Section 4.1 Vectors in ℝ n

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Section 4.1 Vectors in ℝ n

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Section 4.1 Vectors in ℝ n. ℝ n Vectors Vector addition Scalar multiplication. Def. Let v 1 , v 2 , . . . v n be vectors in ℝ n . A linear combination of these vectors is an expression c 1 v 1 + c 2 v 2 + . . . + c n v n where c 1 , c 2 , . . . . , c n are some scalars. - PowerPoint PPT Presentation

Transcript of Section 4.1 Vectors in ℝ n

Page 1: Section  4.1 Vectors in  ℝ n

Section 4.1 Vectors in ℝn

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ℝn  

Vectors

Vector addition

Scalar multiplication.

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Def.Let v1, v2, . . . vn be vectors in ℝn. A linear combination of these vectors is an expression c1v1 + c2v2 + . . . + cnvn where c1, c2, . . . . , cn are some scalars.

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Ex.(a) If possible, write <6, 7> as a linear combination of <2, 3> and <1, 1>.

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Ex.(b) If possible, write <10, 4> as a linear combination of <5, 2> and <3, 1>.

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Ex.(c) If possible, write <7, 10> as a linear combination of <1, 2> and <3, 6>.

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Ex. Verify your answers geometrically for the last example.(a) <6, 7> written as a linear combination of <2, 3> and <1, 1>.

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Ex. Verify your answers geometrically for the last example.(b) <10, 4> written as a linear combination of <5, 2> and <3, 1>.

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Ex. Verify your answers geometrically for the last example.(c) <7, 10> written as a linear combination of <1, 2> and <3, 6>.

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Properties of vectors in ℝn

Let u, v, and w be vectors in ℝn and let c and d be scalars.

1. u + v is in ℝn

2. u + v = v + u

3. (u + v) + w = u + (v + w)

4. There is a vector 0 such that u + 0 = u for all u in ℝn

5. For all u in ℝn there is a vector –u such that u + (–u) = 06. cu is in ℝn

7. c(u + v) = cu + cv8. (c + d)u = cu + du9. c(du) = (cd)u10. 1(u) = u

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Section 4.2Vector Spaces

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Def. Let V be a set on which two operations (called vector addition and scalar multiplication) are defined. V is called a vector space if V satisfies the following properties.

For any u, v, and w be vectors in V and any scalars c and d:

1. u + v is in V

2. u + v = v + u

3. (u + v) + w = u + (v + w)

4. There is a vector 0 such that u + 0 = u for all u in V

5. For all u in V there is a vector –u such that u + (–u) = 06. cu is in V

7. c(u + v) = cu + cv8. (c + d)u = cu + du9. c(du) = (cd)u10. 1(u) = u

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

1. ℝn – the set of all n–tuples.

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

2. C(–∞,∞) – the set of all continuous functions defined on the real line.

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

3. C[a, b] – the set of all continuous functions defined on a closed interval [a, b].

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

4. P – the set of all polynomials.

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

5. Pn – the set of all polynomials of degree ≤ n.

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Examples of various vector spaces (each with the standard operations of addition and scalar multiplication):

6. Mm,n – the set of all mxn matrices.

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Ex. Show that the following are not examples of vector spaces.

(a) The set of all integers (with the standard operations).

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Ex. Show that the following are not examples of vector spaces.

(b) The set of all polynomials with degree 2 (with the standard operations).

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Ex. Show that the following are not examples of vector spaces.

(c) The set of all invertible 2x2 matrices (with the standard operations).

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Ex. Show that the following are not examples of vector spaces.

(d) The set of all matrices (with the standard operations).

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Ex. Let V = ℝ2 with the standard operation of addition and the following nonstandard definition of scalar multiplication: c (x, y) = (cx, 0). Determine whether or not V is a vector space.

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Section 4.3Subspaces of Vector Spaces

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Def. Let V be a vector space and let W be a subset of V. If W is itself a vector space then W is said to be a subspace of V.

Note: Every vector space contains at least two subspaces, the vector space itself and the set consisting only of the zero element. These are called the trivial subspaces.

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Ex. Determine whether or not the following are subspaces.

(a) Let V = ℝ3 with the standard operations. Define W to be { (x,y,0) : x and y are real numbers}.

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Ex. Determine whether or not the following are subspaces.

(b) Let V = ℝ3 with the standard operations. Define W to be { (x,y,1) : x and y are real numbers}.

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Theorem: Test for a Subspace

Let V be a vector space and suppose W is a subset of V. W is a subspace of V if and only if the following hold:

1. If u and v are in W then u + v is in W.2. If u is in W then cu is in W.

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Ex. Let a, b, and c be constants. Show that the set of all solutions to the differential equation ay″ + by′ + cy = 0 is a subspace (a subspace of the vector space of all functions).

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Note: W cannot be a subspace of V unless W contains the zero element.

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Ex. Which of the following are subspaces? (a) V = ℝ2 . W = the solution set of 2x + 3y = 5.

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Ex. Which of the following are subspaces? (b) V = ℝ2 . W = the solution set of -5x + y = 0.

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Ex. Which of the following are subspaces? (c) V = ℝ3 . W = the solution set of 3x + 8y – 4z = 7.

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Ex. Which of the following are subspaces? (d) V = ℝ3 . W = the solution set of -2x + 12y – z = 0.

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All nontrivial subspaces of ℝ2 are lines which pass through the origin.

All nontrivial subspaces of ℝ3 are planes which pass through the origin or lines which pass through the origin.

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Ex. Is W = {(x, x+ z, z) : x and z are real numbers} a subspace of ℝ3?

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Ex. Show that the solution set to x2 + y2 = 1 is not a subspace of ℝ2.

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Ex. (a) Let V = ℝ3 with the standard operations. Define W to be {(x,y,0) : x and y are real numbers}. What can W be identified with?

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Ex. (b) Let V = ℝ3 with the standard operations. Define W to be {(t, t, 3t) : t is a real number}. What can W be identified with?

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Notation:

P0, P1, P2, P3, . . . . , P

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Section 4.4Spanning Sets and Linear Independence

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Recall:

Let v1, v2, . . . vn be vectors in ℝn. A linear combination of these vectors is an expression c1v1 + c2v2 + . . . + cnvn where c1, c2, . . . . , cn are some scalars.

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Ex.(a) If possible, write f (x) = x2 + x + 1 as a linear combination of g1(x) = x2 + 2x + 3, g2(x) = x + 2, and g3(x) = –x2 + 1.

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Ex.(b) If possible, write as a linear combination of , ,

and

0 82 1

2 41 4

1 31 2

2 01 3

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Def.Let S = {v1, v2, . . . vn} be a subset of a vector space V. The span of S is the set of all linear combinations of elements of S. span(S) = {c1v1 + c2v2 + . . . + cnvn : c1, c2, . . . . cn are scalars}

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Def. Let S = {v1, v2, . . . vn} be a subset of a vector space V. If every vector of V can be written as a linear combination of elements of S then we say S spans V.

Put another way, S spans V if span(S) = V.

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Ex. Show that {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spans ℝ3.

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Ex. Does {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} span ℝ3?

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Ex. Does {(1, 0, 0), (0, 1, 0)} span ℝ3?

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Ex. Determine whether or not {(1, 2, 3), (4, 5, 6), (7, 8, 9)} spans ℝ3. If it does, then write a general vector (x, y, z) as a linear combination of these vectors. If it does not, then find a vector in ℝ3 which it does not span.

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Ex. Determine whether or not {(1, 2, 3), (4, 5, 6), (–1, 1, 3)} spans ℝ3. If it does, then write a general vector (x, y, z) as a linear combination of these vectors. If it does not, then find a vector in ℝ3 which it does not span.

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Ex. Give a set which spans . . . (a) ℝ2

(b) M23

(c) P2

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We shall be interested in spanning a vector space with as few vectors as possible.

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Def. Suppose v1, v2, . . . vn are vectors in a vector space V. v1, v2, . . . vn are said to be linearly independent if the only linear combination of these vectors which produces the zero element is the trivial one. (i.e. if the only solution to the equation c1v1 + c2v2 + . . . + cnvn = 0 happens to be the trivial solution c1 = c2 = . . . = cn = 0)

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Ex. Use the definition of linear independence to determine whether the following are linearly independent or linearly dependent.(a) (1,0), (0,1)

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Ex. Use the definition of linear independence to determine whether the following are linearly independent or linearly dependent.(b) (1,2,3), (4,5,6), (7,8,9)

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Ex. Use the definition of linear independence to determine whether the following are linearly independent or linearly dependent.(c) (1,0), (0,1), (2,3)

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Ex. Use the definition of linear independence to determine whether the following are linearly independent or linearly dependent.(d) (1,2), (3,6)

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Ex. Use the definition of linear independence to determine whether the following are linearly independent or linearly dependent.(e) (1,2,3), (4,5,6), (-1,1,3)

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Theorem: v1, v2, . . . vn are linearly dependent if and only if one of the vectors can be written as a linear combination of the others.

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Theorem: Let v1, v2, . . . vn be vectors in a vector space V. Then span({v1, v2, . . . vn}) is a subspace of V.

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Section 4.5Basis and Dimension

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We seek a set of vectors in a vector space which will span the entire vector space, but has as few vectors as possible to do so.

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Def. {v1, v2, . . . vn} is a basis for a vector space V if:1. {v1, v2, . . . vn} spans V.2. {v1, v2, . . . vn} is a linearly independent set.

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Standard bases:

ℝ2

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Standard bases:

ℝ3

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Standard bases:

ℝn

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Standard bases:

M22

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Standard bases:

P2

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Nonstandard bases.

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Ex. Show that {(1,2,3), (0,1,2), (-2,0,1)} is a basis for ℝ3.

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Compare the following three sets to see why we impose these two restrictions in our definition of a basis:

1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )

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Compare the following three sets to see why we impose these two restrictions in our definition of a basis:

1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )2. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} )

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Compare the following three sets to see why we impose these two restrictions in our definition of a basis:

1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )2. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} )3. span( {(1, 0, 0), (0, 1, 0)} )

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Theorem: S = {v1, v2, . . . , vn} is a basis for V if and only if every element of V can be written as a linear combination of v1, v2, . . . , and vn in one and only one way.

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Theorem: If a vector space V has a basis with n vectors, then every basis for V has n vectors.

Def. The dimension of a vector space V, denoted dim(V), is the number of vectors in a basis for V.

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ℝn has dimension __________

Pn has dimension __________

Mmn has dimension _________

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Ex. Quickly show that {(1, 2, 6), (-4, 1, 4), (24, 0, 1), (1, 6, 1)} is not a basis for ℝ3.

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Note: If you know V has dimension n then to ascertain whether or not S = {v1, v2, . . . , vn} is a basis, you only need to check one of the two defining properties of a basis (i.e. check whether S is linearly independent or check whether S spans V, but you do not need to check both.)

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Recall: The solution set to the differential equation ay″ + by′ + cy = 0 is a vector space. Notice what an impact these last theorems have on us trying to find a way to describe the solutions to a differential equation like this.

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Ex. Find all the solutions to y″ – 3y′ + 2y = 0.

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Ex. Find the dimension of the following vector spaces:

(a) V = {(d, c-d, d) : c and d are real numbers}

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Ex. Find the dimension of the following vector spaces:

(b) V = {(2b, b, 0) : b is a real number}

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Ex. Find the dimension of the following vector spaces:

(c) V = : , , and are real numbersa b

a b cb c

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Section 4.6Rank of a Matrix and

Systems of Linear Equations

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Def. Given a matrix A, . . . .

(a) the row space of A is the span of the row vectors of A.

(b) the column space of A is the span of the column vectors of A.

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Ex. Let

(a) List a few vectors in the row space of A.

(b) List a few vectors in the column space of A.

1 2 34 5 6

A

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Ex. Let

(c) Find a basis for the row space of A. Then write a general vector in the row space, (x, y, z) as a linear combination of these

1 2 34 5 6

A

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Ex. Let

(d) Find a basis for the column space of A. Then write a general vector in the column space, (x, y) as a linear combination of these

1 2 34 5 6

A

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Ex. Let

(e) Solve AX = B using B = [11 29]T by using your answer from part (d).

1 2 34 5 6

A

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Ex. Let

(f) The dimension of the row space of A is _______. The dimension of the column space of A is ______.

1 2 34 5 6

A

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Theorem: Let A be a matrix. The dimension of the row space of A equals the dimension of the column space of A. 

Def. The rank of a matrix A, denoted rank(A), is the dimension of the row space of A (which is also the dimension of the column space of A).

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Ex. Give the rank of 1 2 34 5 6

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Ex. Give the rank of

1 3 2 61 2 0 1

0 5 2 71 1 1 1

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Theorem: Let A be a given matrix and A* be a matrix produced after applying elementary row operations on A. Then the row space of A is equal to the row space of A*.

(Note: This does not say that the column space of A is equal to the column space of A*.)

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Ex. Give the rank of

1 3 2 61 2 0 1

0 5 2 71 1 1 1

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Def. Given a matrix A, the nullspace of A is defined to be all of the column matrices X so that AX = O. The dimension of the nullspace of A is called the nullity of A.

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Ex. Verify that the nullspace of a matrix A is indeed a subspace.

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Ex. Let A =

(a) Verify that is in the nullspace of A.

1 2 2 13 6 5 41 2 0 3

1442

2

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Ex. Let A =

(b) Find the nullspace of A. (Give your answer as a span of basis vectors.)

1 2 2 13 6 5 41 2 0 3

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Ex. Let A =

(c) Determine the linear combination of basis vectors that generates

1 2 2 13 6 5 41 2 0 3

14

42

2

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Given a matrix A, we know that the solutions to AX = O form a vector space. The solutions to AX = B do not form a solution space (if B ≠ O), but we can use the solutions to AX = O to help us give a general solution to AX = B.

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Theorem: Suppose we are given an mxn matrix A and an nx1 column matrix B. Further suppose that Xh is the general solution to AX = O, and Xp is one particular solution to AX = B. The general solution to AX = B is then given by Xp + Xh.

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Ex. Use this theorem to describe all the solutions to the following system: x1 – 2x3 + x4 = 53x1 + x2 – 5x3 = 8 x1 + 2x2 – 5x4 = –9

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Recall that the solutions to the differential equation ay″ + by′ + cy = 0 form a vector space. The solutions to the differential equation ay″ + by′ + cy = f (x) do not form a vector space (if f (x) ≠ 0). Compare this with the info in the paragraph at the top of this page. We can use the previous theorem to describe the solution set to ay″ + by′ + cy = f (x).

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Ex. Find all the solutions to y″ – 3y′ + 2y = 14e–x.