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In Excercise 10, find the general form of of the indicated matrices as the span an example 3.17. 10. -span 6. B ✓ (A₁, A₂, A3) in Encercise 6 2 .4 5 3 • A₁ = [10] A₂ = 10 - 17 A.2

In Excercise 10, find the general form of of the indicated matrices as the span an example 3.17. 10. -span 6. B ✓ (A₁, A₂, A3) in Encercise 6 2 .4 5 3 • A₁ = [10] A₂ = 10 - 17 A.2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I have added example 3.17 as per the question
In Excercise 10, find the general form of the span of the indicated matrices ag an example 3.17. 10. span (A₁, A2, A3) in Encercise 6 r 6. B Аз - 3 2 -4 2 11 201 A2 - [₁9] A² = [9-6]₁ 0 1 1 A₁ 10
Transcribed Image Text:In Excercise 10, find the general form of the span of the indicated matrices ag an example 3.17. 10. span (A₁, A2, A3) in Encercise 6 r 6. B Аз - 3 2 -4 2 11 201 A2 - [₁9] A² = [9-6]₁ 0 1 1 A₁ 10
Example 3.17 Y Describe the span of the matrices A₁, A₂, and A3 in Example 3.16. Solution One way to do this is simply to write out a general linear combination of A₁, A₂, and A₁. Thus, 0 [(-; 1] + [1] + [1] -1 0 C₂ + c3 9₂ + =\-9₁ +63 9₂2 +6₁) C₁A₁+C₂A₂C₂A3 = C₁ (which is analogous to the parametric representation of a plane). But suppose we want to know when the matrix is in span (A₁, A2, A3). From the representa- tion above, we know that it is when C₂ + C₂ C₁ + C₂ [-c₁₂₁ + C₂ C₂ + C3] [$] [*] for some choice of scalars C₁, C₂, C3. This gives rise to a system of linear equations whose left-hand side is exactly the same as in Example 3.16 but whose right-hand side 0 1 -1 0 and row reduction produces 1 1 w 0 1 x 0 1 y 1 1 z is general. The augmented matrix of this system is 0 1 1 w 10 1 x 1 y Section 3.2 Matrix Algebra -1 0 01 01 001 0 0 0 1 0 0 x - ly 0-x-y + w x + y w z 157 (Check this carefully.) The only restriction comes from the last row, where clearly we must have w - z = 0 in order to have a solution. Thus, the span of A₁, A₂, and A, con- which =z. That is, span (A₁, A₂, A3) = - sists of all matrices W x y z >= {[***]}· Note If we had known this before attempting Example 3.16, we would have seen immediately that B = the necessary form (take w = 1, x = 4, and y = 2), but C = [13] is a linear combination of A₁, A₂, and A3, since it has cannot be a linear combination of A₁, A₂, and A3, since it does not have the proper form (1 # 4). Linear independence also makes sense for matrices. We say that matrices A₁, A₂,..., Ak of the same size are linearly independent if the only solution of the equation G₁A₁ + G₂A₂+ +₁A₁ = 0 (1) is the trivial one: c₁ = C₂ == Ck = 0. If there are nontrivial coefficients that satisfy (1), then A₁, A₂,..., A are called linearly dependent.
Transcribed Image Text:Example 3.17 Y Describe the span of the matrices A₁, A₂, and A3 in Example 3.16. Solution One way to do this is simply to write out a general linear combination of A₁, A₂, and A₁. Thus, 0 [(-; 1] + [1] + [1] -1 0 C₂ + c3 9₂ + =\-9₁ +63 9₂2 +6₁) C₁A₁+C₂A₂C₂A3 = C₁ (which is analogous to the parametric representation of a plane). But suppose we want to know when the matrix is in span (A₁, A2, A3). From the representa- tion above, we know that it is when C₂ + C₂ C₁ + C₂ [-c₁₂₁ + C₂ C₂ + C3] [$] [*] for some choice of scalars C₁, C₂, C3. This gives rise to a system of linear equations whose left-hand side is exactly the same as in Example 3.16 but whose right-hand side 0 1 -1 0 and row reduction produces 1 1 w 0 1 x 0 1 y 1 1 z is general. The augmented matrix of this system is 0 1 1 w 10 1 x 1 y Section 3.2 Matrix Algebra -1 0 01 01 001 0 0 0 1 0 0 x - ly 0-x-y + w x + y w z 157 (Check this carefully.) The only restriction comes from the last row, where clearly we must have w - z = 0 in order to have a solution. Thus, the span of A₁, A₂, and A, con- which =z. That is, span (A₁, A₂, A3) = - sists of all matrices W x y z >= {[***]}· Note If we had known this before attempting Example 3.16, we would have seen immediately that B = the necessary form (take w = 1, x = 4, and y = 2), but C = [13] is a linear combination of A₁, A₂, and A3, since it has cannot be a linear combination of A₁, A₂, and A3, since it does not have the proper form (1 # 4). Linear independence also makes sense for matrices. We say that matrices A₁, A₂,..., Ak of the same size are linearly independent if the only solution of the equation G₁A₁ + G₂A₂+ +₁A₁ = 0 (1) is the trivial one: c₁ = C₂ == Ck = 0. If there are nontrivial coefficients that satisfy (1), then A₁, A₂,..., A are called linearly dependent.
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