Explanation Procedure used: Finding Inverse of a Square Matrix: 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: Consider the given matrix as A = [ − 1 0 1 1 1 1 2 1 2 − 1 0 1 0 0 0 1 ] . First obtain the determinant of the matrix A to check whether the matrix A is invertible or not. The determinant of the matrix A is as follows. | A | = − 1 | − 1 0 1 1 1 2 2 − 1 0 | [ Expanding by fourth row ] = − 1 [ − 1 ( 0 + 2 ) − 0 ( 0 − 4 ) + 1 ( − 1 − 2 ) ] = − 1 [ − 2 − 3 ] = 5 Since, the determinant | A | ≠ 0 , the matrix A is nonsingular and invertible. Now, use the method of finding inverse of the matrix and obtain A − 1 as follows. The augment matrix [ A | I ] is [ A | I ] = [ − 1 0 1 1 1 0 0 0 1 1 2 1 0 1 0 0 2 − 1 0 1 0 0 1 0 0 0 0 1 0 0 0 1 ] . Perform the row operations R 2 + R 1 → R 2 and R 3 + 2 R 1 → R 3 and obtain the row reduced equivalent matrix as, [ − 1 0 1 1 1 0 0 0 0 1 3 2 1 1 0 0 0 − 1 2 3 2 0 1 0 0 0 0 1 0 0 0 1 ] Perform the row operations R 3 + R 2 → R 3 and obtain the equivalent row reduced matrix as, [ − 1 0 1 1 1 0 0 0 0 1 3 2 1 1 0 0 0 0 5 5 3 1 1 0 0 0 0 1 0 0 0 1 ] Again perform row operations R 3 − 5 R 4 → R 3 , R 2 − 2

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, Problem 24RE

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To find: The inverse of the matrix [−101111212−1010001].

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Chapter 7, Problem 23RE

Chapter 7, Problem 25RE

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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 [ − 1 0 1 1 1 1 2 1 2 − 1 0 1 0 0 0 1 ] .

Solution Summary: The author explains how to find the inverse of a square matrix by forming an augmented matrix left[A|Iright].

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To find: The inverse of the matrix [−101111212−1010001].

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Chapter 7 Solutions