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Chapter 2: Q2.6-4Q (page 93)

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.

4. Determine the production levels needed to satisfy a final demand of 18 units for manufacturing, 18 units for agriculture, and 0 units for services.

Short Answer

Expert verified

The production level needed to satisfy a final demand of 18 units for manufacturing (18 for agriculture and 0 for services) is \(x = \left( {73.333,50,30} \right)\).

Step by step solution

01

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. Also, 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 is needed to satisfy a final demand of 18 units for manufacturing, i.e., 18 for agriculture and 0 for services.

Solve the equation \(x = Cx + {\mathop{\rm d}\nolimits} \) for d, as shown below:

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

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

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

02

Convert the equation into an augmented matrix

The augmented matrix of the system of equation is

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

03

Apply the row operation

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

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

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.

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .6666}&{ - .6666}&{20}\\0&{0.6}&{ - 0.2}&{24}\\0&{ - 0.3}&{0.7}&6\end{array}} \right]\)

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

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&{20}\\0&1&{ - 0.333}&{40}\\0&{ - 0.3}&{0.7}&6\end{array}} \right]\)

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.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - .888}&{46.666}\\0&1&{ - 0.333}&{40}\\0&0&{0.6}&{18}\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - .888}&{46.666}\\0&1&{ - 0.333}&{40}\\0&0&1&{30}\end{array}} \right]\]

At row one, multiply row three by 0.888 and add it to row one. At row two, multiply row three by 0.33 and add it to row two.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{73.333}\\0&1&0&{50}\\0&0&1&{30}\end{array}} \right]\]

Thus, the production level needed to satisfy a final demand of 18 units for manufacturing (18 for agriculture and 0 for services) is \(x = \left( {73.333,50,30} \right)\).

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Most popular questions from this chapter

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

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