An industrial system has two industries with the following input requirements.
(a) 	To produce $1.00 worth of output, Industry A requires $0.10 of its own product and $0.50 of Industry B's product.
(b) 	To produce $1.00 worth of output, Industry B requires $0.40 of its own product and $0.30 of Industry A's product.
Find D, the input-output matrix for this system.
  	  A 	B   	 
D = 	
	Correct: Your answer is correct. 	Correct: Your answer is correct. 	
Correct: Your answer is correct. 	Correct: Your answer is correct.
	A
B

Solve for the output matrix X in the equation
X = DX + E,
where E is the external demand matrix
E = 
	10,000 	
20,000
.
(Round to the nearest whole number.)

Let's break down the problem step-by-step.

### Step 1: Formulate the Input-Output Matrix $ D $
Given the input requirements:
- For Industry A:
  - Requires $0.10 of its own product
  - Requires $0.50 of Industry B's product
- For Industry B:
  - Requires $0.30 of Industry A's product
  - Requires $0.40 of its own product

The input-output matrix $ D $ can be structured as follows:

\[
D = \begin{bmatrix}
0.10 & 0.50 \\
0.30 & 0.40
\end{bmatrix}
\]

Where:
- $ D_{11} = 0.10 $ (A requires 10% of its own output)
- $ D_{12} = 0.50 $ (A requires 50% of B's output)
- $ D_{21} = 0.30 $ (B requires 30% of A's output)
- $ D_{22} = 0.40 $ (B requires 40% of its own output)

### Step 2: Set up the Equation
The equation to find the output matrix $ X $ is:

\[
X = DX + E
\]

Where:
- $ X $ is the output vector
- $ E $ is the external demand matrix, given as $ E = \begin{bmatrix} 10000 \\ 20000 \end{bmatrix} $

### Step 3: Solve for $ X $
Rewriting the equation, we get:

\[
X - DX = E
\]

Factor out $ X $:

\[
(I - D)X = E
\]

Where $ I $ is the identity matrix:

\[
I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\]

So, $ I - D $ is:

\[
I - D = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0.10 & 0.50 \\ 0.30 & 0.40 \end{bmatrix} = \begin{bmatrix} 0.90 & -0.50 \\ -0.30 & 0.60 \end{bmatrix}
\]

### Step 4: Calculate $ X $
To find $ X $, we need to multiply the inverse of $ I - D $ by $ E $:

\[
X = (I - D)^{-1} E
\]

Let's compute this using Python to get the precise numerical values.The output matrix $ X $ for this system, rounded to the nearest whole number, is:

\[
X = \begin{bmatrix} 41026 \\ 53846 \end{bmatrix}
\]

This means that:
- Industry A needs to produce approximately 41,026 units.
- Industry B needs to produce approximately 53,846 units.

Would you like a more detailed explanation of any part of this process or have any questions?

Here are 5 related questions you might find interesting:
1. How would the output change if the external demand $ E $ were different?
2. What happens if one of the industries becomes more self-sufficient (e.g., Industry A requiring more of its own product)?
3. How does the stability of this economic system depend on the values in the input-output matrix $ D $?
4. Can this method be extended to more than two industries?
5. How does this model relate to real-world economic systems?

**Tip:** In linear algebra problems involving economic models, understanding the structure of matrices and their inverses is crucial for interpreting solutions.

An industrial system has two industries with the following input requirements. (a) To produce $1.00 worth of output, Industry A requires $0.10 of its own product and $0.50 of Industry B's product. (b) To produce $1.00 worth of output, Industry B requires $0.40 of its own product and $0.30 of Industry A's product. Find D, the input-output matrix for this system. Solve for the output matrix X in the equation X = DX + E, where E is the external demand matrix E = [10,000; 20,000]. (Round to the nearest whole number.)

Learn how to calculate the output matrix for an industrial economic system with two industries. This problem involves linear algebra concepts such as input-output matrices, matrix inversion, and external demand. The solution provides step-by-step guidance for solving X = DX + E, where E represents external demand. Suitable for undergraduate students studying economics or linear algebra.

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