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)^{-1}D 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.