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Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 6, Problem 29

Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with infinitely many solutions, write the solution with y arbitrary. For systems in three variables with infinitely many solutions, write the solution set with z arbitrary.
(3/8)x - (1/2)y = 7/8
-6x + 8y = -14

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1
Write the system of equations in matrix form as an augmented matrix. For the system: \(\frac{3}{8}x - \frac{1}{2}y = \frac{7}{8}\) and \(-6x + 8y = -14\), the augmented matrix is: \[\left[\begin{array}{cc|c} \frac{3}{8} & -\frac{1}{2} & \frac{7}{8} \\ -6 & 8 & -14 \end{array}\right]\]
Use row operations to create a leading 1 in the first row, first column. You can multiply the first row by the reciprocal of \(\frac{3}{8}\), which is \(\frac{8}{3}\), to make the coefficient of \(x\) equal to 1.
Next, eliminate the \(x\)-term in the second row by adding a suitable multiple of the first row to the second row. Specifically, multiply the first row by 6 (to match the \(-6\) in the second row) and add it to the second row to make the \(x\) coefficient in the second row zero.
Then, create a leading 1 in the second row, second column (the \(y\) coefficient) by dividing the entire second row by the coefficient of \(y\) in that row after elimination.
Finally, eliminate the \(y\)-term in the first row by adding or subtracting a suitable multiple of the second row from the first row to get zeros above and below the leading 1s, resulting in the reduced row echelon form. From this form, write the solution for \(x\) and \(y\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gauss-Jordan Elimination Method

Gauss-Jordan elimination is a systematic procedure to solve systems of linear equations by transforming the augmented matrix into reduced row-echelon form. This method uses row operations to simplify the matrix, making it easier to identify solutions or determine if there are infinitely many or no solutions.
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Types of Solutions for Linear Systems

A system of linear equations can have a unique solution, infinitely many solutions, or no solution. Infinitely many solutions occur when equations are dependent, leading to free variables. Recognizing these cases is essential to express the solution set correctly, often using arbitrary parameters like y or z.
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Expressing Solutions with Arbitrary Variables

When a system has infinitely many solutions, one or more variables are free and can take any value. To describe the solution set, assign an arbitrary parameter (e.g., y or z) to these free variables and express the other variables in terms of this parameter, providing a complete and clear description of all solutions.
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