Explanation Given: The linear equations are: x − 2 y + z = − 4 , 2 x + 2 y − z = 10 and 4 x − y + 2 z = − 1 . Calculation: The system of equations are: x − 2 y + z = − 4 (I) 2 x + 2 y − z = 10 (II) 4 x − y + 2 z = − 1 (III) Use Gaussian elimination with back-substitution to solve the system. Write the augmented matrix for the system, [ 1 − 2 1 2 2 − 1 4 − 1 2 | − 4 10 − 1 ] Use matrix row operations to convert the matrix to row-echelon form. The element in the first row and the first column is already ‘1’. R 2 → R 2 − 2 R 1 To get ‘0’ in the second row, the first column [ 1 − 2 1 2 − 2 × 1 2 − 2 × ( − 2 ) − 1 − 2 × 1 4 − 1 2 | − 4 10 − 2 × ( − 4 ) − 1 ] = [ 1 − 2 1 0 6 − 3 4 − 1 2 | − 4 18 − 1 ] R 2 → R 2 6 To get ‘1’ in the second row, the second column. [ 1 − 2 1 0 ÷ 6 6 ÷ 6 − 3 ÷ 6 4 − 1 2 | − 4 18 ÷ 6 − 1 ] = [ 1 − 2 1 0 1 − 0.5 4 − 1 2 | − 4 3 − 1 ] R 3 → R 3 − 4 R 1 To get ‘0’ in the third row, first column. [ 1 − 2 1 0 1 − 0.5 4 − 4 × 1 − 1 − 4 × ( − 2 ) 2 − 4 × 1 | − 4 3 − 1 − 4 × ( − 4 ) ] = [ 1 − 2 1 0 1 − 0

Algebra & Trigonometry With Additional Material From College Algebra Essentials (custom Edition For Tidewater Community College)

7th Edition

ISBN: 9780536437273

Author: ROBERT BLITZER

Publisher: PEARSON

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Chapter 11.3, Problem 113E

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To calculate: The solution of the system of equations x−2y+z=−42x+2y−z=104x−y+2z=−1 using Gaussian elimination with back substitution or Gauss-Jordan elimination.

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Chapter 11.3, Problem 112E

Chapter 11.3, Problem 114E

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Chapter 11 Solutions

Algebra & Trigonometry With Additional Material From College Algebra Essentials (custom Edition For Tidewater Community College)

Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Prob. 5CVC Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Prob. 8CVC Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E

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The solution of the system of equations x − 2 y + z = − 4 2 x + 2 y − z = 10 4 x − y + 2 z = − 1 using Gaussian elimination with back substitution or Gauss-Jordan elimination.

Solution Summary: The author explains the solution of the system of equations (2,1,-4) using Gaussian elimination with back substitution.

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To calculate: The solution of the system of equations x−2y+z=−42x+2y−z=104x−y+2z=−1 using Gaussian elimination with back substitution or Gauss-Jordan elimination.

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