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.

Answer 1

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