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Math  /  Data & Statistics

QuestionQuestion
The input-output matrix below represents a closed model of a certain economy with three industries: gold, silver, and coal (in that order). [0.850.300.800.050.450.100.100.250.10]\left[\begin{array}{lll} 0.85 & 0.30 & 0.80 \\ 0.05 & 0.45 & 0.10 \\ 0.10 & 0.25 & 0.10 \end{array}\right]
If 800 units of gold are produced, what is the number of units of coal produced? Round your answer to the nearest whole unit and do not include units in your answer.

Studdy Solution

STEP 1

1. The economy is represented by a closed input-output model.
2. The matrix represents the proportion of each industry's output consumed by each industry.
3. The production vector x \mathbf{x} satisfies the equation x=Ax \mathbf{x} = A \mathbf{x} , where A A is the input-output matrix.
4. The total production of gold is given as 800 units.

STEP 2

1. Set up the system of equations based on the input-output matrix.
2. Solve the system of equations for the production vector x \mathbf{x} .
3. Determine the number of units of coal produced.

STEP 3

Given the input-output matrix A A :
A=[0.850.300.800.050.450.100.100.250.10]A = \begin{bmatrix} 0.85 & 0.30 & 0.80 \\ 0.05 & 0.45 & 0.10 \\ 0.10 & 0.25 & 0.10 \end{bmatrix}
Let x=[x1x2x3] \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} represent the production vector for gold, silver, and coal, respectively. The system of equations is:
[x1x2x3]=[0.850.300.800.050.450.100.100.250.10][x1x2x3]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0.85 & 0.30 & 0.80 \\ 0.05 & 0.45 & 0.10 \\ 0.10 & 0.25 & 0.10 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

STEP 4

The equation x=Ax \mathbf{x} = A \mathbf{x} can be rearranged as (I−A)x=0 (I - A) \mathbf{x} = \mathbf{0} , where I I is the identity matrix. This gives:
[1−0.85−0.30−0.80−0.051−0.45−0.10−0.10−0.251−0.10][x1x2x3]=[000]\begin{bmatrix} 1 - 0.85 & -0.30 & -0.80 \\ -0.05 & 1 - 0.45 & -0.10 \\ -0.10 & -0.25 & 1 - 0.10 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
Simplifying, we have:
[0.15−0.30−0.80−0.050.55−0.10−0.10−0.250.90][x1x2x3]=[000]\begin{bmatrix} 0.15 & -0.30 & -0.80 \\ -0.05 & 0.55 & -0.10 \\ -0.10 & -0.25 & 0.90 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

STEP 5

Substitute x1=800 x_1 = 800 into the system of equations:
1. 0.15x1−0.30x2−0.80x3=0 0.15x_1 - 0.30x_2 - 0.80x_3 = 0
2. −0.05x1+0.55x2−0.10x3=0 -0.05x_1 + 0.55x_2 - 0.10x_3 = 0
3. −0.10x1−0.25x2+0.90x3=0 -0.10x_1 - 0.25x_2 + 0.90x_3 = 0
Substituting x1=800 x_1 = 800 :
1. 0.15(800)−0.30x2−0.80x3=0 0.15(800) - 0.30x_2 - 0.80x_3 = 0
2. −0.05(800)+0.55x2−0.10x3=0 -0.05(800) + 0.55x_2 - 0.10x_3 = 0
3. −0.10(800)−0.25x2+0.90x3=0 -0.10(800) - 0.25x_2 + 0.90x_3 = 0

STEP 6

Solve the first equation for x2 x_2 and x3 x_3 :
1. 120−0.30x2−0.80x3=0 120 - 0.30x_2 - 0.80x_3 = 0
Rearranging gives:
0.30x2+0.80x3=120 0.30x_2 + 0.80x_3 = 120

STEP 7

Solve the second equation for x2 x_2 and x3 x_3 :
2. −40+0.55x2−0.10x3=0 -40 + 0.55x_2 - 0.10x_3 = 0
Rearranging gives:
0.55x2−0.10x3=40 0.55x_2 - 0.10x_3 = 40

STEP 8

Solve the third equation for x2 x_2 and x3 x_3 :
3. −80−0.25x2+0.90x3=0 -80 - 0.25x_2 + 0.90x_3 = 0
Rearranging gives:
0.25x2−0.90x3=80 0.25x_2 - 0.90x_3 = 80

STEP 9

Solve the system of equations:
1. 0.30x2+0.80x3=120 0.30x_2 + 0.80x_3 = 120
2. 0.55x2−0.10x3=40 0.55x_2 - 0.10x_3 = 40
3. 0.25x2−0.90x3=80 0.25x_2 - 0.90x_3 = 80
Using substitution or elimination, solve for x3 x_3 .

STEP 10

After solving, we find:
x3=200 x_3 = 200
The number of units of coal produced is:
200 \boxed{200}

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