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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

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

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 a \(n \times n\) 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.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?
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