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File Preview (d) First, we just need one solution to each, then we can form solution sets by adding all the multiples of the basis for the kernel. Since (2,0,0)¯ = 2ā1 +0ã2 + 0ã3 A0 = 0 and so the full solution set to the first is 1 +c 1 : CER For the second equation, notice (0, 1,0)¯ = −ã1 + ã2 + 0ã3 and so A = 0-8 R}. and so the full solution set for the second is + c : CER For the final equation, notice that (2,1,0) is the last column of A and Aе3 and so one solution is (0,0,1) and so the full solution set is +c 1 : CER = ძვ Let A: = 1 1 2 011 0 0 | (a) Find a basis for ran(A). What is dim(ran(A))? (b) Use the Rank-Nullity theorem to find dim (ker (A)). (c) Find a basis for ker(A). (d) Use the basis from (c) to write the solution sets to each of the following: Ax = 0 Ay= |1 Az= Solution. Let ₁ be the first column of A, 2 the second, and a3 the third column of A. (a) Since σ ₁ and 2 are not multiples of one another, they are linearly independent, but ã₁ + a2 = ã3 and so a basis for ran(A) is {ā₁, ā2} and so dim(ran(A)) = 2. (b) The rank-nullity theorem gives 3 = dim(ran(A)) + dim(ker(A)) and since dim(ran(A)) = 2, dim(ker(A)) = 1. (c) Since dim(ker(A)) = 1, we just need one nonzero element of the kernel. Since a1a2a3 = 0, so a basis for ker(A) is (1, 1, −1). = = α1 + α2 − α3 = 0

File Preview (d) First, we just need one solution to each, then we can form solution sets by adding all the multiples of the basis for the kernel. Since (2,0,0)¯ = 2ā1 +0ã2 + 0ã3 A0 = 0 and so the full solution set to the first is 1 +c 1 : CER For the second equation, notice (0, 1,0)¯ = −ã1 + ã2 + 0ã3 and so A = 0-8 R}. and so the full solution set for the second is + c : CER For the final equation, notice that (2,1,0) is the last column of A and Aе3 and so one solution is (0,0,1) and so the full solution set is +c 1 : CER = ძვ Let A: = 1 1 2 011 0 0 | (a) Find a basis for ran(A). What is dim(ran(A))? (b) Use the Rank-Nullity theorem to find dim (ker (A)). (c) Find a basis for ker(A). (d) Use the basis from (c) to write the solution sets to each of the following: Ax = 0 Ay= |1 Az= Solution. Let ₁ be the first column of A, 2 the second, and a3 the third column of A. (a) Since σ ₁ and 2 are not multiples of one another, they are linearly independent, but ã₁ + a2 = ã3 and so a basis for ran(A) is {ā₁, ā2} and so dim(ran(A)) = 2. (b) The rank-nullity theorem gives 3 = dim(ran(A)) + dim(ker(A)) and since dim(ran(A)) = 2, dim(ker(A)) = 1. (c) Since dim(ker(A)) = 1, we just need one nonzero element of the kernel. Since a1a2a3 = 0, so a basis for ker(A) is (1, 1, −1). = = α1 + α2 − α3 = 0

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section: Chapter Questions
Problem 9RQ
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please explain part d because i am confused by the second and third part of part d. 

File Preview (d) First, we just need one solution to each, then we can form solution sets by adding all the multiples of the basis for the kernel. Since (2,0,0)¯ = 2ā1 +0ã2 + 0ã3 A0 = 0 and so the full solution set to the first is 1 +c 1 : CER For the second equation, notice (0, 1,0)¯ = −ã1 + ã2 + 0ã3 and so A = 0-8 R}. and so the full solution set for the second is + c : CER For the final equation, notice that (2,1,0) is the last column of A and Aе3 and so one solution is (0,0,1) and so the full solution set is +c 1 : CER = ძვ
Transcribed Image Text:File Preview (d) First, we just need one solution to each, then we can form solution sets by adding all the multiples of the basis for the kernel. Since (2,0,0)¯ = 2ā1 +0ã2 + 0ã3 A0 = 0 and so the full solution set to the first is 1 +c 1 : CER For the second equation, notice (0, 1,0)¯ = −ã1 + ã2 + 0ã3 and so A = 0-8 R}. and so the full solution set for the second is + c : CER For the final equation, notice that (2,1,0) is the last column of A and Aе3 and so one solution is (0,0,1) and so the full solution set is +c 1 : CER = ძვ
Let A: = 1 1 2 011 0 0 | (a) Find a basis for ran(A). What is dim(ran(A))? (b) Use the Rank-Nullity theorem to find dim (ker (A)). (c) Find a basis for ker(A). (d) Use the basis from (c) to write the solution sets to each of the following: Ax = 0 Ay= |1 Az= Solution. Let ₁ be the first column of A, 2 the second, and a3 the third column of A. (a) Since σ ₁ and 2 are not multiples of one another, they are linearly independent, but ã₁ + a2 = ã3 and so a basis for ran(A) is {ā₁, ā2} and so dim(ran(A)) = 2. (b) The rank-nullity theorem gives 3 = dim(ran(A)) + dim(ker(A)) and since dim(ran(A)) = 2, dim(ker(A)) = 1. (c) Since dim(ker(A)) = 1, we just need one nonzero element of the kernel. Since a1a2a3 = 0, so a basis for ker(A) is (1, 1, −1). = = α1 + α2 − α3 = 0
Transcribed Image Text:Let A: = 1 1 2 011 0 0 | (a) Find a basis for ran(A). What is dim(ran(A))? (b) Use the Rank-Nullity theorem to find dim (ker (A)). (c) Find a basis for ker(A). (d) Use the basis from (c) to write the solution sets to each of the following: Ax = 0 Ay= |1 Az= Solution. Let ₁ be the first column of A, 2 the second, and a3 the third column of A. (a) Since σ ₁ and 2 are not multiples of one another, they are linearly independent, but ã₁ + a2 = ã3 and so a basis for ran(A) is {ā₁, ā2} and so dim(ran(A)) = 2. (b) The rank-nullity theorem gives 3 = dim(ran(A)) + dim(ker(A)) and since dim(ran(A)) = 2, dim(ker(A)) = 1. (c) Since dim(ker(A)) = 1, we just need one nonzero element of the kernel. Since a1a2a3 = 0, so a basis for ker(A) is (1, 1, −1). = = α1 + α2 − α3 = 0
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