Webwork assignment 9 with solutions PDF

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Webwork 9, Solutions.

Assignment 9, #1. Which of the following are subspaces of R3? Solution: Before we start, it may help to remember the Theorem that the only subspaces of R3 are: 1) 2) 3) 4)

all of R3 any plane through the origin any line through the origin the zero vector

A. {(x,y,z) | x,y,z > 0}. Solution: This is not closed under scalar multiplication For example, (1,2,3) is in the set since all coordinates are positive, but -2(1,2,3) = (-2,-4,-6) is not in the set, since not all coordinates are positive. B. {(x,y,z) | x + 3y - 2z = 0}. Solution: This is the equation of a plane through the origin, so it is a subspace of R3. C. {(x,y,z) | 3x + 9y = 0, -6x -7z =0}. Solution: 3x + 9y = 0 and -6x -7z =0 are each planes through the origin, so their intersection is a line through the origin, so this is a subspace of R3. D. {(-3,y,z) | y, z arbitrary numbers }. Solution: This set is not closed under vector addition. For example, (-3, 1,1) and (-3, 2,2) are both in the set, but their sum (-6, 3,3) is not in the set, since its first component is not equal to -3. Hence it is not a subspace of R3.

E. {(-7x + 2y, 4x + 8y, 5x-8y) | x, y arbitrary numbers}. Solution1: E ={x(-7,4,5) +y(2,8,-8)}, so this set consists of all linear combinations of two vectors in R3. This is a plane in R3 which passes through the origin (when x =0 and y=0), so it is a subspace of R3.

Solution2: First form the augmented matrix

Next reduce this matrix to RREF. This gives

This tells us that (5x-8y) =

−9 8

(-7x+2y) +

−7 4 5 ) ( 2 8 −8

−9 8 1 0 −23 ) ( 0 1 32 −23 (4x 32

+ 8y).

In other words, this set consists of all ordered triples of the form (u, v, This is equivalent to the plane in R3 with equation 𝑥 + 8 9

so it is a subspace of R3.

23

32

−9 8

𝑢+

−23 v). 32

y + z = 0. This is a plane through the origin,

F. {(x,y,z) | 4x + 5y +2z = 7}. Solution: This is the equation of a plane, but it does not pass through the origin, so it is not a subspace of R3. It is not closed under vector addition, or scalar multiplication.

Assignment 9, #2. Express the vector v = (26, -7) as a linear combination of the vectors x = (-6,-5) and y = (5,-3).

Solution: The simplest way to do this is to form an augmented matrix using vectors x, y and v as columns. −6 5 26 ( ) −5 −3 −7

Now reduce this matrix to RREF (reduced row-echelon form). 1 0 −1 ( ) 0 1 4

This tells us that (26, -7) = -1(-6,-5) +4(5,-3). (You can check by hand that this final answer is correct.)

Assignment 9, #3. Let H be the set of all vectors of the form (s,s,s). Find a vector v in R3 such that H = Span{v}. Solution: First note that (s,s,s) = s(1,1,1). So any vector in H is a multiple of the vector (1,1,1). So the answer is v = (1,1,1).

Assignment 9, #4. Let W be the set of all vectors of the form (2s – 5t,4s + 5t, 3s+t, 2s – 4t). Find vectors u and v such that W =Span{u, v}. Solution: We can represent W as all vectors of the form s(2,4,3,2) + t(-5,5,1,-4). So if we let u = (2,4,3,2) and v=(-5,5,1-4) then W = Span{u,v}.

Assignment 9, #5. Let v1 = (-3,5,-1), v2=(-6,7,-6) and y = (9,-9,h). For what value of h is y in the plane spanned by v1 and v2. Solution: If y is in the plane spanned by v1 and v2, then it must be expressible as a linear combination of vectors v1 and v2. So the first step is to form the augmented matrix with columns v1, v2, and y. −3 −6 9 (5 7 −9) −1 −6 ℎ

Next we must reduce this matrix to RREF. This gives the following matrix: 1 0 1 (0 1 −2 ) 0 0 ℎ − 11

If you look at the last row of this augmented matrix, you see that there is no solution unless h = 11. If h does equal 11, then we have y = 1 v1 -2 v2. (You can confirm that this answer is correct by checking that (9,-9,11) = 1(-3,5,-1) -2(-6,7,-6). )

Assignment 9, #5. Let v1 = (-1,1,2), v2 = (-3, 4,6) and v3=(-3,5,8), and w = (8,10,3). 1. Is w in {v1, v2, v3}? Solution: This is asking whether w is equal to either v1, v2 , or v3. Clearly it is not, so the answer is no. 2. How many vectors are in {v1, v2, v3}? Solution: This is a set with three vectors, so the answer is 3. 3. How many vectors are in Span{v1, v2, v3}? Solution: The span of a set is the set of all linear combinations of vectors in that set, which in this case is infinitely many. 4. Is w in the subspace spanned by {v1, v2, v3}? Solution: The question is asking whether w can be expressed as a linear combination of vectors v1, v2, and v3 One way to do this is to form the augmented matrix using vectors v1, v2, v3, and w as columns of the augmented matrix.

Next reduce this matrix to RREF.

−1 −3 −3 8 4 5 10 (1 ) 2 6 8 3

1 0 0 −33.5 (0 1 0 −1 ) 0 0 1 9.5

This tells us that w = -33.5v1 -1v2 +9.5v3, so w is in the subspace spanned by v1, v2, and v3.

Assignment 9, #7. Which of the following sets of vectors are linearly independent? A. (6,4,0), (8,2,0), (-3,-5,0). Solution1: One way to do this is to form the augmented matrix using the three vectors as columns.

Next reduce this matrix to RREF.

6 8 −3 (4 2 −5 ) 0 0 0

−17 10 9 0 1 10 (0 0 0 ) 1 0

9 −17 (6,4,0) + 10 (8,2,0). 10

This tells us that (-3,-5,0) can be expressed as a linear combination of the first two vectors, i.e. (-3,-5,0) = Since one of the three vectors can be expressed as a linear combination of the

other two vectors, the set is linearly dependent.

Solution2: Another way to solve this problem is the form the matrix using the three vectors as columns. 6 8 −3 (4 2 −5 ) 0 0 0

Next find the determinant of this matrix. Since it has a row of zeros, the determinant is 0. Therefore, the three vectors must be linearly dependent, according to the equivalence Theorem in the textbook.

B. (-1,8) and (6,-1).

Solution: Since the second vector is not a multiple of the first vector, they are linearly independent.

C. (7, -5), (7, -5). Solution: Since the second vector is a multiple of the first vector, they are linearly dependent.

D. (0,0), (9,4). Solution: We have a Theorem which states that any set which contains the zero vector is linearly dependent. Since this set contains a zero vector, then it is linearly dependent.

E. (3, 0,-4),(6,-1,2), (-7,8,9) Solution: First form the matrix using the three vectors as columns.

3 6 −7 ( 0 −1 8 ) −4 2 9

Next find the determinant of this matrix. Since the determinant equals -239, which is not zero, the vectors are linearly independent. F. (8,6), (9,4), (7,5). Solution: Here we have three vectors in R2. Since 3 > 2, according to the Theorem, we can say, without any calculations, that these vectors are linearly dependent.

Assignment 9, #8. Are the vectors v1=(-2,-2,3,-3),v2=(1,-2,-1,3),v3=(-3,-2,2,3) and v4 =(7,2,2,-27) linearly independent? Solution: First form a matrix using these vectors as the columns of the matrix. −2 1 −3 7 −2 −2 −2 2 ) ( 3 −1 2 2 3 −27 −3 3

Next reduce this matrix to RREF.

1 0 ( 0 0

0 0 4 0 0) 1 0 1 −5 0 0 0

This means that v4 = 4v1 -5v3. Since we can express v4 as a linear combination of the other three vectors, they are linearly dependent. If we rearrange the above equation we obtain -4v1 + 0v2 +5v3 +1v4 = (0,0,0,0). Assignment 9, #9. Find two linearly independent vectors perpendicular to the vector v = (-8,5,7). Solution: Here there any many possible solutions. The vectors (7,0,8) and (5,8,0) are each perpendicular to v, since their dot product with v is zero Since there two vectors are not multiples of each other, they are linearly independent, so this is one possible solution.

Assignment 9, #10. Find a linearly independent set of vectors that spans the same subspace of R3 as that spanned by the vectors v1=(3,-3,2), v2=(5,-6,5) and v3=(1,0,-1). Solution: In this case, we find that (5,-6,5) =2(3,-3,2) -1(1,0,-1). This can be done by finding the RREF of the matrix formed by using the vectors as columns. Hence Span{v1,v2,v3} = Span{v1,v3}. Since v1 and v3 are not multiples of each other, they are linearly independent. Thus a linearly independent set is {v1, v3}. (Of course other answers are possible.)...