# A State of the Art Review of Systems of Linear Inequalities and Related Observability Problems
## 3 Systems of Linear Inequalities
### Complete Systems of Linear Inequalities
#### 3.3.1 Example of Compatibility of the Particular Case of Complete System of Linear Inequalities  
Consider the complete system of inequalities:  
\[ \begin{array}{ccccc}x_{1}&+x_{2}&&-x_{4}&-2x_{5}&=&a\\ x_{2}&-x_{3}&+x_{4}&-x_{5}&=&b\\ 2x_{1}&-x_{2}&+x_{3}&&-x_{5}&=&c\\ x_{2}&+x_{3}&+x_{4}&-x_{5}&=&d\\ &&x_{1},x_{2},x_{3},x_{4},x_{5}&\geq&0.\end{array}  \tag{37}\]  
To see if the system is compatible, it is sufficient to see if the vector of the independent terms belongs to the cone generated by the column vectors of the coefficients of the system, that is, if it belongs to the cone:  
\[\mathbf{A}_{\pi}=\left( \begin{array}{ccccc}1&1&0&-1&-2\\ 0&1&-1&1&-1\\ 2&-1&1&0&-1\\ 0&1&1&1&-1\end{array} \right)_{\pi} \tag{38}\]  
whose dual cone is the cone:  
\[\mathbf{A}_{\pi}^{p}=\left( \begin{array}{c}0&2\\ 1&-2\\ 0&-1\\ -1&-1\end{array} \right)\cdot\pi \tag{39}\]  
Thus, the vector of the independent terms multiplied scalarly by the generators of the cone \(\mathbf{A}_{\pi}^{p}\) must be non-positive, that is to say:  
\[(a,b,c,d)\cdot\left( \begin{array}{c}0&2\\ 1&-2\\ 0&-1\\ -1&-1\end{array} \right)\leq 0\quad\Leftrightarrow\quad\left\{ \begin{array}{c}b-d \leq&0\\ 2a-2b-c-b\ \leq&0.\end{array} \right. \tag{40}\]  
The compatibility of the system of linear inequalities:  
\[ \begin{array}{rrrrrrrr}&&+x_{2}&-x_{3}&-x_{4}&=&-1\\ +2x_{1}&-x_{2}&+x_{3}&=&-1\\ +2x_{1}&+x_{2}&-x_{3}&-2x_{4}&=&-3\\ +x_{1}&-x_{2}&+2x_{4}&=&0\\ +x_{1}&+x_{3}&-2x_{4}&<=&-1\\ &+x_{2}&+x_{3}&+x_{4}&<=&-1\\ -x_{1}&-x_{2}&+x_{4}&<=&2\\ +x_{1}&&-x_{3}&-2x_{4}&<=&-1\end{array}  \tag{48}\]  
can be analyzed by transforming this system into an equivalent one by making \(x_{i}=x_{i}-x_{0};i=1,2,\ldots,n\) and using slack variables to convert the inequalities in equalities  
\[ \begin{array}{rrrrrr}+x_{0}&&+x_{2}&-x_{3}&-x_{4}&=&b_{1}\\ -2x_{0}&+2x_{1}&-x_{2}&+x_{3}&&=&b_{2}\\ +2x_{1}&+x_{2}&-x_{3}&-2x_{4}&=&b_{3}\\ -2x_{0}&+x_{1}&-x_{2}&+2x_{4}&=&b_{4}\\ +x_{1}&+x_{3}&-2x_{4}&+x_{5}&=&b_{5}\\ -3x_{0}&&+x_{2}&+x_{3}&+x_{4}&+x_{6}&=&b_{6}\\ +x_{0}&-x_{1}&-x_{2}&&+x_{4}&+x_{7}&=&b_{7}\\ +2x_{0}&+x_{1}&&-x_{3}&-2x_{4}&+x_{8}&=&b_{8}\\ &&x_{0},x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7},x_{8}&\geq&0\end{array}  \tag{49}\]  
and obtaining the dual cone of the cone generated by its column vectors, that is, the cone:  
\[ \begin{array}{rrrrrrrr}\left(\begin{array}{rrrrrrrrrr}1&0&1&-1&-1&0&0&0&0 \\ -2&2&-1&1&0&0&0&0\\ -0&2&1&-1&-2&0&0&0&0\\ -2&1&-1&0&2&0&0&0\\ -0&1&0&1&-2&1&0&0&0\\ -3&0&1&1&1&0&1&0&0\\ 1&-1&-1&0&1&0&1&0\\ 2&1&0&-1&-2&0&0&0&1\end{array}\right)_{\pi}\end{array}  \tag{50}\]  
which is the following cone:  
\[\begin{array}{r}\left(\begin{array}{r}-2\\ -1\\ 1 To motivate and to understand the main ideas we will use the following example:  
\[ \begin{array}{rrrrrr}&+x_{2}&-x_{3}&-x_{4}&=&b_{1}\\ +2x_{1}&-x_{2}&+x_{3}&-&=&b_{2}\\ +2x_{1}&+x_{2}&-x_{3}&-2x_{4}&=&b_{3}\\ +x_{1}&-x_{2}&+2x_{4}&=&b_{4}\\ +x_{1}&+x_{3}&-2x_{4}&<=&b_{5}\\ &+x_{2}&+x_{3}&+x_{4}&<=&b_{6}\\ -x_{1}&-x_{2}&+x_{4}&<=&b_{7}\\ +x_{1}&-x_{3}&-2x_{4}&<=&b_{8},\end{array}  \tag{53}\]  
which is a system of inequalities in the standard form, which can be converted into a system of equations using additional slack unknowns. This way of representing the system could suggest the \(x\) unknowns as the input and the \(b\) unknowns as the output, that is, given the \(x\) values, the \(b\) values or some bounds for them can be obtained using (53). Conversely, given the \(b\) values, we could think of obtaining the \(x\) values or some bounds for them using the inverse or some type of pseudoinverse for backpropagation. However, this classical partition of the variables into two groups, inputs and outputs, limits the possibilities of the relations implied by (53), and is not always convenient. So, the main idea here consists of considering the right-hand side independent terms as unknowns, and passing them to the left-hand side of the inequalities, as is shown in the following system:  
\[