11.7 Solving systems with inverses

Nancy plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem?

There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.

Finding the inverse of a matrix

We know that the multiplicative inverse of a real number $a$ is $a^{\mathrm{-1}},$ and $a{a}^{\mathrm{-1}}=a^{\mathrm{-1}}a=\left(\frac{1}{a}\right)a=1.$ For example, $2^{\mathrm{-1}}=\frac{1}{2}$ and $\left(\frac{1}{2}\right)2=1.$ The multiplicative inverse of a matrix    is similar in concept, except that the product of matrix $A$ and its inverse $A^{\mathrm{-1}}$ equals the identity matrix    . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by $I_n$ where $n$ represents the dimension of the matrix. [link] and [link] are the identity matrices for a $2\times 2$ matrix and a $3\times 3$ matrix, respectively.

The identity matrix acts as a 1 in matrix algebra. For example, $AI=IA=A.$

A matrix that has a multiplicative inverse has the properties

A matrix that has a multiplicative inverse is called an invertible matrix . Only a square matrix may have a multiplicative inverse, as the reversibility, $A{A}^{\mathrm{-1}}=A^{\mathrm{-1}}A=I,$ is a requirement. Not all square matrices have an inverse, but if $A$ is invertible, then $A^{\mathrm{-1}}$ is unique. We will look at two methods for finding the inverse of a $2\times 2$ matrix and a third method that can be used on both $2\times 2$ and $3\times 3$ matrices.