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Expert-verified Found in: Page 21 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider the economy of Israel in 1958. The three industries considered here are $\begin{array}{l}{\mathbf{I}}_{\mathbf{1}}\mathbf{:}\mathbf{\text{\hspace{0.17em}agriculture,}}\\ {\mathbf{I}}_{\mathbf{2}}\mathbf{:}\mathbf{\text{\hspace{0.17em}manufacturing,}}\\ {\mathbf{I}}_{\mathbf{3}}\mathbf{:}\mathbf{\text{\hspace{0.17em}energy}}\end{array}$Outputs and demands are measured in millions of Israeli pounds, the currency of Israel at that time. We are told that $\begin{array}{l}\stackrel{\mathbf{\to }}{\mathbf{b}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{13}\mathbf{.2}\\ \mathbf{17}\mathbf{.6}\\ \mathbf{1}\mathbf{.8}\end{array}\mathbf{\right]}\mathbf{,}\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{1}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\mathbf{.293}\\ \mathbf{0}\mathbf{.014}\\ \mathbf{0}\mathbf{.044}\end{array}\mathbf{\right]}\mathbf{,}\\ \stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{2}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{0}\mathbf{.207}\\ \mathbf{0}\mathbf{.01}\end{array}\mathbf{\right]}\mathbf{,}\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{3}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{0}\mathbf{.017}\\ \mathbf{0}\mathbf{.216}\end{array}\mathbf{\right]}\end{array}$a. Why do the first components of $\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{2}}}$ and$\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{3}}}$ equal${\mathbf{0}}$ ?b. Find the outputs ${{\mathbit{x}}}_{{\mathbf{1}}}$, ${{\mathbit{x}}}_{{\mathbf{2}}}$, ${{\mathbit{x}}}_{{\mathbf{3}}}$required to satisfy demand.

a. As the sectors of energy and manufacturing do not make demand on the agriculture sector, so, the first components of $\stackrel{\to }{{\upsilon }_{2}}$ and $\stackrel{\to }{{\upsilon }_{3}}$ equals 0.

b. ${x}_{1}\approx 45.05,{x}_{2}\approx 81.95,{x}_{3}\approx 4.63$are the values that satisfy the demand.

See the step by step solution

## (a)Step 1: Consider the relation between the industries of agriculture, manufacturing and energy

The demand vector for the industry ${I}_{i}$ is represented by the vector$\stackrel{\to }{{\upsilon }_{i}}$ .

The vector$\stackrel{\to }{{\upsilon }_{2}}$ represents the manufacturing industry and the vector $\stackrel{\to }{{\upsilon }_{3}}$represents the energy industry.

As manufacturing and the energy industries does not make any kind of demand on agriculture industry, thus, the vectors of manufacturing and the energy industries equal zero.

## (b)Step 2: Consider the values of the vectors of industries

The vector values of the agriculture, manufacturing and the energy industries are,

$\begin{array}{l}\stackrel{\to }{b}=\left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right],\stackrel{\to }{{\upsilon }_{1}}=\left[\begin{array}{c}0.293\\ 0.014\\ 0.044\end{array}\right],\\ \stackrel{\to }{{\upsilon }_{2}}=\left[\begin{array}{c}0\\ 0.207\\ 0.01\end{array}\right],\stackrel{\to }{{\upsilon }_{3}}=\left[\begin{array}{c}0\\ 0.017\\ 0.216\end{array}\right]\end{array}$

The demand function is written as,$\stackrel{\to }{b}={x}_{1}\stackrel{\to }{{\upsilon }_{1}}+{x}_{2}\stackrel{\to }{{\upsilon }_{2}}+{x}_{3}\stackrel{\to }{{\upsilon }_{3}}$$\stackrel{\to }{b}={x}_{1}\stackrel{\to }{{\upsilon }_{1}}+{x}_{2}\stackrel{\to }{{\upsilon }_{2}}+{x}_{3}\stackrel{\to }{{\upsilon }_{3}}$ .

role="math" localid="1659671682712" $\begin{array}{c}\left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right]={x}_{1}\left[\begin{array}{c}0.293\\ 0.014\\ 0.044\end{array}\right]+{x}_{2}\left[\begin{array}{c}0\\ 0.207\\ 0.01\end{array}\right]+{x}_{3}\left[\begin{array}{c}0\\ 0.017\\ 0.216\end{array}\right]\\ \left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right]=\left[\begin{array}{ccc}0.293& 0& 0\\ 0.014& 0.207& 0.017\\ 0.044& 0.01& 0.216\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]\end{array}$

Therefore, the matrix form is,

$\left[\begin{array}{cccc}0.293& 0& 0& 13.2\\ 0.014& 0.207& 0.017& 17.6\\ 0.044& 0.01& 0.216& 1.8\end{array}\right]$

## Step 3: Solve the matrix

Consider the matrix.

$\left[\begin{array}{cccc}0.293& 0& 0& 13.2\\ 0.014& 0.207& 0.017& 17.6\\ 0.044& 0.01& 0.216& 1.8\end{array}\right]$

Using row Echelon form to reduce the matrix.

$\left[\begin{array}{cccc}1& 0& 0& 45.05\\ 0& 1& 0& 81.95\\ 0& 0& 1& -4.63\end{array}\right]$

The values are: ${x}_{1}\approx 45.05,{x}_{2}\approx 81.95,{x}_{3}\approx |-4.63|$

As the output cannot be negative, thus, the values will be,

${x}_{1}\approx 45.05,{x}_{2}\approx 81.95,{x}_{3}\approx 4.63$

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