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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 7, Problem 3

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {5x+8y6z=143x+4y2z=8x+2y2z=3\(\begin{cases}\)5x + 8y - 6z = 14 \\3x + 4y - 2z = 8 \(\x\) + 2y - 2z = 3\(\end{cases}\)

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Write the system of equations as an augmented matrix: \[\left[ \begin{array}{ccc|c} 5 & 8 & -6 & 14 \\ 3 & 4 & -2 & 8 \\ 1 & 2 & -2 & 3 \end{array} \right]\]
Use the first row to eliminate the \(x\)-terms in the second and third rows. For the second row, replace it with (Row 2) - \(\frac{3}{5}\) (Row 1). For the third row, replace it with (Row 3) - \(\frac{1}{5}\) (Row 1).
After these operations, focus on the second row to eliminate the \(y\)-term in the third row. Use the new second row to replace the third row with (Row 3) - (appropriate multiple) \(\times\) (Row 2) to create a zero in the \(y\)-position of the third row.
At this point, the matrix should be in upper triangular form. Use back substitution starting from the third row to express \(z\) in terms of constants, then substitute back into the second row to find \(y\), and finally substitute \(y\) and \(z\) into the first row to find \(x\).
Write the solution as an ordered triple \((x, y, z)\) representing the values found through back substitution, which is the complete solution to the system.

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Key Concepts

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

Systems of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The goal is to find values for the variables that satisfy all equations simultaneously. Understanding how to represent and interpret these systems is fundamental before applying solution methods.
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Introduction to Systems of Linear Equations

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations by transforming the system's augmented matrix into row-echelon form using row operations. This process simplifies the system, making it easier to solve through back-substitution or to determine if no solution exists.
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Row Operations and Row-Echelon Form

Row operations include swapping rows, multiplying a row by a nonzero scalar, and adding multiples of one row to another. These operations are used to convert the augmented matrix into row-echelon form, where the matrix has a staircase pattern of leading coefficients, facilitating the solution of the system.
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