You are given the following transaction matrix for a two-sector economy.
Sector | Sales | Final demand |
Gross output |
|
1 | 2 | |||
1 | 4 | 3 | 13 | 20 |
2 | 5 | 4 | 3 | 12 |
- Write the technology matrix
- Determine the output when the final demand for the output sector 1 alone increases to 23 units.
Solution
a11 = 4, a12 = 3, x1 = 20
a21 = 5, a22 = 4, x2 = 12
`"b"_11 = "a"_11/x_1 = 4/20 = 1/5`, `"b"_12 = "a"_12/x_2 = 3/12 = 1/4`
`"b"_21 = "a"_21/x_1 = 5/20 = 1/4`, `"b"_22 = "a"_22/x_2 = 4/12 = 1/3`
The technology matrix is B = `[(1/5,1/4),(1/4,1/3)]`
I - B = `[(1,0),(0,1)] - [(1/5,1/4),(1/4,1/3)]`
`= [(4/5,-1/4),(-1/4,2/3)]`, elements of main diagonal are positive
|I - B| = `4/5xx2/3-(-1/4)xx(-1/4)`
`= 8/15 - 1/16 = (8 xx 16 - 1 xx 15)/(15 xx 16)`
`= (128 - 15)/(15 xx 16) = 113/240`
The main diagonal elements are positive and |I – B| is positive. Therefore the system is viable.
adj(I - B) = `[(2/3,1/4),(1/4,4/5)]`
`("I - B")^-1 = 1/|"I - B"|` adj (I - B)
`= 1/(113/240) [(2/3,1/4),(1/4,4/5)] = 240/113[(2/3,1/4),(1/4,4/5)]`
X = (I – B)-1D, where
D = `(16 xx 15)/113 [(2/3,1/4),(1/4,4/5)] [(23),(3)]`
`=> 1/113 [(16xx15xx2/3,16xx15xx1/4),(16xx15xx1/4,16xx15xx4/5)] [(23),(3)]`
`= 1/113 [(16xx5xx2,4xx15xx1),(4xx15xx1,16xx3xx4)] [(23),(3)]`
`=> 1/113 [(160,60),(60,192)] [(23),(3)]`
`= 1/113 [(160xx23+60xx3),(60xx23+192xx3)]`
`= 1/113 [(3680+180),(1380+576)]`
`= 1/113 [(3860),(1956)] = [(34.159),(17.3097)]`
X = `[(34.16),(17.31)]`
The output of sector 1 should be 34.16 and sector 2 should be 17.31.