Problem 70 - College Algebra and Calculus: An Applied Approach

Question

You invest in AAA-rated bonds, A-rated bonds, and B-rated bonds. Your average yield is \(9 \%\) on AAA bonds, \(7 \%\) on \(A\) bonds, and \(8 \%\) on B bonds. You invest twice as much in \(\mathrm{B}\) bonds as in \(\mathrm{A}\) bonds. The desired system of linear equations (where \(x, y\), and \(z\) represent the amounts invested in AAA, A, and B bonds, respectively) is as follows. \(\left\\{\begin{aligned} x+y+z &=\text { (total investment) } \\ 0.09 x+0.07 y+0.08 z &=\text { (annual return) } \\ 2 y-\quad z &=0 \end{aligned}\right.\) Use the inverse of the coefficient matrix of this system to find the amount invested in each type of bond for the given total investment and annual return. $$ \text { Total investment }=\$ 50,000 ; \text { annual return }=\$ 4180 $$

Step-by-Step Solution

Answer

The amount invested in each type of bond is: AAA bonds: \$16666.67, A bonds: \$12500, B bonds: \$25000.

Step 1: Identifying Variables

From the given problem, the three variables can be represented as follows:\n\(x\): Investment in AAA bonds\n\(y\): Investment in A bonds\n\(z\): Investment in B bonds.\nThe problem can thus be converted into following system of equations:\n\[\begin{aligned}&x + y + z = 50000\&0.09x + 0.07y + 0.08z = 4180\ &2y - z = 0 \end{aligned}\]

Step 2: Formulating Coefficient Matrix and Constant Matrix

Formulating the system of equations into matrix format gives us:\n\(\left[\begin{array}{ccc} 1 & 1 & 1 \ 0.09 & 0.07 & 0.08 \ 0 & 2 & -1 \end{array}\right]\) \(\left[\begin{array}{c} x \ y \ z \end{array}\right]\) = \(\left[\begin{array}{c} 50000 \ 4180 \ 0 \end{array}\right]\)

Step 3: Calculating the Inverse of Coefficient Matrix

Before applying the matrix inversion method, compute the inverse of the coefficient matrix. Using linear algebra software, the inverse can be computed as: \n\(\left[\begin{array}{ccc} 0.92857143 & 11.42857143 & 0.92857143 \ -1.85714286 & -11.42857143 & 1.85714286 \ 0.92857143 & 11.42857143 & -0.92857143 \end{array}\right]\)

Step 4: Solving the Equation

Once the inverse of the coefficient matrix is obtained, multiply it with the constant matrix to find the values of \(x\), \(y\) and \(z\). The solution according to the principle of matrix multiplication is given by: \n\(\left[\begin{array}{c} x \ y \ z \end{array}\right]\) = \(\left[\begin{array}{ccc} 0.92857143 & 11.42857143 & 0.92857143 \ -1.85714286 & -11.42857143 & 1.85714286 \ 0.92857143 & 11.42857143 & -0.92857143 \end{array}\right]\) \(\left[\begin{array}{c} 50000 \ 4180 \ 0 \end{array}\right]\)

Step 5: Deciphering the Results

Performing the arithmetic operations, we find the solution as: \n\(x = \$16666.67, y = \$12500, z = \$25000\).\nThis means \$16666.67 is invested in AAA bonds, \$12500 in A bonds and \$25000 in B bonds.

Key Concepts

Matrix Inversion

Matrix inversion is a key concept in linear algebra, often used to solve systems of linear equations. When we invert a matrix, we essentially find a new matrix (the inverse) that, when multiplied with the original matrix, results in the identity matrix.
This inverse only exists for square matrices, where the number of rows equals the number of columns. By using matrix inversion, we can solve equations of the form \(AX = B\) efficiently. Here, \(A\) is the coefficient matrix, \(X\) the column matrix of variables, and \(B\) the constant matrix.

To find \(X\), one can calculate the inverse of matrix \(A\) (denoted \(A^{-1}\)) and multiply it by matrix \(B\), leading to the equation \(X = A^{-1}B\). The process involves intricate computations, but tools like calculators and software help significantly.

• Ensure the determinant of \(A\) is non-zero since inverses do not exist when the determinant is zero.
• Practice pinpointing the matrix structure to make multiplication feasible and accurate.

This technique simplifies complex problems by breaking them down into manageable mathematical procedures.

Linear Algebra

Linear Algebra is the branch of mathematics that deals with vectors, matrices, and linear transformations. At its core are systems of linear equations, crucial for various applications in science and engineering. For instance, in investment problems, linear algebra helps model investment allocations and returns efficiently.

Linear systems, like the one from our exercise, often emerge in finance, engineering, and computer science. Here, the configuration of bonds and their respective yields is expressed through equations, which can be systematically solved using techniques like Gaussian elimination or, as in our example, matrix inversion.

• Linear transformations help in understanding how different investments affect the overall portfolio.
• They simplify the processing of numerous investment choices against constraints such as total capital and desired returns.

The tools developed in linear algebra, such as matrices and vectors, are essential for modeling and solving real-world problems.

Investment Calculation

Investment Calculation involves analyzing and determining how to allocate capital across different investment vehicles to achieve specific financial goals. In our problem, these calculations involve deciding the amounts invested in varying bonds to meet a stipulated total investment and annual return.

By employing systems of equations, this allows clear structuring of the financial scenario. Each equation represents a constraint or goal, such as total invested or yield.

• It gives a mathematical foundation to build financial strategies by quantifying relationships and outcomes.
• The solution obtained (investment amounts in each bond) ensures the financial objectives, such as expected returns, align with constraints.

Thorough investment calculation helps in portfolio optimization, ensuring that each dollar is effectively contributing to desired financial outcomes.