Explanation The equation given is x 1 + x 2 = 0 2 x 1 + 3 x 2 = 4 To solve these equations using inverse matrix, first convert these equations into a matrix in the form x = A - 1 b where A = 1 1 2 3 represents the matrix of coefficients. x = x 1 x 2 represents the matrix of variables. b = 0 4 represents the matrix of constants. x = A - 1 b will be written as x 1 x 2 = A - 1 0 4 Finding A - 1 by using the formula, A - 1 = 1 A a d j A Finding A by A = 1 1 2 3 A = 3 - 2 A = 1 ≠ 0 Since A is non singular, A - 1 exists. Normal matrix can be written as A = a b c d where a = 1 , b = 1 , c = 2 , d = 3 . For finding A d j A in 2 × 2 matrix, the above matrix can be written as A d j A = d - b - c a Using the above formula, we can find A d j A

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 1.9, Problem 30E

To determine

To solve:

The given system of equations by forming x=A-1b.

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Chapter 1.9, Problem 29E

Chapter 1.9, Problem 31E

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Chapter 1 Solutions

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...

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