Page 110 - DMTH201_Basic Mathematics-1

Unit 4: Determinants

Notes
Example
For the following market conditions, find the equilibrium quantities and prices by using matrix
inverse method.
D
S
Q = 45 – 2P + 3P – 7P , Q = –5 + 4P 1
1
3
1
1
2
D
S
Q = 16 + 2P – P + 3P , Q = –19 + 5P 2
2
1
3
2
2
S
D
Q = 30 – P + 2P – 8P , Q = –6 + 2P 3
2
1
3
3
3
Solution:
Using the equilibrium condition Q D Q S , we can write
1 1
45 – 2P + 3P – 7P = –5 + 4P
1 2 3 1
or –6P + 3P – 7P = –50
1 2 3
or 6P – 3P + 7P = 50 ...(1)
1 2 3
D
Similarly Q 2 = Q 2 S
16 + 2P – P + 3P = –19 + 5P
1 2 3 2
or 2P – 6P + 3P = –35 ...(2)
1 2 3
D
and Q 3 = Q 3 S
30 – P + 2P – 8P = –6 + 2P
1 2 3 3
or –P + 2P – 10P = –36
1 2 3
or P – 2P + 10P = 36 ...(3)
1 2 3
The system of equations given by (1), (2) and (3) can be written as the matrix equation
6 3 7 P 50 6 3 7
1
2 6 3 P = 35 . Let A 2 6 3
2
1 2 10 P 3 36 1 2 10
|A| = –360 – 9 – 28 + 42 + 36 + 60 = –259.
Since |A| ¹ 0, the solution is unique. Writing the matrix of cofactors as
54 17 2 54 16 33
16 53 9 1
–1
C = A = 17 53 4
33 4 30 259 2 9 30
P 54 16 33 50
1 1
Thus, P 2 = 17 53 4 35
P 259 2 9 30 36
3
50 54 35 16 36 33
Hence P = 8
1 259
50 17 35 53 36 4
P = 11
259
2

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