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Find the inverses of the matrices in Exercises 29-32, if they exist. Use the algorithm introduced in this section. 29. 1 2 [43] 7 30. 5 4 10 7

Find the inverses of the matrices in Exercises 29-32, if they exist. Use the algorithm introduced in this section. 29. 1 2 [43] 7 30. 5 4 10 7

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter2: Matrices
Section2.1: Operations With Matrices
Problem 77E
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29
en X = A-¹B. duction of [A B] is much A¹ B. are nx p matrices and A is true, in general, when nd Care m x n matrices C. xn matrices. Show that g a matrix D such that rtible, and AB is invert- Let C = AB, and solve assuming that A, B, and BP-1. Solve for B in ices, does the equation n, X? If so, find it. rices with A, X, and (3) wert a matrix, explain matrix A are linearly atrix A span R" when 4 in Section 1.4.] x = 0 has only the ot columns and A is s shows that A must e 24 will be cited in ==b has a solution invertible [Hints In 3 x 3 matrix and I = 13. (A general proof would require slightly more notation.) 27. a. Use equation (1) from Section 2.1 to show that row; (A) = row, (I). A, for i = 1, 2, 3. b. Show that if rows 1 and 2 of A are interchanged, then the result may be written as EA, where E is an elementary matrix formed by interchanging rows 1 and 2 of I. c. Show that if row 3 of A is multiplied by 5, then the result may be written as EA, where E is formed by multiplying row 3 of I by 5. 28. Show that if row 3 of A is replaced by row3 (A) - 4. row₁(A), the result is EA, where E is formed from I by replacing row 3 (I) by row 3 (1) 4 row₁ (I). 29. Find the inverses of the matrices in Exercises 29-32, if they exist. Use the algorithm introduced in this section. 4 1 31. -3 0 -2 4 2 -3 4 1 0 0 1 1 A = 33. Use the algorithm from this section to find the inverses of S.S | 0 and 1 correct. 35. Let A = 2 3 30. Let A be the corresponding n x n matrix, and let B be its inverse. Guess the form of B, and then prove that AB = I and BA = I. -2 -7 -9 2 6 1 4 32. 34. Repeat the strategy of Exercise 33 to guess the inverse of 1 0 0 1 2 0 1 2 3 53 3 4 1 1 0 0 0 n 10 7 1-2 4 -7 3 -2 6 -4 0 0 1 0 1 1 azot . Prove that your guess Find the third column of A is.
Transcribed Image Text:en X = A-¹B. duction of [A B] is much A¹ B. are nx p matrices and A is true, in general, when nd Care m x n matrices C. xn matrices. Show that g a matrix D such that rtible, and AB is invert- Let C = AB, and solve assuming that A, B, and BP-1. Solve for B in ices, does the equation n, X? If so, find it. rices with A, X, and (3) wert a matrix, explain matrix A are linearly atrix A span R" when 4 in Section 1.4.] x = 0 has only the ot columns and A is s shows that A must e 24 will be cited in ==b has a solution invertible [Hints In 3 x 3 matrix and I = 13. (A general proof would require slightly more notation.) 27. a. Use equation (1) from Section 2.1 to show that row; (A) = row, (I). A, for i = 1, 2, 3. b. Show that if rows 1 and 2 of A are interchanged, then the result may be written as EA, where E is an elementary matrix formed by interchanging rows 1 and 2 of I. c. Show that if row 3 of A is multiplied by 5, then the result may be written as EA, where E is formed by multiplying row 3 of I by 5. 28. Show that if row 3 of A is replaced by row3 (A) - 4. row₁(A), the result is EA, where E is formed from I by replacing row 3 (I) by row 3 (1) 4 row₁ (I). 29. Find the inverses of the matrices in Exercises 29-32, if they exist. Use the algorithm introduced in this section. 4 1 31. -3 0 -2 4 2 -3 4 1 0 0 1 1 A = 33. Use the algorithm from this section to find the inverses of S.S | 0 and 1 correct. 35. Let A = 2 3 30. Let A be the corresponding n x n matrix, and let B be its inverse. Guess the form of B, and then prove that AB = I and BA = I. -2 -7 -9 2 6 1 4 32. 34. Repeat the strategy of Exercise 33 to guess the inverse of 1 0 0 1 2 0 1 2 3 53 3 4 1 1 0 0 0 n 10 7 1-2 4 -7 3 -2 6 -4 0 0 1 0 1 1 azot . Prove that your guess Find the third column of A is.
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