Tamil Nadu Board of Secondary EducationHSC Commerce Class 11th

# 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. - Business Mathematics and Statistics

Sum

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.

Concept: Input–Output Analysis
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