Explanation Given: 2 x + 3 y − 2 z = 10 3 x − 2 y + 2 z = 0 4 x − y + 3 z = − 1 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: Write the above equations in augmented matrix form. [ 2 3 − 2 3 − 2 2 4 − 1 3 | 10 0 − 1 ] Convert the matrix in row reduced form. Apply operation R 2 → R 2 − R 1 in the above augmented matrix. [ 2 3 − 2 3 − 2 2 4 − 1 3 | 10 0 − 1 ] → R 2 → R 2 − R 1 [ 2 3 − 2 1 − 5 4 4 − 1 3 | 10 − 10 − 1 ] …… ( 1 ) Apply operation R 3 → R 3 − 2 R 1 in the matrix obtained in equation ( 1 ) . [ 2 3 − 2 1 − 5 4 4 − 1 3 | 10 − 10 − 1 ] → R 3 → R 3 − 2 R 1 [ 2 3 − 2 1 − 5 4 0 − 7 7 | 10 − 10 − 21 ] …... ( 2 ) Apply operation R 3 → − 1 7 R 3 in the matrix obtained in equation ( 2 ) . [ 2 3 − 2 1 − 5 4 0 − 7 7 | 10 − 10 − 21 ] → R 3 → − 1 7 R 3 [ 2 3 − 2 1 − 5 4 0 1 − 1 | 10 − 10 3 ] …... ( 3 ) Apply operation R 1 ↔ R 2 in the matrix obtained in equation ( 3 ) . [ 2 3 − 2 1 − 5 4 0 1 − 1 | 10 − 10 3 ] → R 1 ↔ R 2 [ 1 − 5 4 2 3 − 2 0 1 − 1 | − 10 10 3 ] ….... ( 4 ) Apply operation R 2 → R 2 − 2 R 1 in the matrix obtained in equation ( 4 ) . [ 1 − 5 4 2 3 − 2 0 1 − 1 | − 10 10 3 ] → R 2 → R 2 − 2 R 1 [ 1 − 5 4 0 13 − 10 0 1 − 1 | − 10 30 3 ] …

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ISBN: 9780357308615

Author: Tan

Publisher: CENGAGE L

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Chapter 2.2, Problem 48E

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To solve:

The system of linear equations using Gauss-Jordan elimination method.

2x+3y−2z=10

3x−2y+2z=0

4x−y+3z=−1

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Chapter 2.2, Problem 47E

Chapter 2.2, Problem 49E

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The system of linear equations using Gauss-Jordan elimination method. 2 x + 3 y − 2 z = 10 3 x − 2 y + 2 z = 0 4 x − y + 3 z = − 1

Solution Summary: The author explains the Gauss-Jordan elimination method to solve the system of linear equations.

Videos

To solve:

The system of linear equations using Gauss-Jordan elimination method.

2x+3y−2z=10

3x−2y+2z=0

4x−y+3z=−1

Want to see the full answer?

Chapter 2 Solutions

The system of linear equations using Gauss-Jordan elimination method. 2 x + 3 y − 2 z = 10 3 x − 2 y + 2 z = 0 4 x − y + 3 z = − 1

Solution Summary: The author explains the Gauss-Jordan elimination method to solve the system of linear equations.

Videos

To solve: The system of linear equations using Gauss-Jordan elimination method. 2x+3y−2z=10 3x−2y+2z=0 4x−y+3z=−1

Want to see the full answer?

Chapter 2 Solutions

To solve:

The system of linear equations using Gauss-Jordan elimination method.

2x+3y−2z=10

3x−2y+2z=0

4x−y+3z=−1