Two commodities A and B are produced such that 0.4 tonne of A and 0.7 tonne of B are required to produce a tonne of A. Similarly 0.1 tonne of A and 0.7 tonne of B are needed to produce a tonne of B. Write down the technology matrix. If 68 tonnes of A and 10.2 tonnes of B are required, find the gross production of both of them.
Solution
Here the technology matrix is given under
| A | B | Final demand | |
| A | 0.4 | 0.1 | 6.8 |
| B | 0.7 | 0.7 | 10.2 |
The technology matrix is B = `[(0.4,0.1),(0.7,0.7)]`
I - B = `[(1,0),(0,1)] - [(0.4,0.1),(0.7,0.7)]`
`= [(0.6,-0.1),(-0.7,0.3)]`
|I - B| = `|(0.6,-0.1),(-0.7,0.3)|`
= (0.6) (0.3) – (-0.1) (-0.7)
= 0.18 – 0.07
= 0.11
Since the main diagonal elements of I – B are positive and the value of |I – B| is positive, the system is viable.
adj (I - B) = `[(0.3,0.1),(0.7,0.6)]`
(I - B)-1 = `1/|"I" - "B"|` adj (I - B)
`= 1/0.11 [(0.3,0.1),(0.7,0.6)]`
X = (I - B)-1D where D = `[(6.8),(10.2)]`
`= 1/0.11 [(0.3,0.1),(0.7,0.6)][(6.8),(10.2)]`
`= 1/0.11[(2.04 + 1.02),(4.76 + 6.12)]`
`= [(27.8181),(98.9090)] = [(27.82),(98.91)]`
Production of A is 27.82 tonnes and the production of B is 98.91 tonnes.
