Explanation Approach: Conversion of the augmented matrix into diagonal form: STEP 1: Obtaining 1 as the first element of the first column. STEP 2: Using the first row, the remaining entries in the first column are made zero. STEP 3: Obtaining 1 as the second element of the second column. STEP 4: Using the second row, the remaining entries in the second column are made zero. Hence, containing in the above manner the last matrix obtained should be of the form [ A diagonal matrix | A column matrix ] From the above obtained form of the matrix, the solution for the given system can be obtained. CASE 1: Consistent – Single solution A diagonal matrix in the left of the vertical line and any real value in the right of the vertical line against the rows of the diagonal matrix – Consistent and Single solution. CASE 2: Inconsistent – No solution. Any one of the row to the left of the vertical line become zero and a non - zero entry in the right of the vertical line against the zero row in the left – Inconsistent – No solution. CASE 3: Consistent – Infinitely many solution Any one of the row to the left of the vertical line become zero and also the entry in the right of the vertical line against the zero row in the left – Consistent – Infinitely many solution. Matrix Row Transformation: For any augmented matrix of a system of linear equations, the following row transformations interchange any two rows. Multiply or divide the elements of any row by a nonzero real number. Replace any row of the matrix by the sum of the elements of that row and a multiple of the elements of another row. Calculation: Given, { 2 x + 5 y = 7 3 x − y = 2 The above two equations are having its variable terms in the left and the constant terms in the right. Hence, the augmented matrix is given by, [ 2 5 7 3 − 1 2 ] The first entry of the first column in the above matrix is not 1. Divide the first row by 2. [ 2 2 5 2 7 2 3 − 1 2 ] R 1 2 [ 1 5 2 7 2 3 − 1 2 ] Now, let us transform the above matrix. So that the remaining entries of the first column become zero by applying row transformations. Change the row such that second row is added with the first row multiplied by − 3 . [ 1 5 2 7 2 3 + ( − 3 ) ( 1 ) − 1 + ( − 3 ) ( 5 2 ) 2 + ( − 3 ) ( 7 2 ) ] R 2 + ( − 3 ) R 1 [ 1 5 2 7 2 0 − 17 2 − 17 2 ] Now, we have to transform the above matrix

12th Edition

ISBN: 9781305652231

Author: R. David Gustafson, Jeff Hughes

Publisher: Cengage Learning

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Chapter 6.CR, Problem 21E

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The system of equations {2x+5y=73x−y=2 by matrix method if possible.

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Chapter 6.CR, Problem 20E

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The system of equations { 2 x + 5 y = 7 3 x − y = 2 by matrix method if possible.

Solution Summary: The author explains how to solve the system of equations by matrix method.

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The system of equations {2x+5y=73x−y=2 by matrix method if possible.

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Chapter 6 Solutions

The system of equations { 2 x + 5 y = 7 3 x − y = 2 by matrix method if possible.

Solution Summary: The author explains how to solve the system of equations by matrix method.

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To solve: The system of equations {2x+5y=73x−y=2 by matrix method if possible.

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Chapter 6 Solutions

To solve:

The system of equations {2x+5y=73x−y=2 by matrix method if possible.