Solve using techniques of matrix algebra
A manufacturing company Ltd. Produces two types of drugs D1 and D2 with the help of three chemicals C1 , C2 and C3 . The quantity of chemical requirements per kg of each D1 and D2 are given below in suitable units.
| D1 | D2 | |
| C1 | 10 | 12 |
| C2 | 16 | 15 |
| C3 | 8 | 16 |
The price of chemicals in three different markets M1, M2 and M3 are as follows:
| C1 | C2 | C3 | |
| M1 | 20 | 19 | 16 |
| M2 | 8 | 9 | 7 |
| M3 | 6 | 7 | 8 |
Find the price per kg of each drug in each market by using matrix algebra.
Solution
Let matrix A represents prices of chemical in different market.
A =
20
8
6
19
9
7
16
7
8
20
8
6
19
9
7
16
7
8
Let matrix B represents quantity of chemical required for 1 kd of each drugs.
B =
10
16
8
12
15
16
10
16
8
12
15
16
Multiplication of A and B gives price per kg of each drugs in each market.
=
20
8
6
19
9
7
16
7
8
×
10
16
8
12
15
16
=
20×10+19×16+16×8
8×10+9×16+7×8
6×10+7×16+8×8
20×12+19×15+16×16
8×12+9×15+7×16
6×12+7×15+8×16
=
200+304+128
80+144+56
60+112+64
240+285+256
96+135+112
72+105+128
=
632
280
236
781
343
305
20
8
6
19
9
7
16
7
8
×
10
16
8
12
15
16
=
20×10+19×16+16×8
8×10+9×16+7×8
6×10+7×16+8×8
20×12+19×15+16×16
8×12+9×15+7×16
6×12+7×15+8×16
=
200+304+128
80+144+56
60+112+64
240+285+256
96+135+112
72+105+128
=
632
280
236
781
343
305
| D1 | D2 | |
| M1 | 632 | 781 |
| M2 | 280 | 343 |
| M3 | 236 | 305 |
The per kg price of drug D1 in Market M1, M2 and M3 are 632, 280 and 236 respectively.
The per kg price of drug D2 in Market M1, M2 and M3 are 781, 343 and 305 respectively.
The per kg price of drug D2 in Market M1, M2 and M3 are 781, 343 and 305 respectively.
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