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Question
The data are about an economy of two industries A and B. The values are in crores of rupees.
| Producer | User | Final demand | Total outout | |
| A | B | |||
| A | 50 | 75 | 75 | 200 |
| B | 100 | 50 | 50 | 200 |
Find the output when the final demand changes to 300 for A and 600 for B.
Solution
a11 = 50, a12 = 75, x1 = 200
a21 = 100, a22 = 50, x2 = 200
`"b"_11 = "a"_11/x_1 = 50/200 = 1/4`, `"b"_12 = "a"_12/x_2 = 75/200 = 3/8`
`"b"_21 = "a"_21/x_1 = 100/200 = 1/2`, `"b"_22 = "a"_22/x_2 = 50/200 = 1/4`
The technology matrix is B = `[(1/4,3/8),(1/2,1/4)]`
I - B = `[(1,0),(0,1)] [(1/4,3/8),(1/2,1/4)]`
`= [(3/4,(-3)/8),(-1/2,3/4)]`
|I - B| = `|(3/4,(-3)/8),(-1/2,3/4)|`
`= [3/4][3/4] - [1/2][3/8]`
`= 9/16 - 3/16 = 6/16 = 3/8 > 0`
Since the diagonal elements of (I - B) are positive and |I - B| is positive, the system is viable
`("I - B")^-1 = 1/|"I - B"|` adj (I - B)
adj (I - B) = `1/(3/8)[(3/4,3/8),(1/2,3/4)]`
adj (I - B) = `8/3 [(3/4,3/8),(1/2,3/4)]`
adj (I - B) = `1/3 [(6,3),(4,6)]`
Now, X = (I - B)-1D Where D = `[(300),(600)]`
∴ X = `1/3[(6,3),(4,6)][(300),(600)]`
`= 1/3[(1800 + 1800),(1200 + 3600)]`
`= 1/3[(3600),(4800)]`
`= [(1200),(1600)]`
∴ The output is 1200 for A and 1600 for B.
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