Explanation Procedure used: 1. Form an augmented matrix [ A | I ] where a matrix A is placed along with a same dimension identity matrix I separated by a line. 2. Apply row operations on [ A | I ] in order to obtain another augmented matrix [ I | B ] where I is again an identity matrix and B is an inverse of A. 3. If the reduction process yields a row of zeros in the left half of the augmented matrix, then the inverse of A does not exist. Calculation: The given matrix is A = [ 1 2 − 1 1 1 − 2 0 1 2 − 1 ] . Obtain the augmented matrix as follows. [ 1 2 − 1 1 1 0 0 1 − 2 0 0 1 0 1 2 − 1 0 0 1 ] Apply the row operation 2 R 1 → R 1 and reduce the matrix. [ 1 − 2 2 2 0 0 1 − 2 0 0 1 0 1 2 − 1 0 0 1 ] Apply the row operations − R 1 + R 2 → R 2 and − R 1 + R 3 → R 3 . [ 1 − 2 2 2 0 0 0 0 − 2 − 2 1 0 0 4 − 3 − 2 0 1 ] Apply the row operation R 2 ↔ R 3 . [ 1 − 2 2 2 0 0 0 4 − 3 − 2 0 1 0 0 − 2 − 2 1 0 ] Apply the row operation 2 R 1 + R 2 → R 1

College Algebra in Context with Integrated Review and Worksheets Plus MyLab Math with Pearson eText-- Access Card Package

1st Edition

ISBN: 9780134380155

Author: Ronald J. Harshbarger, Lisa S. Yocco

Publisher: PEARSON

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Chapter 7.4, Problem 8E

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To find: The inverse of the matrix A=[12−111−2012−1].

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Chapter 7.4, Problem 7E

Chapter 7.4, Problem 9E

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Chapter 7 Solutions

College Algebra in Context with Integrated Review and Worksheets Plus MyLab Math with Pearson eText-- Access Card Package

Ch. 7.1 - In Exercises 14, solve the systems of equations....Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The inverse of the matrix A = [ 1 2 − 1 1 1 − 2 0 1 2 − 1 ] .

Solution Summary: The author explains how to obtain the inverse of the matrix A. The augmented matrix is left[A|Iright], where A is placed along with a same dimension identity matrix.

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To find: The inverse of the matrix A=[12−111−2012−1].

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Chapter 7 Solutions