Question

Exercises 1-4 refer to an economy that is divided into three sectors - manufacturing, agriculture, and services. For each unit of output, manufacturing requires .10 unit from other companies in that sector, .30 unit from services. For each unit of output, agriculture uses .20 unit of its own output, .60 unit from manufacturing, and .10 unit from services. For each unit of output, the services sector consumes .10 unit from services, .60 unit from manufacturing, but no agricultural products.

2. Determine the production levels needed to satisfy a final demand of 18 units for agriculture, with no final demand for the other sectors. (Do not compute an inverse matrix.)

Short answer

\(x = \left( {33.33,35,15} \right)\)

Step-by-step solution

Solve the equation \(x = Cx + {\mathop{\rm d}\nolimits} \) for d

Theorem 11 states that C is the consumption matrix for an economy. Let d be the final demand. If C and d have non-negative entries and each column sum of C is less than 1, then \({\left( {I - C} \right)^{ - 1}}\) exists, and the production vector \(x = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} \) has non-negative entries and is the unique solution of \(x = Cx + {\mathop{\rm d}\nolimits} \).

The consumption matrix is \(C = \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\).

The production level needed to satisfy a final demand of 18 units for agriculture and no demand for the other sectors.

For d, solve the equation \(x = Cx + {\mathop{\rm d}\nolimits} \).

\[\begin{array}{c}{\mathop{\rm d}\nolimits} = x - Cx\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10{x_1}}&{.60{x_2}}&{.60{x_3}}\\{.30{x_1}}&{.20{x_2}}&0\\{.30{x_1}}&{.10{x_2}}&{.10{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{.9{x_1}}&{ - .60{x_2}}&{ - .60{x_3}}\\{ - .30{x_1}}&{.8{x_2}}&0\\{ - .30{x_1}}&{.1{x_2}}&{.9{x_3}}\end{array}} \right]\end{array}\]

Write the matrix in the system of equations, as shown below:

\(\begin{array}{c}.9{x_1} - .6{x_2} - .6{x_3} = 0\\ - .3{x_1} - .8{x_2} = 18\\ - .3{x_1} - .1{x_2} + .9{x_3} = 0\end{array}\)

Convert the equation into an augmented matrix

The augmented matrix of the system of the equation is

\(\left[ {\begin{array}{*{20}{c}}{.90}&{ - .60}&{ - .60}&0\\{ - .30}&{.80}&{.00}&{18}\\{ - .30}&{ - .10}&{.90}&0\end{array}} \right]\).

Apply the row operation

At row one, multiply row one by \( - \frac{1}{{0.90}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&0\\{ - .30}&{.80}&{.00}&{18}\\{ - .30}&{ - .10}&{.90}&0\end{array}} \right]\)

At row two, multiply row one by 0.3 and add it to row two. At row three, multiply row one by 0.30 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .6666}&{ - .6666}&0\\0&{0.6}&{ - 0.20}&{18}\\0&{ - 0.3}&{0.7}&0\end{array}} \right]\)

At row two, multiply row two by \(\frac{1}{{0.6}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&0\\0&1&{ - 0.333}&{30}\\0&{ - 0.3}&{0.7}&0\end{array}} \right]\)

At row one, multiply row two by 0.666 and add it to row one. At row three, multiply row two by 0.3 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 0.888}&{20}\\0&1&{ - 0.33}&{30}\\0&0&{0.6}&9\end{array}} \right]\)

At row two, multiply row three by 0.333 and add it to row two.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{33.33}\\0&1&0&{35}\\0&0&1&{15}\end{array}} \right]\)

Thus, \(x = \left( {33.33,35,15} \right)\).