Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 4E from Chapter 2.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 4E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We have to construct the consumption matrix for this economy, and determine the production levels needed to satisfy a final demand of 18 units for manufacturing, 18 units for agriculture, and 0 units for services.
Step-1: Tabulated data
Step-2:The consumption matrix matrix
Generate the consumption matrix: \[ C = \left[ \begin{array} { c c c } { .10 } & { .60 } & { .60 } \\ { .30 } & { .20 } & { 0 } \\ { .30 } & { .10 } & { .10 } \end{array} \right] \]
Step-3: The production levels needed
The solution of the system $X = C X + d$ gives production levels needed to meet a final demand. \[\begin{array}{l} X = CX + d\\ X = CX + d\\ X = CX + d \end{array}\] \[\begin{array}{l} x = {(I - C)^{ - 1}}d\\ x = {\left[ {\begin{array}{*{20}{c}} {0.9}&{ - 0.6}&{ - 0.6}\\ { - 0.3}&{0.8}&0\\ { - 0.3}&{ - 0.1}&{0.9} \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} {18}\\ {18}\\ 0 \end{array}} \right] \end{array}\]
Step-4: The Augmented Matrix
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {I - C}&b \end{array}} \right] \to \left[ {\begin{array}{*{20}{c}} {0.9}&{ - 0.6}&{ - 0.6}&{18}\\ { - 0.3}&{0.8}&0&{18}\\ { - 0.3}&{ - 0.1}&{0.9}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 9&{ - 6}&{ - 6}&{180}\\ { - 3}&8&0&{180}\\ { - 3}&{ - 1}&9&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to 10{R_1}}\\ {{R_2} \to 10{R_2}}\\ {{R_3} \to 10{R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 9&{ - 6}&{ - 6}&{180}\\ { - 3}&8&0&{180}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_3} \to {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ { - 3}&8&0&{180}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_1} \to \dfrac{1}{3}{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&6&{ - 2}&{240}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_2} \to {R_2} + {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&{ - 3}&3&{ - 60} \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to \dfrac{1}{2}{R_2}}\\ {{R_3} \to \dfrac{1}{3}{R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&0&2&{60} \end{array}} \right]::\left\{ {{R_3} \to {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_3} \to \dfrac{1}{2}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&0&{150}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_2} \to {R_2} + {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&0&{120}\\ 0&3&0&{150}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to {R_1} + 2{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&0&{120}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_2} \to \dfrac{1}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&0&0&{220}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to {R_1} + 2{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{\dfrac{{220}}{3}}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to \dfrac{1}{3}{R_1}} \right\} \end{array}\] Therefore the solution is:
Manufacturing must produce units, agriculture must produce 50 units and service must produce 30 units.
We have to construct the consumption matrix for this economy, and determine the production levels needed to satisfy a final demand of 18 units for manufacturing, 18 units for agriculture, and 0 units for services.
Step-1: Tabulated data
| Purchased From | Input Consumed per unit output | ||
|---|---|---|---|
| Manufacturing | Agriculture | Services | |
| Manufacturing | 0 10 | 0 60 | 0 60 |
| Agriculture | 0 30 | 0 20 | 0.00 |
| Services | 0.30 | 010 | 0.10 |
Step-2:The consumption matrix matrix
Generate the consumption matrix: \[ C = \left[ \begin{array} { c c c } { .10 } & { .60 } & { .60 } \\ { .30 } & { .20 } & { 0 } \\ { .30 } & { .10 } & { .10 } \end{array} \right] \]
Step-3: The production levels needed
The solution of the system $X = C X + d$ gives production levels needed to meet a final demand. \[\begin{array}{l} X = CX + d\\ X = CX + d\\ X = CX + d \end{array}\] \[\begin{array}{l} x = {(I - C)^{ - 1}}d\\ x = {\left[ {\begin{array}{*{20}{c}} {0.9}&{ - 0.6}&{ - 0.6}\\ { - 0.3}&{0.8}&0\\ { - 0.3}&{ - 0.1}&{0.9} \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} {18}\\ {18}\\ 0 \end{array}} \right] \end{array}\]
Step-4: The Augmented Matrix
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {I - C}&b \end{array}} \right] \to \left[ {\begin{array}{*{20}{c}} {0.9}&{ - 0.6}&{ - 0.6}&{18}\\ { - 0.3}&{0.8}&0&{18}\\ { - 0.3}&{ - 0.1}&{0.9}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 9&{ - 6}&{ - 6}&{180}\\ { - 3}&8&0&{180}\\ { - 3}&{ - 1}&9&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to 10{R_1}}\\ {{R_2} \to 10{R_2}}\\ {{R_3} \to 10{R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 9&{ - 6}&{ - 6}&{180}\\ { - 3}&8&0&{180}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_3} \to {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ { - 3}&8&0&{180}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_1} \to \dfrac{1}{3}{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&6&{ - 2}&{240}\\ 0&{ - 9}&9&{ - 180} \end{array}} \right]::\left\{ {{R_2} \to {R_2} + {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&{ - 3}&3&{ - 60} \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to \dfrac{1}{2}{R_2}}\\ {{R_3} \to \dfrac{1}{3}{R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&0&2&{60} \end{array}} \right]::\left\{ {{R_3} \to {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&{ - 1}&{120}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_3} \to \dfrac{1}{2}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 2}&{60}\\ 0&3&0&{150}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_2} \to {R_2} + {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&0&{120}\\ 0&3&0&{150}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to {R_1} + 2{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&0&{120}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_2} \to \dfrac{1}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&0&0&{220}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to {R_1} + 2{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{\dfrac{{220}}{3}}\\ 0&1&0&{50}\\ 0&0&1&{30} \end{array}} \right]::\left\{ {{R_1} \to \dfrac{1}{3}{R_1}} \right\} \end{array}\] Therefore the solution is:
Manufacturing must produce units, agriculture must produce 50 units and service must produce 30 units.