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Chapter 2: Q2.6-3Q (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.

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

Short Answer

Expert verified

The production level needed to satisfy a final demand of 18 units for manufacturing is \(x = \left( {40,15,15} \right)\).

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 needed to satisfy a final demand of 18 units for manufacturing and no demand for the other sectors.

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

\[\begin{array}{c}{\mathop{\rm d}\nolimits} = x - Cx\\\left[ {\begin{array}{*{20}{c}}{18}\\0\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{18}\\0\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10{x_1}}&{.60{x_2}}&{.60{x_3}}\\{.30{x_1}}&{.20{x_2}}&0\\{.30{x_1}}&{.10{x_2}}&{.10{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{18}\\0\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{.90{x_1}}&{ - .60{x_2}}&{ - .60{x_3}}\\{ - .30{x_1}}&{.80{x_2}}&0\\{ - .30{x_1}}&{ - .10{x_2}}&{.90{x_3}}\end{array}} \right]\end{array}\]

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

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

02

Convert the equation into an augmented matrix

The augmented matrix of the system of equations

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

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.

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

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

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&{20}\\0&1&{ - 0.333}&{10}\\0&{ - 0.3}&{0.7}&6\end{array}} \right]\)

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.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - .888}&{26.666}\\0&1&{ - 0.333}&{10}\\0&0&{0.6}&9\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - .888}&{26.666}\\0&1&{ - 0.333}&{10}\\0&0&1&{15}\end{array}} \right]\]

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.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{40}\\0&1&0&{15}\\0&0&1&{15}\end{array}} \right]\]

Thus, the production level needed to satisfy a final demand of 18 units for manufacturing is \(x = \left( {40,15,15} \right)\).

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Most popular questions from this chapter

In exercise 11 and 12, the matrices are all \(n \times n\). Each part of the exercise is an implication of the form 鈥淚f 鈥渟tatement 1鈥� then 鈥渟tatement 2鈥�.鈥滿ark the implication as True if the truth of 鈥渟tatement 2鈥漚lways follows whenever 鈥渟tatement 1鈥� happens to be true. An implication is False if there is an instance in which 鈥渟tatement 2鈥� is false but 鈥渟tatement 1鈥� is true. Justify each answer.

a. If the equation \[A{\bf{x}} = {\bf{0}}\] has only the trivial solution, then \(A\) is row equivalent to the \(n \times n\) identity matrix.

b. If the columns of \(A\) span \({\mathbb{R}^n}\), then the columns are linearly independent.

c. If \(A\) is an \(n \times n\) matrix, then the equation \(A{\bf{x}} = {\bf{b}}\) has at least one solution for each \({\bf{b}}\) in \({\mathbb{R}^n}\).

d. If the equation \[A{\bf{x}} = {\bf{0}}\] has a non trivial solution, then \[A\] has fewer than \(n\) pivot positions.

e. If \({A^T}\) is not invertible, then \(A\) is not invertible.

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Unless otherwise specified, assume that all matrices in these exercises are \(n \times n\). Determine which of the matrices in Exercises 1-10 are invertible. Use a few calculations as possible. Justify your answer.

10. [M] \[\left[ {\begin{array}{*{20}{c}}5&3&1&7&9\\6&4&2&8&{ - 8}\\7&5&3&{10}&9\\9&6&4&{ - 9}&{ - 5}\\8&5&2&{11}&4\end{array}} \right]\]

If the equation \(Hx = c\) is inconsistent for some c in \({\mathbb{R}^{\bf{n}}}\), what can you say about the equation \(Hx = {\bf{0}}\)? Why?

In Exercises 13 and 14, find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace?

14. \(\left[ {\begin{array}{*{20}{c}}1\\{ - {\bf{1}}}\\{ - 2}\\{\bf{5}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{3}}}\\{ - {\bf{1}}}\\{\bf{6}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{2}}\\{ - {\bf{6}}}\\{\bf{8}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{4}}\\{ - {\bf{7}}}\\7\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\{ - 8}\\9\\{ - 5}\end{array}} \right]\)
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