Question

A man goes to buy fruits from the market. The shopkeeper informs him that 4 apples, 3 oranges and 2 bananas cost ₹ 60; 2 apples, 4 oranges and 6 bananas cost ₹ 90; whereas 6 apples, 2 oranges and 3 bananas cost ₹ 70. Using matrix method, find the cost of one fruit of each kind.

Answer 1

Let the cost of one apple be x, one orange be x, and one banana be z.

Write the following system of linear equations:

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

This can be written in the matrix form AX = B:

`[(4, 3, 2),(2, 4, 6),(6, 2, 3)] [(x),(y),(z)] = [(60),(90),(70)]`

Find the determinant of matrix A:

|A| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24)

= 4(0) − 3(−30) + 2(−20)

= 0 + 90 − 40

= 50

Since |A| ≠ 0, the system has a unique solution given by X = A⁻¹ B.

The cofactors C_ij are:

C₁₁ = +(12 − 12) = 0, C₁₂ = −(6 − 36), C₁₃ = +(4 − 24) = −20

C₂₁ = −(9 − 4) = −5, C₂₂ = +(12 − 12) = 0, C₂₃ = −(8 − 18) = 10

C₃₁ = +(18 − 8) = 10, C₃₂ = −(24 − 4) = −20, C₃₃ = +(16 − 6) = 10

adj A = `[(0, -5, 10),(30, 0, -20),(-20, 10, 10)]`

Using X = `1/|A|`(adj A) B:

`[(x),(y),(z)] = 1/50 [(0, -5, 10),(30, 0, -20),(-20, 10, 10)] [[(60),(90),(70)]`

= `1/50 [((0 xx 60) + (-5 xx 90) + (10 xx 70)),((30 xx 60) + (0 xx 90) + (-20 xx 70)),((-20 xx 60) + (10 xx 90) + (10 xx 70))]`

= `1/50 [(-450 + 700),(1800 - 1400),(-1200 + 900 + 700)]`

= `1/50 [(250),(400),(400)]`

= `[(5),(8),(8)]`

The cost per fruit is ₹ 5 per apple, ₹ 8 per orange and ₹ 8 per banana.