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Question
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.
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Solution
Let the price of onion be Rs. per kg = x
The price of wheat is Rs. per kg = y
The price of rice be Rs. per kg = z
Then according to the given conditions,
4x + 3y + 2z = 60;
2x + 4y + 6z = 90;
6x + 2y + 3z = 70
This system of equations can be written as AX = B
`[(4,3,2),(2,4,6),(6,2,3)] [(x),(y),(z)] = [(60),(90),(70)]`
A = `[(4,3,2),(2,4,6),(6,2,3)]`, X = `[(x),(y),(z)]`, B = `[(60),(90),(70)]`
|A| = `|(4,3,2),(2,4,6),(6,2,3)|`
= 4(12 − 12) − 3(2 × 3 − 6 × 6) + 2(2 × 2 − 6 × 4)
= 0 + 90 − 40
= 50 ≠ 0
∴ A−1 can be found,
Cofactors of the elements of |A|:
A11 = 0, A12 = 30, A13 = −20
A21 = −5, A22 = 0, A23 = 10
A31 = 10, A32 = −20, A33 = 10
∴ adj A = `[(0,30,-20),(-5,0,10),(10,-20,10)] = [(0,-5,10),(30,0,-20),(-20,10,10)]`
A−1 = `1/|A|` adj A
= `1/50 [(0,-5,10),(30,0,-20),(-20,10,10)]`
AX = B
⇒ X = A−1B
∴ `[(x),(y),(z)] = 1/50 [(0,-5,10),(30,0,-20),(-20,10,10)] [(60),(90),(70)]`
= `1/50 [(0 - 450 + 700),(1800 + 0 - 1400),(-1200 + 900 + 700)]`
= `1/50 [(250),(400),(400)]`
= `[(5),(8),(8)]`
⇒ x = 5, y = 8, z = 8
Hence, the cost of 1 kg onion = Rs. 5
Price of 1 kg wheat = Rs. 8
Price of 1 kg rice = Rs. 8
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