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[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

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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

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

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 \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

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