# Thread: Linear Algebra Word Problem

1. ## Linear Algebra Word Problem

I am really stuck on this problem....

An economy consists of five sectors-agriculture,manufacturing, services, entertainment, and mining. For each unit of output, agriculture used .25 unit of its own output, .40 unit from manufacturing, .10 unit from services, .02 unit from entertainment, and .03 unit from mining. For each unit of output, manufacturing uses .40 unit of its own output, .15 unit from agriculture, .10 unit from services, .03 unit from entertainment, and .20 unit from mining. For each unit of output, services use .25 unit of its own output, .10 unit from agriculture. .15 from manufacturing, .06 unit from entertainment, and .02 from mining. For each unit of output, entertainment uses .15 unit of its own output, .06 unit from agriculture, .45 unit from manufacturing, .10 unit from services, and .05 unit from mining. For each unit of output, mining uses .08 of its own output, .05 unit from agriculture, .55 unit from manufacturing, .20 unit from services, and .05 unit from entertainment.

1. construct a table that illustrates the inputs consumed per unit of output.
2. A) determine what the individual intermediate demands are created if agriculture plans to produce 240 units, manufacturing 160 units, services 125 units, entertainment 90 units, and mining 120 units.
B) construct the consumption matrix C for this economy and use the matrix to compute the total intermediate demand from all 5 sectors if the amounts produced for each sector are as given from part A).
3. Develop a system of equations that can be used to determine the production levels needed to satisfy a final demand of 60 units for agriculture, 20 units for manufacturing, 10 units for services, 30 units for entertainment, and 10 units for mining. Round-off answer to 2 decimal places (calculator).
4. Use your calculator to obtain an inverse matrix and use the inverse to determine the production levels needed to satisfy a final demand as specified in question 3. Round-off your answer to two decimal places. How do the answers in question 3 and question 4 compare?
5. Give the inverse matrix that you used in question 4. Round-off its entires to four decimal places. What matrix operation did you use in question 4 to compute production levels? Specify.

2. Originally Posted by chris25
I am really stuck on this problem....

An economy consists of five sectors-agriculture,manufacturing, services, entertainment, and mining. For each unit of output, agriculture used .25 unit of its own output, .40 unit from manufacturing, .10 unit from services, .02 unit from entertainment, and .03 unit from mining. For each unit of output, manufacturing uses .40 unit of its own output, .15 unit from agriculture, .10 unit from services, .03 unit from entertainment, and .20 unit from mining. For each unit of output, services use .25 unit of its own output, .10 unit from agriculture. .15 from manufacturing, .06 unit from entertainment, and .02 from mining. For each unit of output, entertainment uses .15 unit of its own output, .06 unit from agriculture, .45 unit from manufacturing, .10 unit from services, and .05 unit from mining. For each unit of output, mining uses .08 of its own output, .05 unit from agriculture, .55 unit from manufacturing, .20 unit from services, and .05 unit from entertainment.
This is the classic "Leontief Input-Output Model" problem. Please see your textbook or search on line to find more details about this economic model and how Linear Algebra played an important role in solving it. I am just going to give you the basic setup for your problem, you might need to fill in some details.

Originally Posted by chris25
1. construct a table that illustrates the inputs consumed per unit of output.
From the given, I can construct the following table that illustrates the inputs consumed per unit of output, please see the attachment.

Realize that the first column in above table is the consumption vector of Agriculture, to simplify the notation, let's denote it as $\vec{c}_1$ (i.e. $\vec{c}_1=\begin{bmatrix}0.25\\0.4\\0.1\\0.02\\0.0 3\end{bmatrix}$); similarly, we can denote the consumption vectors of Manufacturing, Service, Entertainment and Mining as $\vec{c}_2,\;\vec{c}_3,\;\vec{c}_4$ and $\vec{c}_5$ respectively. And they are the 2nd, 3rd, 4th and 5th columns of above table respectively.

Originally Posted by chris25
2. A) determine what the individual intermediate demands are created if agriculture plans to produce 240 units, manufacturing 160 units, services 125 units, entertainment 90 units, and mining 120 units.
The intermediate demands of an economic sector can be found through multiplying the number of units by its corresponding consumption vector. For instance, in order to find the intermediate demand of Agriculture if it plans to produce 240 units, we calculate

$240\cdot \vec{c}_1=240\cdot\begin{bmatrix}0.25\\0.4\\0.1\\0 .02\\0.03\end{bmatrix}=\begin{bmatrix}60\\96\\24\\ 4.8\\7.2\end{bmatrix}$

Similarly, the intermediate demand of manufacturing 160 units is:

$160\cdot \vec{c}_2=160\cdot\begin{bmatrix}0.15\\0.4\\0.1\\0 .03\\0.2\end{bmatrix}$

the intermediate demand of services 125 units is:

$125\cdot \vec{c}_3=125\cdot\begin{bmatrix}0.1\\0.15\\0.25\\ 0.06\\0.02\end{bmatrix}$

I think you get the idea and can finish the last two ...

Originally Posted by chris25
2. B) construct the consumption matrix C for this economy and use the matrix to compute the total intermediate demand from all 5 sectors if the amounts produced for each sector are as given from part A).
Here the consumption matrix is just $C=[\vec{c}_1\;\vec{c}_2\;\vec{c}_3\;\vec{c}_4\;\vec{c }_5]$, namely,

$C=\begin{bmatrix}0.25 & 0.15 & 0.1 & 0.06 & 0.05\\0.4 & 0.4 & 0.15 & 0.45 & 0.55\\0.1 & 0.1 & 0.25 & 0.1 & 0.2\\0.02 & 0.03 & 0.06 & 0.15 & 0.05\\0.03 & 0.2 & 0.02 & 0.05 & 0.08\\\end{bmatrix}$

Let $\vec{x}$ be the production vector, form (A), we have

$\vec{x}=\begin{bmatrix}240\\160\\125\\90\\120\end{ bmatrix}$

The total intermediate demand from all five sectors is given by

$C\vec{x}=\begin{bmatrix}0.25 & 0.15 & 0.1 & 0.06 & 0.05\\0.4 & 0.4 & 0.15 & 0.45 & 0.55\\0.1 & 0.1 & 0.25 & 0.1 & 0.2\\0.02 & 0.03 & 0.06 & 0.15 & 0.05\\0.03 & 0.2 & 0.02 & 0.05 & 0.08\\\end{bmatrix}\begin{bmatrix}240\\160\\125\\9 0\\120\end{bmatrix}$

Originally Posted by chris25
3. Develop a system of equations that can be used to determine the production levels needed to satisfy a final demand of 60 units for agriculture, 20 units for manufacturing, 10 units for services, 30 units for entertainment, and 10 units for mining. Round-off answer to 2 decimal places (calculator).
In order to answer this part, we need to know the Leontief Input-Output Model (Called Production Equation), which states that

Amt. Produced = Intermediate Demand + Final Demand

Mathematically, we have:

$\vec{x}=C\vec{x}+\vec{d}$

where $\vec{x}$ is the production amount (level) vector, $C$ is the consumption matrix and $\vec{d}$ is the final demand vector. Back to our problem, we have found out $C$ in part 2(B) and given $\vec{d}=\begin{bmatrix}60\\20\\10\\30\\10\end{bmat rix}$

In order to determine the production levels needed, we just need to solve the following system for $\vec{x}$

$\vec{x}=C\vec{x}+\vec{d}$

Or equivalently,

$(I_5-C)\vec{x}=\vec{d}$, where $I_5$ is a 5x5 identity matrix. Note here we know the matrix $I_5-C$ and the vector $\vec{d}$, then we have a system of equations to solve for $\vec{x}$.

Originally Posted by chris25
4. Use your calculator to obtain an inverse matrix and use the inverse to determine the production levels needed to satisfy a final demand as specified in question 3. Round-off your answer to two decimal places. How do the answers in question 3 and question 4 compare?
Realize that the system we obtained in (3) can also be solved by finding out the inverse matrix to $I_5-C$ (i.e. $(I_5-C)^{-1}$) provided the inverse exists. If the inverse exists, we can determine the production level vector by

$\vec{x}=(I_5-C)^{-1}\vec{d}$

How do we know the matrix $I_5-C$ has an inverse or not? We have a nice theorem to use here. The theorem says: If $C$ and $\vec{d}$ have nonnegative entries and it each column sum of $C$ is less than 1, then $(I-C)^{-1}$ exists and the production vector $\vec{x}=(I-C)^{-1}\vec{d}$ has nonnegative entries and is the unique solution of $\vec{x}=C\vec{x}+\vec{d}$.

If we apply this theorem to our problem, we can conclude that $(I_5-C)^{-1}$ exists and the production vector $\vec{x}=(I_5-C)^{-1}\vec{d}$ has nonnegative entries and is the unique solution of our system $\vec{x}=C\vec{x}+\vec{d}$. Thus he answers you obtained in question 3 and question 4 should be same.

Originally Posted by chris25
5. Give the inverse matrix that you used in question 4. Round-off its entires to four decimal places. What matrix operation did you use in question 4 to compute production levels? Specify.
It is your turn to finish it...

Roy

3. ## question to linear algebra word problem

hi im still confussed on parts 4 and 5. i received my answer for part three and when i solve part for they dont match up so i think i have the wrong matrix???