Respuesta :
Let the production matrix = X
[tex]X\text{ = }(I-A)^{-1}D[/tex]Next
[tex]\begin{gathered} A\text{ = }\begin{bmatrix}{0.1} & {} & {0.2} \\ {} & {} & {} \\ {0.45} & {} & {0.6}\end{bmatrix} \\ D\text{ = }\begin{bmatrix}{2} & {} & {} \\ {} & {} & {} \\ {4} & & \end{bmatrix} \end{gathered}[/tex][tex]I\text{ = }\begin{bmatrix}{1} & {} & {0} \\ {} & {} & {} \\ {0} & {} & {1}\end{bmatrix}[/tex]Next, evaluate X
[tex]\begin{gathered} I\text{ - A = }\begin{bmatrix}{1} & {} & {0} \\ {} & {} & {} \\ {0} & {} & {1}\end{bmatrix}\text{ - }\begin{bmatrix}{0.1} & {} & {0.2} \\ {} & {} & {} \\ {0.45} & {} & {0.6}\end{bmatrix} \\ =\text{ }\begin{bmatrix}{0.9} & {} & {-0.2} \\ {} & {} & {} \\ {-0.45} & {} & {0.4}\end{bmatrix} \end{gathered}[/tex]Next, you find the inverse of I - A which is adjount of I - A divided by it determinant.
Determinant of (I - A) = 0.9 x 0.4 - (-0.45 x -0.2) = 0.36 - 0.09 = 0.27
Adjount of (I - A) is the transpose of it co-factor
[tex]\begin{gathered} Co-\text{factor of (I - A) = }\begin{bmatrix}{0.4} & {} & {0.45} \\ {} & {} & {} \\ {0.2} & {} & {0.9}\end{bmatrix} \\ \text{Adjount = }\begin{bmatrix}{0.4} & {} & {0.2} \\ {} & {} & {} \\ {0.45} & {} & {0.9}\end{bmatrix} \end{gathered}[/tex]Therefore
[tex]\begin{gathered} (I-A)^{-1}\text{ = }\frac{1}{0.27}\begin{bmatrix}{0.4} & {} & {0.2} \\ {} & {} & {} \\ {0.45} & {} & {0.9}\end{bmatrix} \\ =\text{ }\begin{bmatrix}{1.48} & {} & {0.74} \\ {} & {} & {} \\ {1.67} & {} & {3.33}\end{bmatrix} \end{gathered}[/tex]Now, we find x
[tex]\begin{gathered} X=(I-A)^{-1}D \\ =\text{ }\begin{bmatrix}{1.48} & {} & {0.74} \\ {} & {} & {} \\ {1.67} & {} & {3.33}\end{bmatrix}\text{ X }\begin{bmatrix}{} & {2} & {} \\ {} & {} & {} \\ {} & {4} & {}\end{bmatrix} \\ X=\text{ }\begin{bmatrix}{} & {5.92} & {} \\ {} & {} & {} \\ {} & {16.66} & {}\end{bmatrix} \end{gathered}[/tex][tex]\text{The production matrix = }\begin{bmatrix}{} & {5.92} & {} \\ {} & {} & {} \\ {} & {16.66} & {}\end{bmatrix}[/tex]