Step 1
We are given with following system of Equations\[\begin{array}{l}8{x_1} + 6{x_2} = 2\\\\5{x_1} + 4{x_2} = - 1\end{array}\]We have to find the solution of above system of equations. For a system of equations written in the form of $A x = b$, the solution is given by: \[x = {A^{ - 1}}b\]So, first we will convert the system of equations in matrix form and then find the inverse of the coefficient matrix to solve for $x$. To find the inverse of a matrix $A = \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right]$, we will use the following equation:\[{A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]


Step 2: The matrix form of the equations
The coefficient matrix $A$ is governed by the coefficients of the variables $x_i$.\[\begin{array}{l}Ax = b\\\\\left[ {\begin{array}{*{20}{l}}8&6\\5&4\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\end{array}\]

Step 3: The inverse of the coefficient matrix
For the matrix $A = \left[ {\begin{array}{*{20}{l}}8&6\\5&4\end{array}} \right]$,\[\begin{array}{l}{A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\\\\ = \dfrac{1}{{(8)(4) - (5)(6)}}\left[ {\begin{array}{*{20}{c}}4&{ - 6}\\{ - 5}&8\end{array}} \right]\\\\ = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}4&{ - 6}\\{ - 5}&8\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - \dfrac{5}{2}}&4\end{array}} \right]\end{array}\]

Step 4: The Solution of the equations is given by
\[\begin{array}{l}x = {A^{ - 1}}b\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - \dfrac{5}{2}}&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{l}}7\\{ - 9}\end{array}} \right]\end{array}\]

ANSWER
\[{x_1} = 7\,\,\,{\rm{and}}\,\,\,{x_2} = - 9\]