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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 cost of each item per kg by matrix method.
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)]`
`abs A = [(4,3,2),(2,4,6),(6,2,3)]`
`= 4(12 - 12) - 3(2 xx 3 - 6 xx 6) + 2 (2 xx 2 - 6 xx 4)`
`= 0 + 90 - 40 = 50 ne 0`
`therefore A^-1` can be found
Cofactors of the elements of `abs A`
`A_11 = 0, A_12 = 30, A_13 = - 20`
`A_21 = - 5, A_22 = 0, A_23 = 10`
`A_31 = 10, A_32 = - 20, A_33 = 10`
`therefore adj A = [(0,30,-20),(-5,0,10),(10,-20,10)] = [(0,-5,10),(30,0,-20),(-20,10,10)]`
`A^-1 = adj A/abs A = 1/50 [(0,-5,10),(30,0,-20),(-20,10,10)]`
AX = B ⇒ X = A-1 B
`therefore [(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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