Problem

Input-output analysis can also be used to model how changes in one city can affect cities that are connected with it in some way. For example, if a large manufacturing company shuts down in one city, it is very likely that the economic welfare of all of the cities around it will suffer. Consider the three Pennsylvania communities Sharon, Farrell, and Hermitage. Due to their proximity to each other, residents of these three communities regularly spend time and money in the other communities. Suppose that we have gathered information in the form of an input-output matrix $\left[\begin{array}{lll}0.3 & 0.1 & 0.1 \\ 0.1 & 0.3 & 0.3 \\ 0.6 & 0.5 & 0.6\end{array}\right]$. This matrix can be thought of as the likelihood that a person from each particular community will spend money in each of the communities. Complete parts (a) and (b).
(a) Treat this matrix like an input-output matrix and calculate $(I-A)^{-1}$, where $A$ is the given input-output matrix.
\[
(1-A)^{-1}=
\]
(Round to two decimal places as needed.)

Answer

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Answer

So, \((\mathbf{I} - \mathbf{A})^{-1} = \boxed{\begin{pmatrix} 2.19 & 0.10 & 0.87 \\ 0.10 & 1.73 & 2.09 \\ 0.31 & 0.31 & 2.45 \end{pmatrix}}\).

Steps

Step 1 :First, we identify the matrix \(\mathbf{A}\) as \(\begin{pmatrix} 0.3 & 0.1 & 0.1 \\ 0.1 & 0.3 & 0.3 \\ 0.6 & 0.5 & 0.6 \end{pmatrix}\).

Step 2 :Next, we calculate the matrix \(\mathbf{I} - \mathbf{A}\), where \(\mathbf{I}\) is the identity matrix. This gives us \(\begin{pmatrix} 0.7 & -0.1 & -0.1 \\ -0.1 & 0.7 & -0.3 \\ -0.6 & -0.5 & 0.4 \end{pmatrix}\).

Step 3 :Then, we calculate the inverse of the matrix \(\mathbf{I} - \mathbf{A}\). This can be done by using the formula for the inverse of a 3x3 matrix, which is \(\frac{1}{\text{det}(\mathbf{I} - \mathbf{A})}\) times the adjugate of \(\mathbf{I} - \mathbf{A}\).

Step 4 :First, we calculate the determinant of \(\mathbf{I} - \mathbf{A}\), which is \(0.7(0.7*0.4 - (-0.3)*(-0.5)) - (-0.1)(-0.1*0.4 - (-0.3)*(-0.6)) - (-0.1)(-0.1*0.7 - (-0.1)*(-0.6)) = 0.196\).

Step 5 :Next, we calculate the adjugate of \(\mathbf{I} - \mathbf{A}\). This is done by calculating the cofactor matrix and then taking its transpose. The cofactor matrix is \(\begin{pmatrix} 0.7*0.4 - (-0.3)*(-0.5) & -(-0.1*0.4 - (-0.3)*(-0.6)) & -(-0.1*0.7 - (-0.1)*(-0.6)) \\ -(-0.1*0.4 - (-0.3)*0.4) & 0.7*0.4 - (-0.1)*(-0.6) & -(-0.1*0.7 - (-0.1)*(-0.5)) \\ -(-0.6*0.3 - (-0.1)*(-0.1)) & -(-0.6*0.7 - (-0.1)*(-0.1)) & 0.7*0.7 - (-0.1)*(-0.1) \end{pmatrix} = \begin{pmatrix} 0.43 & 0.02 & 0.06 \\ 0.02 & 0.34 & 0.06 \\ 0.17 & 0.41 & 0.48 \end{pmatrix}\). The transpose of this matrix is \(\begin{pmatrix} 0.43 & 0.02 & 0.17 \\ 0.02 & 0.34 & 0.41 \\ 0.06 & 0.06 & 0.48 \end{pmatrix}\).

Step 6 :Finally, we multiply the adjugate by \(\frac{1}{\text{det}(\mathbf{I} - \mathbf{A})}\) to get the inverse of \(\mathbf{I} - \mathbf{A}\). This gives us \(\begin{pmatrix} 2.19 & 0.10 & 0.87 \\ 0.10 & 1.73 & 2.09 \\ 0.31 & 0.31 & 2.45 \end{pmatrix}\).

Step 7 :So, \((\mathbf{I} - \mathbf{A})^{-1} = \boxed{\begin{pmatrix} 2.19 & 0.10 & 0.87 \\ 0.10 & 1.73 & 2.09 \\ 0.31 & 0.31 & 2.45 \end{pmatrix}}\).

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