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 = [ 2 3 1 3 4 − 1 2 5 1 ] . Obtain the augmented matrix as follows. [ 2 3 1 1 0 0 3 4 − 1 0 1 0 2 5 1 0 0 1 ] Apply the row operation 3 R 1 − 2 R 2 → R 2 and − R 1 + R 3 → R 3 and reduce the matrix. [ 2 3 1 1 0 0 0 1 5 3 − 2 0 0 2 0 − 1 0 1 ] Apply the row operation 2 R 2 − R 3 → R 2 . [ 2 3 1 1 0 0 0 0 10 7 − 4 − 1 0 2 0 − 1 0 1 ] Apply the row operation R 3 ↔ R 2 . [ 2 3 1 1 0 0 0 2 0 − 1 0 1 0 0 10 7 − 4 − 1 ] Apply the row operation 2 R 1 − 2 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

See similar textbooks

Videos

Question

Chapter 7.4, Problem 11E

To determine

To find: The inverse of the matrix A=[23134−1251].

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.4, Problem 10E

Chapter 7.4, Problem 12E

Students have asked these similar questions

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter | (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) ☐ 1. For all n > 1, seriesΣ In(n) In(n) converges. 2, 1, arctan(n) the series arctan(n) n³ ☐ 4. For all n > 1, 123 converges. 1 n ln(n) series In(n) diverges. 2n . and the seriesΣconverges, so by the Comparison Test, 2, 3, and the series converges, so by the Comparison Test, the series-3 1 converges. ☐ 6. For all n > 2, In(n) >, and the series Σ converges, so by the Comparison Test, the seriesΣ In(n) converges.

In 2012, the employees of Radcliff Ltd. agreed to purchase 5% of the share capital of 10 million shares of $2 each. There are 20 employees in the plan, and each purchased an equal number of shares. Johnson works at Radcliff Ltd. What would be his ESOP share deduction? $45,000 $25,000 $75,000 $50,000.

Instructions. "I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."

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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The inverse of the matrix A = [ 2 3 1 3 4 − 1 2 5 1 ] .

Solution Summary: The author explains how to obtain the inverse of the matrix A.

Videos

To find: The inverse of the matrix A=[23134−1251].

Want to see the full answer?

Chapter 7 Solutions