Explanation Approach: A matrix is said to be an augmented matrix if it consists of a coefficient matrix and column matrix of constants separated by a vertical line. The solution of a system of linear equations, using the Gauss Jordan elimination method, can be found using the following steps: 1. From the linear system of equations, write the augmented matrix. 2. Apply the operations on the first row, such that first entry of the first row is non-zero and pivot the matrix about this entry. 3. Apply the operations of the second row such that the second entry of the second row is non zero and pivot the matrix about this entry. 4. Repeat the same process for further rows, until the row-reduced form of augmented matrix is obtained. Calculation: The augmented matrix for the system of given equations is given by, [ 2 − 1 1 3 − 3 2 3 2 − 6 3 − 3 | − 4 − 6 12 ] Multiply row 2 by 2 in the above matrix. [ 2 − 1 1 6 − 3 3 − 6 3 − 3 | − 4 − 12 12 ] Add row 2 and row 3 in the above matrix. [ 2 − 1 1 6 − 3 3 0 0 0 | − 4 − 12 0 ] Multiply row 1 by 6 and subtract it from row 2 in the above matrix

Finite Mathematics for the Managerial, Life, and Social Sciences

12th Edition

ISBN: 9781337405782

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 2.3, Problem 26E

To determine

To find:

The solution of given system of linear equations, using the Gauss-Jordan elimination method.

2x1−x2+x3=−43x1−32x2+32x3=−6−6x1+3x2−3x3=12

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Chapter 2 Solutions

Finite Mathematics for the Managerial, Life, and Social Sciences

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The solution of given system of linear equations, using the Gauss-Jordan elimination method. 2 x 1 − x 2 + x 3 = − 4 3 x 1 − 3 2 x 2 + 3 2 x 3 = − 6 − 6 x 1 + 3 x 2 − 3 x 3 = 12

Solution Summary: The author explains how the Gauss-Jordan elimination method can be used to find the solution of a given system of linear equations.

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To find:

The solution of given system of linear equations, using the Gauss-Jordan elimination method.

2x1−x2+x3=−43x1−32x2+32x3=−6−6x1+3x2−3x3=12

Chapter 2 Solutions

The solution of given system of linear equations, using the Gauss-Jordan elimination method. 2 x 1 − x 2 + x 3 = − 4 3 x 1 − 3 2 x 2 + 3 2 x 3 = − 6 − 6 x 1 + 3 x 2 − 3 x 3 = 12

Solution Summary: The author explains how the Gauss-Jordan elimination method can be used to find the solution of a given system of linear equations.

Videos

To find: The solution of given system of linear equations, using the Gauss-Jordan elimination method. 2x1−x2+x3=−43x1−32x2+32x3=−6−6x1+3x2−3x3=12

Chapter 2 Solutions

To find:

The solution of given system of linear equations, using the Gauss-Jordan elimination method.

2x1−x2+x3=−43x1−32x2+32x3=−6−6x1+3x2−3x3=12