Explanation Approach: If a square matrix A of dimension n × n has an inverse A − 1 , and if X is a variable matrix and B is a known matrix both with n rows, then the solution of the matrix equation A X = B is given by: X = A − 1 B . Calculation: Consider the system of equations { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 . The given system of linear equations can be expressed as a matrix equation as following: A X = B where, A = [ 2 1 5 1 2 2 1 0 3 ] , X = [ x y z ] , and B = [ 1 3 1 4 1 6 ] . The inverse of the coefficient matrix A − 1 is calculated below: Now the solution of the matrix equation is given by: Consider a matrix, [ 2 1 5 1 2 2 1 0 3 | 1 0 0 0 1 0 0 0 1 ] Perform the transformation R 1 ↔ R 2 . [ 2 1 5 1 2 2 1 0 3 | 1 0 0 0 1 0 0 0 1 ] → [ 1 2 2 2 1 5 1 0 3 | 0 1 0 1 0 0 0 0 1 ] Perform the transformation R 2 → R 2 − 2 R 1 . [ 1 2 2 2 − 2 ( 1 ) 1 − 2 ( 2 ) 5 − 2 ( 2 ) 1 0 3 | 0 1 0 1 − 2 ( 0 ) 0 − 2 ( 1 ) 0 0 0 1 − 2 ( 0 ) ] → [ 1 2 2 0 − 3 1 1 0 3 | 0 1 0 1 − 2 0 0 0 1 ] Perform the transformation R 3 → R 3 − R 1 . [ 1 2 2 0 − 3 1 1 − 1 0 − 2 3 − 2 | 0 1 0 1 − 2 0 0 − 0 0 − 1 1 − 0 ] → [ 1 2 2 0 − 3 1 0 − 2 1 | 0 1 0 1 − 2 0 0 − 1 1 ] Perform the transformation R 2 → − 1 3 R 2 . [ 1 2 2 0 ( − 1 3 ) − 3 ( − 1 3 ) 1 ( − 1 3 ) 0 − 2 1 | 0 1 0 1 ( − 1 3 ) − 2 ( − 1 3 ) 0 ( − 1 3 ) 0 − 1 1 ] → [ 1 2 2 0 1 − 1 3 0 − 2 1 | 0 1 0 − 1 3 2 3 0 0 − 1 1 ] Perform the transformation R 1 → R 1 − 2 R 2 . [ 1 − 2 ( 0 ) 2 − 2 ( 1 ) 2 − 2 ( − 1 3 ) 0 1 − 1 3 0 − 2 1 | 0 − 2 ( − 1 3 ) 1 − 2 ( 2 3 ) 0 − 2 ( 0 ) − 1 3 2 3 0 0 − 1 1 ] → [ 1 0 8 3 0 1 − 1 3 0 − 2 1 | 2 3 − 1 3 0 − 1 3 2 3 0 0 − 1 1 ] Perform the transformation R 3 → R 3 + 2 R 2 . [ 1 0 8 3 0 1 − 1 3 0 + 2 ( 0 ) − 2 + 2 ( 1 ) 1 + 2 ( − 1 3 ) | 2 3 − 1 3 0 − 1 3 2 3 0 0 + 2 ( − 1 3 ) − 1 + 2 ( 2 3 ) 1 + 2 ( 0 ) ] → [ 1 0 8 3 0 1 − 1 3 0 0 1 3 | 2 3 − 1 3 0 − 1 3 2 3 0 − 2 3 1 3 1 ] Perform the transformation R 3 → 3 R 3

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN: 9781305071742

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 11.CR, Problem 63E

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To solve:

The system of a linear equation {2x+y+5z=13x+2y+2z=14x+3z=16.

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Chapter 11.CR, Problem 62E

Chapter 11.CR, Problem 64E

Chapter 11 Solutions

Algebra and Trigonometry (MindTap Course List)

Ch. 11.1 - If a system of linear equations has infinitely...Ch. 11.1 - Write the augmented matrix of the following system...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - 5-10 Dimension of a Matrix State the dimension of...Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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The system of a linear equation { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 .

Solution Summary: The author explains the solution of a linear equation c2x+y+5z=

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To solve:

The system of a linear equation {2x+y+5z=13x+2y+2z=14x+3z=16.

Want to see the full answer?

Chapter 11 Solutions

The system of a linear equation { 2 x + y + 5 z = 1 3 x + 2 y + 2 z = 1 4 x + 3 z = 1 6 .

Solution Summary: The author explains the solution of a linear equation c2x+y+5z=

Videos

To solve: The system of a linear equation {2x+y+5z=13x+2y+2z=14x+3z=16.

Want to see the full answer?

Chapter 11 Solutions

To solve:

The system of a linear equation {2x+y+5z=13x+2y+2z=14x+3z=16.