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Problem 10

# You are given a technology matrix A and an external demand vector $$D$$. Find the corresponding production vector $$X .$$ [HINT: See Quick Example $$1 .]$$ $$A=\left[\begin{array}{lll} 0.5 & 0.1 & 0 \\ 0 & 0.5 & 0.1 \\ 0 & 0 & 0.5 \end{array}\right], D=\left[\begin{array}{l} 3,000 \\ 3,800 \\ 2,000 \end{array}\right]$$

Expert verified
The short answer to this question is: The production vector $$X$$ is found to be: $X=\left[\begin{array}{l} 7,680 \\ 8,400 \\ 4,000 \end{array}\right]$
See the step by step solution

## Step 1: Identify the given matrices and vector.

We are given the technology matrix $$A$$, and external demand vector $$D$$: $A=\left[\begin{array}{lll} 0.5 & 0.1 & 0 \\\ 0 & 0.5 & 0.1 \\\ 0 & 0 & 0.5 \end{array}\right], D=\left[\begin{array}{l} 3,000 \\\ 3,800 \\\ 2,000 \end{array}\right]$

## Step 2: Calculate the matrix (I - A).

Here $$I$$ represents the identity matrix. The size of the identity matrix should match the size of the technology matrix A. Since A is a 3x3 matrix, the identity matrix is also a 3x3 matrix: $I = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ Now, let's calculate $$(I - A)$$: $(I - A) = \left[\begin{array}{lll} 1-0.5 & 0-0.1 & 0, \\ 0 & 1-0.5 & 0-0.1 \\ 0 & 0 & 1-0.5 \end{array}\right] = \left[\begin{array}{lll} 0.5 & -0.1 & 0 \\ 0 & 0.5 & -0.1 \\ 0 & 0 & 0.5 \end{array}\right]$

## Step 3: Find the inverse of (I - A).

Now we need to find the inverse of the matrix $$(I - A)$$, denoted as $$(I - A)^{-1}$$. We can use the formula for the inverse of a 3x3 matrix or use software tools such as MATLAB or Python to compute the inverse. In this case, we find: $(I - A)^{-1}= \left[\begin{array}{ccc} 2 & 0.4 & 0.08 \\ 0 & 2 & 0.4 \\ 0 & 0 & 2 \end{array}\right]$

## Step 4: Calculate X by multiplying (I - A)^{-1} with D.

Finally, let's calculate the production vector X by multiplying $$(I - A)^{-1}$$ with D: $X = (I - A)^{-1} D = \left[\begin{array}{ccc} 2 & 0.4 & 0.08 \\ 0 & 2 & 0.4 \\ 0 & 0 & 2 \end{array}\right] \left[\begin{array}{c} 3,000 \\ 3,800 \\ 2,000 \end{array}\right]$ $X = \left[\begin{array}{c} 6,000 + 1,520 + 160 \\ 0 + 7,600 + 800 \\ 0 + 0 + 4,000 \end{array}\right] = \left[\begin{array}{c} 7,680 \\ 8,400 \\ 4,000 \end{array}\right]$ The production vector X is found to be: $X=\left[\begin{array}{l} 7,680 \\ 8,400 \\ 4,000 \end{array}\right]$

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