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Chapter 2: Q2.6-13Q (page 93)

[M] The consumption matrix C below is based on input-output data for the U.S. economy in 1958, with data for 81 sectors grouped into 7 larger sectors: (1) nonmetal household and personal products, (2) final metal products (such as motor vehicles), (3) basic metal products and mining, (4) basic nonmetal products and agriculture, (5) energy, (6) services, and (7) entertainment and miscellaneous products. Find the production levels needed to satisfy the final demand d. (Units are in millions of dollars.)

\(\left[ {\begin{array}{*{20}{c}}{.{\bf{1588}}}&{.{\bf{0064}}}&{.{\bf{0025}}}&{.{\bf{0304}}}&{.{\bf{0014}}}&{.{\bf{0083}}}&{.{\bf{1594}}}\\{.{\bf{0057}}}&{.{\bf{2645}}}&{.{\bf{0436}}}&{.{\bf{0099}}}&{.{\bf{0083}}}&{.{\bf{0201}}}&{.{\bf{3413}}}\\{.{\bf{0264}}}&{.{\bf{1506}}}&{.{\bf{3557}}}&{.{\bf{0139}}}&{.{\bf{0142}}}&{.{\bf{0070}}}&{.{\bf{0236}}}\\{.{\bf{3299}}}&{.{\bf{0565}}}&{.{\bf{0495}}}&{.{\bf{3636}}}&{.{\bf{0204}}}&{.{\bf{0483}}}&{.{\bf{0649}}}\\{.{\bf{0089}}}&{.{\bf{0081}}}&{.{\bf{0333}}}&{.{\bf{0295}}}&{.{\bf{3412}}}&{.{\bf{0237}}}&{.{\bf{0020}}}\\{.{\bf{1190}}}&{.{\bf{0901}}}&{.{\bf{0996}}}&{.{\bf{1260}}}&{.{\bf{1722}}}&{.{\bf{2368}}}&{.{\bf{3369}}}\\{.{\bf{0063}}}&{.{\bf{0126}}}&{.{\bf{0196}}}&{.{\bf{0098}}}&{.{\bf{0064}}}&{.{\bf{0132}}}&{.{\bf{0012}}}\end{array}} \right]\), \({\bf{d}} = \left[ {\begin{array}{*{20}{c}}{{\bf{74,000}}}\\{{\bf{56,000}}}\\{{\bf{10,500}}}\\{{\bf{25,000}}}\\{{\bf{17,500}}}\\{{\bf{196,000}}}\\{{\bf{5,000}}}\end{array}} \right]\)

Short Answer

Expert verified

\({\bf{x}} = \left( {10000,98000,51000,132000,49000,330000,14000} \right)\)

Step by step solution

01

Find the matrix \[I - C\]

Use the following MATLAB code to calculate \(I - C\).

\( > > I = \left[ \begin{array}{l}\begin{array}{*{20}{c}}1&0&0&0&0&0&{0;\,\,}\end{array}\begin{array}{*{20}{c}}0&1&0&0&0&0&{0;\,\,}\end{array}\begin{array}{*{20}{c}}0&0&1&0&0&0&{0;\,\,}\end{array}\\\begin{array}{*{20}{c}}0&0&0&1&0&0&{0;\,\,}\end{array}\begin{array}{*{20}{c}}0&0&0&0&1&0&{0;\,\,}\end{array}\begin{array}{*{20}{c}}0&0&0&0&0&1&{0;\,\,}\end{array}\\\begin{array}{*{20}{c}}0&0&0&0&0&0&{1;\,\,}\end{array}\end{array} \right]\)

\[ > > C = \left[ \begin{array}{l}\begin{array}{*{20}{c}}{.1588}&{.0064}&{.0025}&{.0304}&{.0014}&{.0083}&{.1594;\,\,}\end{array}\\\begin{array}{*{20}{c}}{.0057}&{0.2645}&{0.0436}&{.0099}&{.0083}&{.0201}&{0.3413;}\end{array}\\\begin{array}{*{20}{c}}{.0264}&{.1506}&{.3557}&{.0139}&{.0142}&{.0070}&{.0236;\,\,\,}\end{array}\\\begin{array}{*{20}{c}}{.3299}&{.0565}&{.0495}&{.3636}&{.0204}&{.0483}&{.0649;}\end{array}\\\begin{array}{*{20}{c}}{.0089}&{.0081}&{.0333}&{.0295}&{0.3412}&{.0237}&{.0020;\;}\end{array}\\\;\begin{array}{*{20}{c}}{.1190}&{0.0901}&{.09966}&{.1260}&{.1722}&{.2368}&{.3369;}\end{array}\\\begin{array}{*{20}{c}}{.0063}&{.0126}&{.0196}&{.0098}&{.0064}&{.0132}&{.0012}\end{array}\end{array} \right]\]

\( > > I - C\)

\[I - C = \left[ {\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}} \right]\]

02

Write the augmented matrix \(\left[ {\begin{array}{*{20}{c}}{I - C}&{\bf{d}}\end{array}} \right]\)

\[\left[ {\begin{array}{*{20}{c}}{I - C}&{\bf{d}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}}&{\begin{array}{*{20}{c}}{74000}\\{56000}\\{10500}\\{25000}\\{17500}\\{196000}\\{5000}\end{array}}\end{array}\,} \right]\]

03

Convert the matrix into row-reduced echelon form

Consider the following:

\(A = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}}&{\begin{array}{*{20}{c}}{74000}\\{56000}\\{10500}\\{25000}\\{17500}\\{196000}\\{5000}\end{array}}\end{array}\,} \right]\)

Use the code in MATLAB to obtain the row-reducedechelon form, as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }} > > C = \left[ \begin{array}{l}\begin{array}{*{20}{c}}{.8412}&{ - .0064}&{ - .0025}&{ - .0304}&{ - .0014}&{ - .0083}&{ - .1594\,\,\,74000;\,\,}\end{array}\\\begin{array}{*{20}{c}}{ - .0057}&{0.7355}&{ - 0.0436}&{ - .0099}&{ - .0083}&{ - .0201}&{ - 0.3413\,\,56000;}\end{array}\\ - \begin{array}{*{20}{c}}{.0264}&{ - .1506}&{.6443}&{ - .0139}&{ - .0142}&{ - .0070}&{ - .0236\,\,\,\,10500;\,\,\,}\end{array}\\\begin{array}{*{20}{c}}{ - .3299}&{ - .0565}&{ - .0495}&{.6364}&{ - .0204}&{ - .0483}&{ - .0649\,\,\,\,\,25000;}\end{array}\\\begin{array}{*{20}{c}}{ - .0089}&{ - .0081}&{ - .0333}&{ - .0295}&{0.6588}&{ - .0237}&{ - .0020\,\,\,17500;\;}\end{array}\\\;\begin{array}{*{20}{c}}{ - .1190}&{ - 0.0901}&{ - .09966}&{ - .1260}&{ - .1722}&{.7632}&{ - .3369\,\,\,196000;}\end{array}\\\begin{array}{*{20}{c}}{ - .0063}&{ - .0126}&{ - .0196}&{ - .0098}&{ - .0064}&{ - .0132}&{0.9988\,\,\,\,\,5000;}\end{array}\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&0&0&{99576}\\0&1&0&0&0&0&0&{97703}\\0&0&1&0&0&0&0&{51231}\\0&0&0&1&0&0&0&{131570}\\0&0&0&0&1&0&0&{49488}\\0&0&0&0&0&1&0&{329554}\\0&0&0&0&0&0&1&{13835}\end{array}} \right]\)

04

Find the production level

The production level for the seven categories is

\(x\left( {\begin{array}{*{20}{c}}{99576}&{97703}&{51231}&{131570}&{49488}&{329554}&{13835}\end{array}} \right)\).

The entries of \({\bf{x}}\) suggest greater precision when approximated to the nearest thousand. So, the realistic answer is

\({\bf{x}} = \left( {100000,98000,51000,132000,49000,330000,14000} \right)\).

So, the production level of seven categories is \({\bf{x}} = \left( {100000,98000,51000,132000,49000,330000,14000} \right)\).

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).
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