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Problem 10

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

Expert verified
The production vector $$X$$ can be found by applying the formula $$X = (I - A)^{-1}D$$. After computing the inverse of the matrix $$(I - A)$$ and multiplying it by the demand vector $$D$$, we find that the production vector $$X=\begin{bmatrix} 6200\\ 6400\\ 8000 \end{bmatrix}$$.
See the step by step solution

## Step 1: Calculate the matrix (I - A)

First, we need to calculate the matrix $$(I - A)$$. Here, $$I$$ is the identity matrix of the same dimensions as the technology matrix $$A$$. The identity matrix is: $$I = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ Now, subtract the matrix $$A$$ from $$I$$: $$I - A = \left[ \begin{array}{ccc} 1-0.5 & 0-0.1 & 0-0 \\ 0-0.1 & 1-0.5 & 0-0.1 \\ 0-0 & 0-0 & 1-0.5 \end{array}\right]$$ So, we have: $$I - A = \left[ \begin{array}{ccc} 0.5 & -0.1 & 0 \\ -0.1 & 0.5 & -0.1 \\ 0 & 0 & 0.5 \end{array} \right]$$

## Step 2: Find the inverse of (I - A)

Next, we need to find the inverse of the matrix $$(I - A)$$. Let's call this inverse matrix $$B$$ such that $$B = (I - A)^{-1}$$. We can find the inverse of a $$3 \times 3$$ matrix by using the following formula: $$B = \frac{1}{\text{det}(I-A)} \left[\begin{array}{ccc}C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{array} \right]^{T}$$ where $$\text{det}(I-A)$$ is the determinant of the matrix $$(I - A)$$, $$C_{ij}$$ is the cofactor of the element in the ith row and jth column of the matrix $$(I - A)$$, and $$T$$ denotes transposition. Since the matrix is small, we can find the inverse matrices manually. However, for larger matrices, it is better to use software or calculator tools. First, we calculate the determinant of $$(I - A)$$: $$\text{det}(I-A)=\left | \begin{array}{ccc}0.5 & -0.1 & 0 \\ -0.1 & 0.5 & -0.1 \\ 0 & 0 & 0.5 \end{array} \right| =0.5(0.5(0.5) - 0 -0.1(0)) = 0.5(0.5*0.5) = 0.5*0.25 = 0.125$$ Now, we find the cofactors: $$C_{11} = \left| \begin{array}{cc} 0.5 & -0.1 \\ 0 & 0.5 \end{array}\right|= 0.5(0.5)$$ $$C_{12} = -\left| \begin{array}{cc} -0.1 & -0.1 \\ 0 & 0.5 \end{array}\right|= -(-0.1 (0.5))$$ $$C_{13} = \left| \begin{array}{cc} -0.1 & 0.5 \\ 0 & 0 \end{array}\right|= (-0.1*0)$$ $$C_{21} = -\left| \begin{array}{cc} -0.1 & 0 \\ 0 & 0.5 \end{array}\right|= -((-0.1)(0.5))$$ $$C_{22} = \left| \begin{array}{cc} 0.5 & 0 \\ 0 & 0.5 \end{array}\right|= (0.5)(0.5)$$ $$C_{23} = -\left| \begin{array}{cc} 0.5 & -0.1 \\ 0& 0 \end{array}\right|= -(0.5 (0))$$ $$C_{31} = \left| \begin{array}{cc} -0.1 & 0.5 \\ 0 & 0 \end{array}\right| = (-0.1*0)$$ $$C_{32} = -\left| \begin{array}{cc} 0.5 & -0.1 \\ -0.1 & 0.5 \end{array}\right|= -(0.5(0.5)-0.1(-0.1))$$ $$C_{33} = \left| \begin{array}{cc} 0.5 & -0.1 \\ -0.1 & 0.5 \end{array}\right|= (0.5)(0.5)-(0.1)(-0.1)$$ By using the formula mentioned above and substituting the cofactors and determinant: $$B = \frac{1}{0.125} \left[ \begin{array}{ccc} 0.25 & -0.05 & 0 \\ -0.05 & 0.25 & -0 \\ 0 & -0 & 0.3 \end{array}\right]^{T} = \left[ \begin{array}{ccc} 2 & 0.4 & 0 \\ 0.4 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]$$

## Step 3: Calculate the production vector X

Finally, multiply the demand vector $$D$$ by the inverse matrix $$B$$ to get the production vector $$X$$: $$X = BD = \left[ \begin{array}{ccc} 2 & 0.4 & 0 \\ 0.4 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right] \left[ \begin{array}{c} 3000 \\ 3800 \\ 2000 \end{array}\right] = \left[ \begin{array}{c} 6200 \\ 6400 \\ 8000 \end{array}\right]$$ So, the production vector $$X=\begin{bmatrix} 6200\\ 6400\\ 8000 \end{bmatrix}$$.

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