Question

[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

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

Step-by-step solution

Find the matrix \[I - C\]

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

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

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

\( > > I - C\)

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

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

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

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:

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

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

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