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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.
Solution
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.
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