Solve using matrices. Investments. Elena receives 112...

Solve using matrices.
Investments. Elena receives 112 dollars per year in simple interest from three investments totaling 2500 dollars. Part is invested at 3%, part at 4%, and part at 5%. There is $1100 more invested at 5% than at 4%. Find the amount invested at each rate.

Key Concepts

System of Linear Equations

This concept involves setting up multiple equations that describe the relationships between different unknown quantities. In many applied problems, such as investment or mixture problems, several conditions are given that translate naturally into equations. The goal is to find values for the variables that satisfy all the equations simultaneously.

Matrix Representation

Matrices provide a compact and systematic way to represent and solve systems of linear equations. By organizing coefficients into an array format, one can apply various matrix operations, such as finding an inverse or performing Gaussian elimination, to efficiently solve for the unknowns.

Solving Systems Using Inverse Matrices

When a system of linear equations is represented in matrix form as Ax = b, if the matrix A is invertible, the unique solution can be found using the formula x = A?¹b. This method is particularly useful in problems with multiple equations and variables, providing a straightforward algorithmic approach.

Simple Interest

Simple interest is calculated only on the initial principal, and it provides a linear relationship between the principal amount, the interest rate, and the interest earned. Understanding this concept is important for formulating equations that relate different investments and their respective yields.

Constraint Equations in Applied Problems

Many practical problems include additional conditions or constraints that link the variables, such as a specified difference between amounts or a total sum. Recognizing and incorporating these constraints into the system of equations is crucial for accurately modeling and solving the problem.

Transcript

00:01 For this question, let us assume that the money invested at 3 % is x.

00:15 Similarly, the money invested at 4%, let it be y, and let us call the money invested at 5 % as z.

00:26 Now, from the data given in the question, we can write the following equations.

00:31 The first thing we can write, we know the total money is $2 ,500.

00:35 So we can write x plus y plus z is equal to 2500.

00:41 Now we also know the total interest earned, so we can write 3 % of x or 0 .03x plus 0 .04y, that is 4 % of y, similarly plus 0 .05 z is equal to 112.

01:00 Now we also know that the money invested at 5%, that is, is $1 ,100 greater than the money invested at 4%, that is y.

01:14 On rearranging we can write y minus z is equal to minus 110.

01:23 So now from these equations we can write them as a matrix.

01:28 So we write in matrix form we can write those equations as 1 -1 -1 and then the constant 2500 then 0 .03, 0 .04 and 0 .05 and here we have 112.

01:47 Finally, the coefficient of x here is 0 and the coefficient of y and z are 1 and minus 1 is minus 1100.

01:58 We'll draw our dotted line to separate the coefficients and the constant terms.

02:04 Next we have to do a few row operations on this matrix to make it into a row echelon form.

02:10 So the first row operation we will do is row 2 prime, that is new row 2 is equal to existing row 2 minus 0 .03 r1...

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