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Question
A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was ₹ 19,800 per month at the end of the first month after 3 years of service and ₹ 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method to solve the problem.)
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Solution
Let the man starting the salary be Rs x and his annual increment be Rs y.
Given x + 3y = 19,800
x + 9y = 23,400
The equation can be written as
`[(1, 3),(1, 9)][(x),(y)] = [(19800),(23400)]`
AX = B
X = A-1B
A = `[(1, 3),(1, 9)]`
To find A–1
|A| = 9 – 3
= 6 ≠ 0 A-1 exists.
adj(A) = `[(9, -3),(-1, 1)]`
`"A"^-1 = 1/|"A"|* "adj"("A")`
= `1/6[(9, -3),(-1, 1)]`
X = `"A"^-1"B"`
`[(x),(y)] = 1/6[(9, -3),(-1, 1)][(19800),(23400)]`
= `1/6[(178200 - 70200),(- 19800 + 23400)]`
= `1/6[(10800)/(3600)]`
`[(x),(y)] = [(18000),(3600)]`
⇒ `[(x = 18000),(y = 600)]`
Monthly salary = ₹ 18000
Annual increment = ₹ 1800
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