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Chapter 2: Q2.6-14Q (page 93)

[M] The demand vector in Exercise 13 is reasonable for 1958 data, but Leontief’s discussion of the economy in the reference cited there used a demand vector closer to 1964 data:

\({\bf{d}} = \left( {{\bf{99640}},\,{\bf{75548}},\,{\bf{14444}},\,{\bf{33501}},\,{\bf{23527}},\,{\bf{263985}},\,{\bf{6526}}} \right)\)

Find the production level needed to satisfy the demand.

Short Answer

Expert verified

\({\bf{x}} = \left( {134000,132000,69000,177000,67000,444000,18000} \right)\)

Step by step solution

01

Write the matrix in the \[I - C\] form

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

02

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

\[\left[ {\begin{array}{*{20}{c}}{I - C}&{\bf{d}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}}&{\begin{array}{*{20}{c}}{99640}\\{75548}\\{14444}\\{33501}\\{23527}\\{263985}\\{6526}\end{array}}\end{array}\,} \right]\]

03

Convert the matrix into row-reduced echelon form

Consider the following:

\(A = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{.8412}&{ - 0.0064}&{ - 0.0025}&{ - 0.0304}&{ - 0.0014}&{ - 0.0083}&{ - 0.1594}\\{ - .0057}&{0.7355}&{ - 0.0436}&{ - 0.0099}&{ - 0.0083}&{ - 0.0201}&{ - 0.3413}\\{ - .0264}&{ - 0.1506}&{0.6443}&{ - 0.0139}&{ - 0.0142}&{ - 0.0070}&{ - 0.0236}\\{ - .3299}&{ - 0.0565}&{ - 0.0495}&{0.6364}&{ - 0.0204}&{ - 0.0483}&{ - 0.0649}\\{ - .0089}&{ - 0.0081}&{ - 0.0333}&{ - 0.0295}&{0.6588}&{ - 0.0237}&{ - 0.0020}\\{ - 0.1190}&{ - 0.0901}&{ - 0.0996}&{ - 0.1260}&{ - 0.1722}&{0.7632}&{ - 0.3369}\\{ - 0.0063}&{ - 0.0126}&{ - 0.0196}&{ - 0.0098}&{ - 0.0064}&{ - 0.0132}&{0.9988}\end{array}}&{\begin{array}{*{20}{c}}{99640}\\{75548}\\{14444}\\{33501}\\{23527}\\{263985}\\{6526}\end{array}}\end{array}\,} \right]\)

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

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }} > > C = \left[ \begin{array}{l}\begin{array}{*{20}{c}}{.8412}&{ - .0064}&{ - .0025}&{ - .0304}&{ - .0014}&{ - .0083}&{ - .1594\,\,\,99640;\,\,}\end{array}\\\begin{array}{*{20}{c}}{ - .0057}&{0.7355}&{ - 0.0436}&{ - .0099}&{ - .0083}&{ - .0201}&{ - 0.3413\,\,75548;}\end{array}\\ - \begin{array}{*{20}{c}}{.0264}&{ - .1506}&{.6443}&{ - .0139}&{ - .0142}&{ - .0070}&{ - .0236\,\,\,\,14444;\,\,\,}\end{array}\\\begin{array}{*{20}{c}}{ - .3299}&{ - .0565}&{ - .0495}&{.6364}&{ - .0204}&{ - .0483}&{ - .0649\,\,\,\,\,33501;}\end{array}\\\begin{array}{*{20}{c}}{ - .0089}&{ - .0081}&{ - .0333}&{ - .0295}&{0.6588}&{ - .0237}&{ - .0020\,\,\,23527;\;}\end{array}\\\;\begin{array}{*{20}{c}}{ - .1190}&{ - 0.0901}&{ - .09966}&{ - .1260}&{ - .1722}&{.7632}&{ - .3369\,\,\,263985;}\end{array}\\\begin{array}{*{20}{c}}{ - .0063}&{ - .0126}&{ - .0196}&{ - .0098}&{ - .0064}&{ - .0132}&{0.9988\,\,\,\,\,6526;}\end{array}\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&0&0&{134034}\\0&1&0&0&0&0&0&{131687}\\0&0&1&0&0&0&0&{69472}\\0&0&0&1&0&0&0&{176912}\\0&0&0&0&1&0&0&{66596}\\0&0&0&0&0&1&0&{443773}\\0&0&0&0&0&0&1&{18431}\end{array}} \right]\)

04

Find the production level

The production level for the seven categories is

\(x\left( {\begin{array}{*{20}{c}}{134034}&{131687}&{69472}&{176912}&{66596}&{443773}&{18431}\end{array}} \right)\).

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

\({\bf{x}} = \left( {134000,132000,69000,177000,67000,444000,18000} \right)\).

So, the production level of seven categories is \({\bf{x}} = \left( {134000,132000,69000,177000,67000,444000,18000} \right)\).

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Most popular questions from this chapter

[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.]

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.
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