First, write down all the equations at each node. Since the flows are non-negative, the value of the flow will always be positive.

We know that the incoming flow at each node will be equal to the outgoing flow.

For node A:

\({x_1} + {x_3} = 20\)

For node B:

\({x_2} = {x_4} + {x_3}\)

For node C:

\({x_1} + {x_2} = 80\)

The total flow of the system:

\({x_4} + 20 = 80\)

Re-arrange all the above equations to get the augmented matrix.

\(\begin{array}{c}{x_1} + {x_3} = 20\\{x_2} - {x_3} - {x_4} = 0\\{x_1} + {x_2} = 80\\{x_4} = 60\end{array}\)

Write down the equations in an augmented matrix form.

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0&{20}\\0&1&{ - 1}&{ - 1}&0\\1&1&0&0&{80}\\0&0&0&1&{60}\end{array}} \right]\)

Reduce the augmented matrix into an echelon matrix.

Apply row operation \({R_3} \to {R_3} - {R_1}\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0&{20}\\0&1&{ - 1}&{ - 1}&0\\0&1&{ - 1}&0&{60}\\0&0&0&1&{60}\end{array}} \right]\)

Apply row operation \({R_3} \to {R_3} - {R_2}\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0&{20}\\0&1&{ - 1}&{ - 1}&0\\0&0&0&1&{60}\\0&0&0&1&{60}\end{array}} \right]\)

Apply row operation \({R_4} \to {R_4} - {R_3}\) .

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0&{20}\\0&1&{ - 1}&{ - 1}&0\\0&0&0&1&{60}\\0&0&0&0&0\end{array}} \right]\)

The general flow pattern of the system is given by:

\[\left\{ \begin{array}{c}{x_1} = 20 - {x_3}\\{x_2} = 60 + {x_3}\\{x_3}\,{\rm{is}}\,{\rm{free}}\\{x_4} = 60\end{array} \right.\]

Since \({x_1}\) cannot be negative as all the flows are non-negative, the value of \({x_1}\) is 0.

Hence, the largest value of \({x_3}\) is 20.