An investor deposited Rs. 100000 in a saving bank. Part of the money is invested at a yearly rate of 5% and the remaining at an annual rate of 12%. At the end of the year, he recei... An investor deposited Rs. 100000 in a saving bank. Part of the money is invested at a yearly rate of 5% and the remaining at an annual rate of 12%. At the end of the year, he received an annual interest of Rs. 11,600. Using Matrix method, find out how much he deposited at 5% half-yearly rate?
Understand the Problem
The question is asking to determine how much money was deposited at a half-yearly rate of 5% using the Matrix method. It specifies that the investor deposited Rs. 100,000, part of which is invested at a rate of 5% and the remainder at a rate of 12%, receiving an annual interest of Rs. 11,600 by the end of the year.
Answer
The amount deposited at 5% half-yearly rate is approximately Rs. 5,714.29.
Answer for screen readers
The amount deposited at 5% half-yearly rate is approximately Rs. 5,714.29.
Steps to Solve
- Define the Variables
Let ( x ) be the amount deposited at 5% per annum, and ( y ) be the amount deposited at 12% per annum.
- Set Up the Equations
From the problem, we know two things:
-
The total deposit is Rs. 100,000: $$ x + y = 100000 $$
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The total interest earned is Rs. 11,600: $$ 0.05x + 0.12y = 11600 $$
- Express the System of Equations in Matrix Form
We can express the above equations in matrix form: $$ \begin{bmatrix} 1 & 1 \ 0.05 & 0.12 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}
\begin{bmatrix} 100000 \ 11600 \end{bmatrix} $$
- Calculate the Inverse of the Coefficient Matrix
First, we find the determinant of the coefficient matrix: $$ \text{det} = (1)(0.12) - (1)(0.05) = 0.12 - 0.05 = 0.07 $$
Now, find the inverse of the coefficient matrix: $$ \begin{bmatrix} 1 & 1 \ 0.05 & 0.12 \end{bmatrix}^{-1}
\frac{1}{0.07} \begin{bmatrix} 0.12 & -1 \ -0.05 & 1 \end{bmatrix} $$
- Multiply the Inverse Matrix by the Constants Matrix
Now we multiply the inverse of the coefficient matrix by the constants matrix: $$ \begin{bmatrix} x \ y \end{bmatrix}
\frac{1}{0.07} \begin{bmatrix} 0.12 & -1 \ -0.05 & 1 \end{bmatrix} \begin{bmatrix} 100000 \ 11600 \end{bmatrix} $$
- Perform the Multiplication
Calculate the product: $$ \begin{bmatrix} 0.12 \cdot 100000 - 1 \cdot 11600 \ -0.05 \cdot 100000 + 1 \cdot 11600 \end{bmatrix}
\begin{bmatrix} 12000 - 11600 \ -5000 + 11600 \end{bmatrix}
\begin{bmatrix} 400 \ 6600 \end{bmatrix} $$
- Find ( x ) and ( y )
Thus, $$ x = \frac{400}{0.07} \approx 5714.29 $$ $$ y = \frac{6600}{0.07} \approx 94285.71 $$
- Identify the Amount at 5%
Since ( x ) represents the amount deposited at 5%, the final answer can be concluded.
The amount deposited at 5% half-yearly rate is approximately Rs. 5,714.29.
More Information
This problem utilizes the matrix method to solve a system of linear equations. It demonstrates how to set up and solve a real-world financial scenario using matrices, which is a valuable technique in linear algebra.
Tips
- Not setting up the equations correctly with the right coefficients.
- Forgetting to calculate the determinant before finding the inverse of the matrix.
- Miscalculating the multiplication of matrices, especially when transferring the operations to constants.
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