Problem 59 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use matrices to solve the system of equations, if possible. Use Gaussian elimination with back-substitution. $$\left\\{\begin{array}{l} -x+y=-22 \\ 3 x+4 y=4 \\ 4 x-8 y=32 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -18\) and \(y = 4\)

Step 1: Conversion to Augmented Matrix

The first step is to convert the system of equations into an augmented matrix. This results in the following matrix:\[\begin{bmatrix} -1 & 1 & | & -22 \\ 3 & 4 & | & 4 \\ 4 & -8 & | & 32 \end{bmatrix}\]

Step 2: Gaussian Elimination to Row-Echelon Form

Next, the augmented matrix is transformed into row-echelon form using the following row operations: - Swap Row 1 and Row 2- Multiply Row 1 by -1- Add 3 times Row 1 to Row 3This results in:\[\begin{bmatrix} 1 & 1 & | & -22 \\ -1 & 4 & | & 4 \\ 0 & 1 & | & -30 \end{bmatrix}\] Then, add Row 1 to Row 2 to yield the following matrix in row-echelon form: \[\begin{bmatrix} 1 & -1 & | & -22 \\ 0 & 3 & | & -18 \\ 0 & 1 & | & -30 \end{bmatrix}\]

Step 3: Reduced Row-Echelon Form and Back-Substitution

Now, subtract Row 3 from Row 2 then multiply by 1/3 to obtain the reduced row-echelon form. \[\begin{bmatrix} 1 & -1 & | & -22 \\ 0 & 1 & | & 4 \\ 0 & 0 & | & 0 \end{bmatrix}\] Reading off the matrix, we obtain the solution as \(x = -22 + y\) and \(y = 4\). Substitute \(y = 4\) into the equation for \(x\), we get \(x = -18\).

Key Concepts

Matrices

Matrices are an essential component when solving systems of equations through methods like Gaussian Elimination. They are simply rectangular arrays of numbers or expressions arranged in rows and columns. Each element in a matrix can represent a coefficient from an equation in a system. By converting a system of equations into a matrix form, we structure the problem in a way that makes it easier to apply systematic elimination methods. This converted form, known as the augmented matrix, includes the coefficients of variables and the constants from the equations. It's initially daunting but remember that a matrix is just another way of organizing information efficiently.

Systems of Equations

A system of equations is a set of multiple equations containing multiple variables that you want to solve simultaneously. For example, the given system involves three equations with two variables, which can be quite complex to solve by substitution or elimination manually. By employing matrices and Gaussian Elimination, we simplify the manipulation of these equations. The goal is to find the variable values that satisfy all equations in the system simultaneously. Think of the system as a mathematical puzzle where each equation is a piece that must fit within the larger solution framework formed by all the equations together.

Back-Substitution

Back-substitution is a crucial step in the process of solving systems of equations that have been transformed into row-echelon form or reduced row-echelon form using Gaussian Elimination. Once the matrix is in this simpler form, it usually resembles an upper triangular matrix. The idea is to start solving from the bottommost equation upwards. You find the value of one variable and then substitute it back into the other equations. For example, if you have determined the value of \( y \), you substitute \( y \) back into another equation to find \( x \). This step-by-step method provides a clear path to discovering the values of all the unknowns in the system.

Row-Echelon Form

Achieving row-echelon form is key to making systems of equations easier to solve. In this form, all elements below the main diagonal are zeros, allowing each equation to represent a simpler linear system. We use row operations to transform matrices step-by-step into this form. These operations can include swapping rows, multiplying a row by a constant, or adding and subtracting rows from one another. The ultimate goal is to simplify the matrix to a stage where back-substitution becomes straightforward. By isolating variables at each step, solving these equations becomes much more manageable.