Consider an economy with three sectors, Chemicals & Metals, Fuels & Power and Machinery. Chemicals sell \(30\% \) of its output to fuels and \(50\% \) to Machinery and retains the rest. Fuels sells \(80\% \) of its output to chemicals and \(10\% \) to Machinery and retains the rest. Machinery sells \(40\% \) to chemicals and \(40\% \) to Fuels and retains the rest.

a. Construct the exchange table for the economy

b. Develop a system of equations that leads to prices at which each sector’s income matches its expanses. Then write the augmented matrix that can be row reduced to find these prices.

c. \(\left[ M \right]\)Find a set of equilibrium prices when the price for the machinery output is 100 units.

The table below represents the economy:

To express a problem in the form of a linear system,assume variables for it.

Let \({p_c}\), \({p_f}\), and \({p_m}\) be the outputs of the Chemicals, Fuels, and Machinery. From the first row of the table, the total input to the chemicals is \(0.2{p_c} + 0.8{p_f} + 0.4{p_m}\). So, the equilibrium prices must satisfy:

\(\begin{array}{c}{p_c} = 0.2{p_c} + 0.8{p_f} + 0.4{p_m}\\0.8{p_c} - 0.8{p_f} - 0.4{p_m} = 0\end{array}\)

From the second row, the total input to the Fuels is \(0.3{p_c} + 0.1{p_f} + 0.4{p_m}\). So, the equilibrium prices must satisfy:

\(\begin{array}{c}{p_f} = 0.3{p_c} + 0.1{p_f} + 0.4{p_m}\\ - 0.3{p_c} + 0.9{p_f} - 0.4{p_m} = 0\end{array}\)

From the third row of the economic table, the total input to Machinery is \(0.5{p_c} + 0.1{p_f} + 0.2{p_m}\). So, the equilibrium prices must satisfy:

\(\begin{array}{c}{p_m} = 0.5{p_c} + 0.1{p_f} + 0.2{p_m}\\ - 0.5{p_c} - 0.1{p_f} + 0.8{p_m} = 0\end{array}\)

The augmented matrix for a linear system is formed using the coefficients of variables of the linear system.

For the linear system of equations \(0.8{p_c} - 0.8{p_f} - 0.4{p_m} = 0\), \( - 0.3{p_c} + 0.9{p_f} - 0.4{p_m} = 0\), and \( - 0.5{p_c} - 0.1{p_f} + 0.8{p_m} = 0\), the augmented matrix can be formed as:

\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\{ - 0.3}&{0.9}&{ - 0.4}&0\\{ - 0.5}&{ - 0.1}&{0.8}&0\end{array}} \right]\)

Row operations\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\{ - 0.3}&{0.9}&{ - 0.4}&0\\{ - 0.5}&{ - 0.1}&{0.8}&0\end{array}} \right]\)do not change the linear system.

Perform an elementary row operation on the matrix as shown below.

For row 3, add row 1 and row 2 with row 3, i. e. \({R_3} \to {R_1} + {R_2} + {R_3}\).

\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\{ - 0.3}&{0.9}&{ - 0.4}&0\\{0.8 - 0.3 - 0.5}&{ - 0.8 + 0.9 - 0.1}&{ - 0.4 - 0.4 + 0.8}&0\end{array}} \right]\)

After the row operation, the augmented matrix becomes:

\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\{ - 0.3}&{0.9}&{ - 0.4}&0\\0&0&0&0\end{array}} \right]\)

For row 2, multiply row 1 with and row 2 with and add them, i.e., \({R_2} \to 0.3{R_1} + 0.8{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\{0.24 - 0.24}&{ - 0.24 + 0.72}&{ - 0.12 - 0.32}&0\\0&0&0&0\end{array}} \right]\)

After the row operation, the augmented matrix becomes:

\(\left[ {\begin{array}{*{20}{c}}{0.8}&{ - 0.8}&{ - 0.4}&0\\0&{0.48}&{ - 0.44}&0\\0&0&0&0\end{array}} \right]\)

The diagonal elements of row 1 and row 2 are to be reduced to 1.

For row 1, divide row 1 with 0.8 and row 2 with 0.48 i.e., \({R_1} \to \frac{1}{{0.8}}{R_1}\) and \({R_2} \to \frac{1}{{0.48}}{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}{\frac{{0.8}}{{0.8}}}&{\frac{{ - 0.8}}{{0.8}}}&{\frac{{ - 0.4}}{{0.8}}}&{\frac{0}{{0.8}}}\\0&{\frac{{0.48}}{{0.48}}}&{\frac{{ - 0.44}}{{0.48}}}&{\frac{0}{{0.48}}}\\0&0&0&0\end{array}} \right]\)

After the row operation, the augmented matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 0.5}&0\\0&1&{ - 0.917}&0\\0&0&0&0\end{array}} \right]\)

For row 1, add row 1 and row 2, i.e., \({R_1} \to {R_1} + {R_2}\).

\(\left[ {\begin{array}{*{20}{c}}{1 + 0}&{ - 1 + 1}&{ - 0.5 - 0.917}&0\\0&1&{ - 0.917}&0\\0&0&0&0\end{array}} \right]\)

After row operations, the augmented matrix will become:

\(0.3\)\(0.8\)

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1.417}&0\\0&1&{ - 0.917}&0\\0&0&0&0\end{array}} \right]\)

Using the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1.417}&0\\0&1&{ - 0.917}&0\\0&0&0&0\end{array}} \right]\), the general solution can be written as:

\(\begin{array}{c}{p_c} - 1.417{p_m} = 0\\{p_f} - 0.971{p_m} = 0\end{array}\)

As Machinery’s output is 100 units, i.e., , substitute 100 for \({p_m}\) in the equation \({p_f} - 0.971{p_m} = 0\).

\(\begin{array}{c}{p_f} - 0.971\left( {100} \right) = 0\\{p_f} - 97.1 = 0\\{p_f} = 97.1\end{array}\)

As \({p_m} = 100\), so by the equation \({p_c} - 1.417{p_m} = 0\), determine the output of Chemicals.

\(\begin{array}{c}{p_c} - 1.417\left( {100} \right) = 0\\{p_c} - 141.7 = 0\\{p_c} = 141.7\end{array}\)

Thus, the outputs of Chemical and Fuels for 100 outputs of Machinery are 141.7 and 91.7, respectively.