E}c_{j}x_{i-j}+\bar{B}u_ {i},\quad i\in\mathbb{Z}_{+}, \tag{6}\]  
where  
\[\bar{A}_{\alpha}=\bar{A}+\bar{E}\alpha. \tag{7}\]  
and  
\[c_{j}=(-1)^{j}\binom{\alpha}{j+1}. \tag{8}\]  
Observe that the values of the coefficients \(c_{j}\) determined by (8) highly decrease for increasing \(j\). Therefore, in many cases the upper bound of the summation can be limited by some natural number \(L\), which is called the length of practical implementation [24]. Hence, we can write Equation (6) in the form  
\[\bar{E}x_{i+1}=\bar{A}_{\alpha}x_{i}+\sum_{j=1}^{L}\bar{E}c_{j}x_{i-j}+\bar{B}u _{i} \tag{9}\]  
with \(x_{-k}=0\), \(k=1,2,\ldots\). **Definition 2** ([3,7]). : _The Drazin inverse of \(\bar{E}\in\mathbb{R}^{n\times n}\), denoted by \(\bar{E}^{D}\in\mathbb{R}^{n\times n}\), is a matrix satisfying the following conditions_  
\[\bar{E}\bar{E}^{D}=\bar{E}^{D}\bar{E},\quad\bar{E}^{D}\bar{E}\bar{E}^{D}=\bar {E}^{D},\quad\bar{E}^{D}\bar{E}^{q+1}=\bar{E}^{q}, \tag{10}\]  
_where \(q\) is the index of \(\bar{E}\), i.e., the smallest nonnegative integer such that_  
\[\mathrm{rank}\bar{E}^{q}=\mathrm{rank}\bar{E}^{q+1}. \tag{11}\]  
Every square matrix has its own unique Drazin inverse [3,5,7]. For a nonsingular matrix the Drazin inverse is equivalent to the standard matrix inverse. Some methods for the computation of the Drazin inverse can be found in [3]. **Lemma 1** ([3,7]). : _The properties of the matrices \(\bar{E}\) and \(\bar{A}\) given by (5) are as follows:_  
\[\bar{A}\bar{E}=\bar{E}\bar{A},\quad\bar{E}^{D}\bar{A}=\bar{A}\bar{E}^{D},\quad \bar{A}^{D}\bar{E}=\bar{E}\bar{A}^{D},\quad\bar{A}^{D}\bar{E}^{D}=\bar{E}^{D} \bar{A}^{D}, \tag{12}\]  
\[\mathrm{ker}\bar{A}\cap\mathrm{ker}\bar{E}=\{0\}, \tag{13}\]  
\[\bar{E}=[Ec-A]^{-1}E,\quad\bar{A}=[Ec-A]^{-1}A,\quad\bar{B}=[Ec-A]^{-1}B. \tag{5}\]0  
\[\bar{E}=T\left[ \begin{array}{cc}J&0\\ 0&N\end{array} \right]T^{-1},\quad\bar{E}^{D}=T\left[ \begin{array}{cc}J^{-1}& 0\\ 0&0\end{array} \right]T^{-1},\quad\bar{A}=T\left[ \begin{array}{cc}A_{1}&0\\ 0&A_{2}\end{array} \right]T^{-1}, \tag{15}\]  
_where \(T\in\mathbb{R}^{n\times n}\) and \(J\in\mathbb{R}^{n_{1}\times n_{1}}\) are nonsingular matrices, \(N\in\mathbb{R}^{n_{2}\times n_{2}}\) is a nilpotent matrix, i.e., for some \(\mu\) we have \(N^{\mu-1}\
eq 0\), \(N^{\mu}=0\) and \(A_{1}\in\mathbb{R}^{n_{1}\times n_{1}}\), \(A_{2}\in\mathbb{R}^{n_{2}\times n_{2}}\), \(n_{1}+n_{2}=n\)._  
Let \(\mathbb{U}\subset\mathbb{R}^{m}\) be the set of admissible inputs \(u_{i}\in\mathbb{U}\) and \(\mathbb{X}_{0}\subset\mathbb{R}^{n}\) be the set of consistent initial conditions \(x_{0}\in\mathbb{X}_{0}\) for which Equation (1) has a solution \(x_{i}\) for \(u_{i}\in\mathbb{U}\). **Theorem 1** ([3]).