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Question
The prices of three commodities A, B and C are ₹ x, y and z per units respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B . Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn ₹ 15,000, ₹ 1,000 and ₹ 4,000 respectively. Find the prices per unit of A, B and C. (Use matrix inversion method to solve the problem.)
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Solution
Let x, y, z are commodities of A, B, C
2x – 4y + 5z = 15000 ..........(1)
3x + y – 2z = 1000 ..........(2)
– x + 3y + z = 4000 ..........(3)
Matrix form `[(2, -4, 5),(3, 1, -2),(-1, 3, 1)] [(x),(y),(z)] = [(15000),(1000),(4000)]`
AX = B
X = `"A"^-1"B"`
A = `[(2, -4, 5),(3, 1, -2),(-1, 3, 1)]`
|A| = 2(1 + 6)+ 4(3 – 2) + 5(9 + 1)
= 2(7) + 4(1) + 5(10)
= 14 + 4 + 50
= 68
≠ 0 A-1 exists.
adj A = `[(+(1 + 6), -(3 - 2), +(9 + 1)),(-(-4 - 15), +(2 + 5), -(6 - 4)),(+(8 - 5), -(-4 - 15), +(2 + 12))]^"T"`
= `[(7, -1, 10),(19, 7, -2),(3, 19, 14)]^"T"`
= `[(7, 19, 3),(-1, 7, 19),(10, -2, 14)]`
A–1 = `1/|"A"|`
adj A = `1/68 [(7, 19, 3),(-1, 7, 19),(10, -2, 14)]`
X = `"A"^-1"B"`
`[(x),(y),(z)] = 1/68[(7, 19, 3),(-1, 7, 19),(10, -2, 14)][(15000),(1000),(4000)]`
= `1/68[(105000 + 19000 + 12000),(-15000 + 7000 + 76000),(150000 - 2000 + 56000)]`
= `1/68[(136000),(68000),(204000)]`
`[(x),(y),(z)] = [(2000),(1000),(3000)]`
x = ₹ 2000, y = ₹ 1000, z = ₹ 3000
The prices per unit of A, B, and C are ₹ 2000, ₹ 1000, ₹ 3000
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