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.)

Answer 1

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