Question

An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from these investments is ₹ 350. If the total annual income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment using matrix method.

Answer 1

Let the amounts in three investments by ₹ x, ₹ y, and ₹ z respectively.

Then x + y + z = 5000

Since the rate of interest in these investments are 6%, 7%, and 8% respectively, the annual income of the three investments are `"6x"/100, "7y"/100,  "and"  "8z"/100` respectively.

According to the given conditions,

`"6x"/100 + "7y"/100 + "8z"/100 = 350`

i.e. 6x + 7y + 8z = 35000

Also, `"6x"/100 + "7y"/100 = "8z"/100 + 70`

i.e. 6x + 7y - 8z = 7000

Hence, the system of linear equation is

x + y + z = 5000

6x + 7y + 8z = 35000

6x + 7y - 8z = 7000

These equations can be written in matrix form as:

`[(1,1,1),(6,7,8),(6,7,-8)] [("x"),("y"),("z")] = [(5000),(35000),(7000)]`

By R₃ - R₂, we get,

`[(1,1,1),(6,7,8),(0,0,-16)] [("x"),("y"),("z")] = [(5000),(35000),(-28000)]`

By R₂ - 6R₁, we get,

`[(1,1,1),(0,1,2),(0,0,-16)] [("x"),("y"),("z")] = [(5000),(5000),(-28000)]`

∴ `[("x" + "y" + "z"),(0 + "y" + "2z"),(0 + 0 - "16z")] = [(5000),(5000),(-28000)]`

By equality of matrices,

x + y + z = 5000    ...(1)

y + 2z = 5000        ....(2)

- 16z = - 28000      ...(3)

From (3), z = 1750

Substituting z = 1750 in (2), we get, 

y + 2(1750) = 5000

∴ y = 5000 - 3500 = 1500

Substituting y = 1500, z = 1750 in (1), we get,

x + 1500 + 1750 = 5000

∴ x = 5000 - 3250 = 1750

∴ x = 1750, y = 1500, z = 1750

Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.