Problem 53 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{rr} -x+2 y= & 3 \\ 2 x-4 y= & -6 \end{array}$$

Step-by-Step Solution

Answer

The system has infinite solutions: \(x = 2t - 3\), \(y = t\) for any real number \(t\).

Step 1: Write the System in Augmented Matrix Form

First, write the given system of equations as an augmented matrix. The system is:\[\begin{array}{rr} -x+2y= & 3 \ 2x-4y= & -6 \end{array}\]The corresponding augmented matrix is:\[\begin{bmatrix}-1 & 2 & | & 3 \2 & -4 & | & -6 \end{bmatrix}\]

Step 2: Forward Elimination to Achieve Row Echelon Form

Use row operations to transform the matrix into row echelon form. First, add two times row 1 (\(R_1\)) to row 2 (\(R_2\)) to eliminate the \(-1\) in the first column of the second row:\[R_2 = R_2 + 2R_1 \]\[\begin{bmatrix}-1 & 2 & | & 3 \0 & 0 & | & 0 \end{bmatrix}\]Now the matrix is in row echelon form.

Step 3: Interpret the Result

The system now appears as two equations:\[-x + 2y = 3 \]\[0 = 0 \]The second row, \(0 = 0\), is a tautology, indicating that this row can be ignored, and the system has an infinite number of solutions depending on one parameter.

Step 4: Solve for the Free Variable

Let's assign \(y = t\) where \(t\) is a parameter, then substitute it into the first equation:\[-x + 2t = 3 \]Solve for \(x\):\[x = 2t - 3 \]This represents the solution set for the system in terms of a parameter \(t\).

Key Concepts

System of Linear Equations

A system of linear equations is simply a set of linear equations that you work with at the same time. These equations have the same variables, and your goal is to find a common solution. This means finding values for the variables that satisfy all the equations simultaneously. In the given exercise, we work with a system of two equations:\[\begin{array}{rr} -x+2y &= 3 \ 2x-4y &= -6 \end{array}\]Here, the task is to solve for the variables \(x\) and \(y\). By utilizing Gaussian elimination, the system can be simplified, making it easier to identify solutions that satisfy both equations at once.

Augmented Matrix

The augmented matrix is a convenient tool for solving systems of linear equations. It combines both the coefficients of the variables and the constants from each equation into a rectangular array. This approach simplifies calculations and helps in applying row operations. For the system given in the exercise, the augmented matrix is constructed as follows:\[\begin{bmatrix}-1 & 2 & | & 3 \2 & -4 & | & -6 \end{bmatrix}\]Here, each row corresponds to an equation, and each column corresponds to a coefficient of the respective variables or the constant on the right side. The bar '|' separates the coefficients from the constants, allowing an easy view of the system that we are working with.

Row Echelon Form

Row echelon form is a special arrangement for matrices that makes it easier to solve a system of equations. The main characteristic is that it has a staircase-like pattern of zeros. This form allows for straightforward back-substitution to solve the system. To reach row echelon form using Gaussian elimination, we apply a series of row operations on the augmented matrix:- We aim to have ones (leading coefficients) and zero out the entries below.- This requires adding, subtracting, or multiplying rows by constants.In our example, we transform the matrix to:\[\begin{bmatrix}-1 & 2 & | & 3 \0 & 0 & | & 0 \end{bmatrix}\]The second row becomes \(0 = 0\), indicating no further information is provided for checking solutions. This suggests an infinite number of solutions, typically involving a free parameter. Consequently, we can express our variables in terms of this parameter to find the solution set.