Explanation Given information: The given matrix A is [ 1 0 2 − 1 2 3 1 − 1 0 ] . The given matrix C is [ 5 − 1 0 2 − 2 1 − 3 0 4 ] . Calculation: Use Gauss Jordan method to find the A − 1 value. Write the given matrix with identity matrix. A = [ 1 0 2 − 1 2 3 1 − 1 0 | 1 0 0 0 1 0 0 0 1 ] (1) The vertical line in Equation (1) denotes the separation of two matrices. The first element in the given matrix should be 1. The value of element a 11 is 1. The elements a 21 and a 31 in Equation (1) should be 0. Replace the elements of row 2 by adding the element of row 1 from the corresponding element in row 2 and replace the elements of row 3 by subtracting element of row 1 from the corresponding element in row 3. A = [ 1 0 2 − 1 2 3 1 − 1 0 | 1 0 0 0 1 0 0 0 1 ] R 2 ⇒ R 2 + R 1 R 3 ⇒ R 3 − R 1 = [ 1 0 2 − 1 + 1 2 + 0 3 + 2 1 − 1 − 1 − 0 0 − 2 | 1 0 0 0 + 1 1 + 0 0 + 0 0 − 1 0 − 0 1 − 0 ] = [ 1 0 2 0 2 5 0 − 1 − 2 | 1 0 0 1 1 0 − 1 0 1 ] (2) The bottom row element a 22 in Equation (2) should be 1. Replace the row 2 by dividing 2 with each element in row 2. A = [ 1 0 2 0 2 5 0 − 1 − 2 | 1 0 0 1 1 0 − 1 0 1 ] R 2 ⇒ R 2 2 = [ 1 0 2 0 2 2 2 5 2 0 − 1 − 2 | 1 0 0 1 2 1 2 0 2 − 1 0 1 ] = [ 1 0 2 0 1 5 2 0 − 1 − 2 | 1 0 0 1 2 1 2 0 − 1 0 1 ] (3) The elements a 12 and a 32 in Equation (3) should be zero. The element a 12 is already 1. Replace the elements of row 3 by adding each element of row 2 from the corresponding element in row 3. A = [ 1 0 2 0 1 5 2 0 − 1 − 2 | 1 0 0 1 2 1 2 0 − 1 0 1 ] R 3 ⇒ R 3 + R 2 = [ 1 0 2 0 1 5 2 0 + 0 − 1 + 1 − 2 + 5 2 | 1 0 0 1 2 1 2 0 − 1 + 1 2 0 + 1 2 1 + 0 ] = [ 1 0 2 0 1 5 2 0 0 1 2 | 1 0 0 1 2 1 2 0 − 1 2 1 2 1 ] (4) The bottom row element a 33 in Equation (4) should be 1. Replace the row 3 elements by dividing 1 2 with each element. A = [ 1 0 2 0 1 5 2 0 0 1 2 | 1 0 0 1 2 1 2 0 − 1 2 1 2 1 ] R 3 ⇒ R 3 1 2 = [ 1 0 2 0 1 5 2 0 1 2 0 1 2 1 2 1 2 | 1 0 0 1 2 1 2 0 − 1 2 1 2 1 2 1 2 1 1 2 ] A = [ 1 0 2 0 1 5 2 0 0 1 | 1 0 0 1 2 1 2 0 − 1 1 2 ] (5) The top row element a 13 and second row element a 23 in Equation (5) should be 0

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 16.3, Problem 34E

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Find the matrix value of CA−1.

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Chapter 16.3, Problem 33E

Chapter 16.3, Problem 35E

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Chapter 16 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 16.1 - For matrices A and B, find A + B. Ch. 16.1 - Prob. 2PE Ch. 16.1 - Prob. 3PE Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - In Exercises 3–10, determine the value of the...Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E

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Chapter 16 Solutions