Ryan Ren Assignment Homework 2 due 09/14/2017 at 02:15pm PDT 1. (1 pt) Compute the following product. \u0014 \u0015\u0014 \u0015 \u0014 5 0 4 7 4 = 0 7 1 7 7 Math4A-02-M17-Wen \u0015 Answer(s) submitted: 20 (correct) 2. (1 pt) Compute the following product. 3 0 0 0 3 3 6 1 0 9 0 0 7 6 5 2 0 6 0 7 1 5 9 0 0 0 0 3 7 3 1 1 = 6. (1 pt) Compute the following products. 2 3 6 1 1 5 7 5 = 9 8 9 6 2 3 6 8 1 5 7 8 = 9 8 9 4 2 3 6 1 8 5 8 = 1 5 7 9 8 9 6 4 Answer(s) submitted: -19 16 -19 Answer(s) submitted: 9 (correct) (correct) 7. (1 pt) If \u0014 A= 3. (1 pt) Compute the following product. 8 0 0 9 1 5 0 1 0 6 9 4 = 0 0 6 3 7 8 1 2 8 3 \u0015 , then \u0014 A1 = \u0015 Answer(s) submitted: Answer(s) submitted: -72 (correct) 4. (1 pt) Solve for X. 3/19 8/19 2/19 -1/19 (correct) \u0014 4 3 5 8 7 4 \u0014 \u0015 \u0014 = 4X + 5 3 6 4 3 7 8 8. (1 pt) If \u0015 . 3 A= 1 1 \u0015 X= 2 0 , 1 then Answer(s) submitted: 19/4 (correct) 5. (1 pt) Compute the following product. \u0015 6 6 \u0014 3 9 9 0 = 0 6 5 2 6 1 2 A1 = Answer(s) submitted: Answer(s) submitted: -54 (correct) 1 -1 2 2 -1 1 2 -1 0 3 (correct) 5 -10 9. (1 pt) If \u0014 A= 1 8 8 9 \u0015 (correct) 12. (1 pt) If then \u0015 \u0014 1 A = 1 A = 3 1 Answer(s) submitted: 9/73 -8/73 -8/73 -1/73 then (correct) 1 7 4 21 , 1 6 A1 = . 4 Given ~b = 1 , solve A~x = ~b. 2 10. (1 pt) Consider the following two systems. (a) \u001a x + y 3x + 3y = 3 = 2 x + y 3x + 3y = 3 = 4 ~x = . (b) \u001a Answer(s) submitted: -3 -1 (i) Find the inverse of the (common) coefficient matrix of the two systems. A1 (correct) 13. (1 pt) The 2 2 elementary matrix E can be gotten from the identity matrix using the row operation R1 = r1 + 2r2 . Find EA if \u0014 \u0015 1 2 A= . -4 5 \u0015 \u0014 = (ii) Find the solutions to the two systems by using the inverse, 1 i.e. by evaluating \u0014 \u0015A B where B represents\u0014the right \u0015 hand side 3 3 (i.e. B = for system (a) and B = for system 2 4 (b)). Solution to system (a): x = ,y= Solution to system (b): x = ,y= Answer(s) submitted: Answer(s) submitted: -1/2 1/6 1/2 1/6 11/6 -7/6 -13/6 5/6 (incorrect) \u0014 \u0015 1 8 . 14. (1 pt) Let X = 9 9 Give an example of two 2 2 matrices A and B, neither of which is the identy matrix I, such that AB = X. \u0014 \u0015 A= , \u0014 \u0015 B= (correct) 11. (1 \u0014pt) Find the \u0015 inverse of AB \u0014 if \u0015 -1 2 5 4 1 1 A = and B = . 1 -5 -5 0 (AB)1 \u0014 = \u0015 \u0014 EA = \u0015 Answer(s) submitted: 1 Answer(s) submitted: -1 -10 (incorrect) 2 17. (1 pt) Given the augmented matrix 1 3 6 1 2 3 , A = 2 5 3 5 1 5 15. (1 pt) On the augmented matrix A below , perform all three row operations in the order given, ((a) followed by (b) followed by (c)) and then write the resulting augmented matrix. 1 2 3 A = 4 7 5 2 1 3 perform each row operation in the order specified and enter the final result. 5 1 5 First: 2R1 + R2 R2 , Second: 3R1 + R3 R3 , Third: 4R2 + R3 R3 . (a)R2 = 4r1 + r2 (b)R3 = 2r1 + r3 (c)R3 = 3r2 + r3 Answer(s) submitted: Answer(s) submitted: 1 1 -2 -3 -5 0 1 7 19 0 0 12 42 (correct) 18. (1 pt) If \u0014 A= then A1 = 1 0 3 3 \u0015 , \u0015 . \u0014 Answer(s) submitted: -1 (correct) (correct) 19. (1 pt) If 4 A= 0 0 16. (1 pt) Given the augmented matrix 1 3 5 3 13 4 3 , A= 4 3 11 2 1 then perform each row operation in the order specified and enter the final result. 9 6 6 8 , 0 8 A1 = . Answer(s) submitted: 1/4 First: R2 = 4r1 + r2 , Second: R3 = 3r1 + r3 , Third: R3 = 2r2 + r3 . (correct) 20. (1 pt) If 8 A = 3 7 then A1 = . Answer(s) submitted: Answer(s) submitted: 1 1/8 (correct) (correct) 3 0 0 4 0 , 5 4 \u0014 \u0015 2 5 25. (1 pt) Given the matrix 3 7 (a) does the inverse of the matrix exist? Your answer is (input Yes or No) : (b) if your answer is yes, write the inverse here: \u0014 \u0015 21. (1 pt) If \u0014 A= then A1 1 8 8 9 \u0015 , \u0015 . \u0014 = . Answer(s) submitted: Answer(s) submitted: Yes -7 5 3 -2 9/73 (correct) 22. (1 pt) If \u0014 A= then A1 = 1 1 6 7 \u0015 (correct) , 4 4 26. (1 pt) If A = 2 3 1 1 3 2 1 2 4 4 , then 2 2 2 \u0015 . \u0014 Given ~b = \u0014 ~x = \u0014 \u0015 1 , solve A~x = ~b. 2 \u0015 . A+B = 0 3 and B = 0 Answer(s) submitted: Answer(s) submitted: 7 -7 -19 (correct) (correct) 27. (1 pt) Evaluate the following matrix product. \u0015 2 4 \u0014 1 3 4 = 3 4 2 23. (1 pt) If 8 0 A= 0 0 then 0 0 0 3 0 0 , 0 5 0 0 0 9 Answer(s) submitted: 10 (correct) A1 = . 28. (1 pt) Evaluate the following matrix product. \u0014 \u0015 3 3 4 2 1 1 = 1 1 4 0 Answer(s) submitted: Answer(s) submitted: 9 -1/8 (correct) (correct) 24. (1 pt) Find a 2 2 matrix A such that \u0014 2 2 5 4 \u0015\u0014 \u0015 29. (1 pt) Let A be a 5 by 9 , B be a 9 by 9 and C be a 9 by 5 matrix. Determine the size of the following matrices (if they do not exist, type N in both answer boxes): \u0014 = 1 0 0 1 \u0015 . AB: BA: AT B: BC: ABC: Answer(s) submitted: -2/9 (correct) 4 by by by by by by by by CA: BT A: BCT : 31. (1 pt) Determine whether the product Ax is defined or undefined. Answer(s) submitted: 8 10 6 ? 1. A = 5 7 , x = 6 1 3 7 6 7 1 1 9 6 5 3 8 , x = ? 2. A = 2 9 7 6 7 7 2 2 3 7 9 3 0 7 , x = 3 ? 3. A = 3 6 1 6 1 6 5 6 2 \u0002 \u0003 ? 4. A = 6 6 2 4 4 , x = 3 1 6 8 \u0014 \u0015 6 10 1 2 ? 5. A = ,x= 4 7 10 1 8 5 9 N N N N 9 5 5 5 9 9 N N N N (correct) 30. (1 pt) Determine whether the product Ax is defined or undefined. 7 10 10 10 10 10 8 , x = 8 A= 8 5 6 4 1 1 10 8 0 2 3 1 6 A = 6 5 3 6 , x = 6 3 8 8 5 4 2 3 6 A = 4 0 , x = 3 0 3 5 10 \u0014 \u0015 8 1 8 10 A= ,x= 4 5 2 2 2 7 3 \u0002 \u0003 A = 4 9 7 4 2 , x = 9 9 4 ? 1. ? 2. ? 3. ? 4. ? 5. Answer(s) submitted: (correct) 32. (1 pt) Compute the following product. 9 7 7 5 6 4 8 7 = 4 10 4 4 Answer(s) submitted: -24 (correct) 33. (1 pt) Reduce the matrix 1 3 5 A = 4 3 20 2 0 10 to reduced row-echelon form. Answer(s) submitted: Undefined Defined Defined Defined Undefined Defined Defined Undefined Undefined Defined Answer(s) submitted: (correct) 1 5 (correct) 34. (1 pt) Reduce the matrix \u0014 2 3 A= 3 1 6 5 37. (1 pt) Given the matrix 2 1 \u0015 7 , 3 to reduced row-echelon form. (a) Does the inverse of the matrix exist? ? (b) If your answer is yes, write the inverse here: \u0014 \u0015 \u0015 \u0014 Answer(s) submitted: Answer(s) submitted: 1 (correct) 35. (1 pt) Reduce the matrix 3 3 A = 2 1 2 1 1 0 3 (correct) 26 4 2 38. (1 pt) Solve for X. to reduced row-echelon form. \u0014 3 4 9 4 \u0015 \u0014 9 2 X+ 1 9 \u0015 \u0014 5 7 = \u0015 X. X= Answer(s) submitted: (incorrect) 1 39. (1 pt) Solve for X. (correct) \u0014 6 , 5 9 4 2 8 \u0015 \u0014 X+ \u0014 6 4 4 8 \u0015 \u0014 = \u0015 X= Answer(s) submitted: (a) does the inverse of the matrix exist? ? (b) if your answer is yes, write the inverse here: \u0014 \u0015 (incorrect) 40. (1 pt) If \u0014 A= Answer(s) submitted: 5 6 \u0015 \u0014 Answer(s) submitted: 36. (1 pt) Given the matrix 5 5 Yes -3 7 1 -2 Yes 1 6/5 -1 -1 then A1 = \u0014 Answer(s) submitted: -1/2 (correct) (correct) 6 \u0015 . 2 0 0 7 \u0015 , 5 5 3 9 \u0015 . (correct) 41. (1 pt) If 3 0 A= 0 0 then A1 4 7 1 2 0 4 0 0 42. (1 pt) If 2 2 , 4 3 0 A= 0 1 then = . A1 = . Answer(s) submitted: 1 Answer(s) submitted: 1/3 (correct) c Generated by WeBWorK, http://webwork.maa.org, Mathematical Association of America 7 1 1 , 0 0 1 1 Ryan Ren Assignment Homework 3 due 09/14/2017 at 02:15pm PDT 1. (1 pt) Given the matrix 3 3 4 1 2 1 3 Math4A-02-M17-Wen 4. (1 pt) Given the matrix 2 0 3 0 0 2 5 3 1 3 1 find its determinant. The determinant is : find its determinant. Do not use a calculator. (WeBWorK will know if you do!) The determinant is : Answer(s) submitted: 6 Answer(s) submitted: (correct) 36 (correct) 2. (1 pt) Given the matrix 2 a a A= 4 8 0 5. (1 pt) If A = 1 0 then det (A) = 2 0 0 1 Answer(s) submitted: 9 a 6 1 0 0 -1 (correct) 1 6. (1 pt) If A = 0 k then det (A) = find all values of a that make the |A| = 0. Give your answers in increasing order. a can be , , or . 0 0 1 0 0 1 Answer(s) submitted: 1 Note: Leave any unneeded answer spaces blank. (correct) Answer(s) submitted: 1/8-(sqrt(2113)/8) 1/8+(sqrt(2113)/8) 7. (1 pt) If A = (correct) 3. (1 pt) Given the matrix 4 0 0 0 0 2 9 0 0 0 4 6 2 0 0 3 6 6 1 0 then det (A) = 3 1 A= Answer(s) submitted: -648 3 (correct) 3 k 8. (1 pt) If A = 0 0 then det (A) = find its determinant; The determinant of A is Answer(s) submitted: Answer(s) submitted: -12 k (correct) (correct) 1 0 0 1 0 0 1 8 4 7 6 9 (correct) 1 9. (1 pt) If A = 0 0 then det (A) = 0 k 0 0 0 1 5 14. (1 pt) If A = 5 2 Answer(s) submitted: k 3 9 0 3 10. (1 pt) If A = 0 0 0 0 then det (A) = 4 1 then det (A) = 5 Is A invertible? (correct) 2 2 0 9 8 2 0 A. Yes B. No 4 6 3 7 Answer(s) submitted: -120 A Answer(s) submitted: -126 (correct) (correct) a b c 15. (1 pt) If det d e f = 1 g h i a b c then det 2d + a 2e + b 2 f + c = g h i 11.(1 pt) Find the determinantof the matrix 1 0 0 2 0 1 0 1 0 0 0 3 M = 0 3 0 0 0 0 1 2 0 3 1 0 0 det (M) = . Answer(s) submitted: -2 Answer(s) submitted: -3 (correct) (incorrect) 12. (1 pt) Consider the following Gauss elimination: a b c (1 pt) If det d e f = 5 16. 1 0 6 0 0 1 1 0 0 1 0 0 2 4 A 0 1 0 A 0 1 0 E1 A 0 7 0 E2 E1 A 0 1 0 E3 Eg2 E1hA =i 0 9 c 0 0 1 1 0 0 0 0 1 0 0 5a b 0 0 | {z } | {z } | {z } | det{z 6d } 6e 6 f = then E1 E2 E3 E4 g h i 6 5 7 Answer(s) submitted: What is the determinant of A? det(A) = -30 Answer(s) submitted: -1 (correct) (incorrect) 17. (1 pt) The determinant of the matrix 13. (1 pt) Given the matrix 2 0 3 0 4 3 5 0 2 2 1 A= 0 5 6 0 6 5 0 0 0 8 5 0 0 8 is . Hint: Find a good row or column and expand by minors. (a) its determinant is: Answer(s) submitted: (b) does the matrix have an inverse? ? 240 Answer(s) submitted: -76 yes (correct) 2 A 5 -4 -4 -6 19. (1 pt) Let x = -6 and y = -4 . 0 -5 Findthe vectors v = 2x, u = x + y, and w = 2x + y. v = (correct) 22. (1 pt) The general solution to a linear system is given. Express this as a linear combination of vectors. , u = x1 = 3 1s1 , w = , Answer(s) submitted: \u0015 \u0014 \u0015 \u0014 x1 = x2 \u0014 + x2 = 7 + 1s1 \u0015 s1 Answer(s) submitted: -8 -12 0 -10 -10 -5 -14 -18 -5 3 -7 -1 1 (correct) 5 20 36 23. (1 pt) Let v1 = -4 , v2 = -17 , v3 = -29 4 16 28 7 and w = 10 . 7 1. Is w in {v1 , v2 , v3 }? Type \"yes\" or \"no\". (score 0.8888888955116272) 20. (1 pt) 2 4 22 Let a1 = 3 , a2 = 1, and b = 13 . 3 4 24 Is b a linear combination of a1 and a2 ? A. Yes, b is a linear combination. B. No, b is not a linear combination. C. We cannot tell if b is a linear combination. Either fill in the coefficients of the vector equation, or enter \"NONE\" if no solution is possible. b= a1 + a2 2. How many vectors are in {v1 , v2 , v3 }? Enter \"inf\" if the answer is infinitely many. 3. How many vectors are in Span {v1 , v2 , v3 }? Enter \"inf\" if the answer is infinitely many. 4. Is w in the subspace spanned by {v1 , v2 , v3 }? Type \"yes\" or \"no\". Answer(s) submitted: Answer(s) submitted: B NONE NONE (correct) no 3 inf yes (correct) 21. (1 pt) 1 2 3 Let a1 = 5, a2 = 12, and b = 23 . 1 5 15 Is b a linear combination of a1 and a2 ? A. Yes b is a linear combination. B. b is not a linear combination. C. We cannot tell if b is a linear combination. Either fill in the coefficients of the vector equation, or enter \"NONE\" if no solution is possible. b= a1 + a2 24. (1 pt) Let H be the set of all vectors of the form 5t 0 . Find a vector ~v in R3 such that H = span {~v}. 4t ~v = . Answer(s) submitted: -5 (correct) Answer(s) submitted: 3 pt) Let W be the set of all vectors of the form 25. (1 2a + 2b a . Find vectors ~u and ~v in R3 such that W = b span {~u,~v}. , ~v = ~u = 29. (1 pt) Find a set of vectors {~u,~v} in R4 that spans the solution set of the equations \u001a w x + y 2z = 0, 2w + 2x + y + 2z = 0. . Answer(s) submitted: 2 ~u = (correct) , ~v = , ~v = . 1 (incorrect) 30. (1 pt) Thevectors 5 -3 -2 v = 3 , u = 0 , and w = -1 . -12 1 +k 5 are linearly independent if and only if k 6= . . Answer(s) submitted: 4 (correct) Answer(s) submitted: -1 (1 pt) Let H be the set of all vectors of the form 27. 5s s . Find a vector ~v in R3 such that H = span {~v}. 5s ~v = Answer(s) submitted: 26. (1 pt) Let W be the set of all vectors of the form 4s 2t 5t 3s . Find vectors ~u and ~v in R3 such that W = 2s + 4t span {~u,~v}. ~u = (incorrect) 31. (1 pt) Let u, v, w be three linearly independent vectors in R7 . Determine a value of k, . k= , so that the set S = {u 4v, v 5w, w ku} is linearly dependent. Answer(s) submitted: -5 (correct) Answer(s) submitted: 0 28. (1 pt) Show that the vectors 1 1 1 2 , 3 , 4 1 1 1 (incorrect) 32. (1 pt) Let u, v, w be three linearly independent vectors in R7 . Determine a value of k, do not span R3 by giving a vector not in their span. k= , so that the set S = {u 5v, v 8w, w ku} is linearly dependent. Answer(s) submitted: Answer(s) submitted: 0 (incorrect) (incorrect) 4 1 33. (1 pt) If A = 0 0 then det (A) = 0 1 k A. The equation has nontrivial solutions. B. The equation has no nontrivial solutions. C. We cannot tell if the equation has nontrivial solutions or not. 0 0 1 Answer(s) submitted: Answer(s) submitted: 1 3/2 B (correct) 7 34. (1 pt) If A = 2 7 then det (A) = 0 8 6 (correct) 0 0 2 39. (1 pt)\u0014 \u0015 3 3 Let A = . 5 5 We want to determine if the equation Ax = 0 has nontrivial solutions. To do that we row reduce A. times the first row to the second. To do this we add We conclude that Answer(s) submitted: 112 (correct) 35.(1 pt) Find the determinant of the matrix 3 3 3 3 1 3 3 0 C= 3 2 2 2 2 1 2 1 det (C) = . A. The equation has nontrivial solutions. B. The equation has no nontrivial solutions. C. We cannot tell if the equation has nontrivial solutions or not. Answer(s) submitted: -219 Answer(s) submitted: (correct) -5/3 A 36. of the matrix (1 pt) Find the determinant 6 0 0 0 2 6 0 0 A= 9 5 5 0 4 5 5 3 det (A) = . (correct) 40. (1 pt)\u0014 \u0015 \u0014 \u0015 2 10 Let u = , and v = . 2 10 We want to determine by inspection (with minimal computation) if {u, v} is linearly dependent or independent. Choose the best answer. Answer(s) submitted: -540 (correct) 37. (1 pt) A square matrix is called a permutation matrix if each row and each column contains exactly one entry 1, with all otherentries being0. An example is 1 0 0 P= 0 0 1 0 1 0 Find the determinant of this matrix. det (P) = . A. The set is linearly dependent because the number of vectors in the set is greater than the dimension of the vector space. B. The set is linearly dependent because one of the vectors is the zero vector. C. The set is linearly dependent because one of the vectors is a scalar multiple of another vector. D. The set is linearly independent because we only have two vectors and they are not scalar multiples of each other. E. The set is linearly dependent because two of the vectors are the same. F. We cannot easily tell if the set is linearly dependent or linearly independent. Answer(s) submitted: -1 (correct) 38. (1 pt)\u0014 \u0015 2 10 . 3 14 We want to determine if the equation Ax = 0 has nontrivial solutions. To do that we row reduce A. To do this we add times the first row to the second. We conclude that Let A = Answer(s) submitted: C (correct) 5 c Generated by WeBWorK, http://webwork.maa.org, Mathematical Association of America 6 Ryan Ren Assignment Homework 4 due 09/14/2017 at 02:15pm PDT 1. (1 pt) Consider the subspace 1 3 U = span{ 2 , 0 5 3 1 4 , 7 8 1 4 4. (1 pt) Find a basis of the subspace of R3 defined by the equation 6x1 + 5x2 + 8x 3 = 0. } 1 1 2 0 3 { 3 , 1 3 3 x= . (correct) , Answer(s) submitted: of R4 . Create a basis for U. Math4A-02-M17-Wen 5. (1 pt) Find a linearly independent set of vectors that spans 4 the by the vectors same subspace of R as that spanned -4 -2 -4 2 -1 -1 -3 2 11 , 3 , 1 , 2 . 2 0 -2 2 , x} Linearly independent set: , . Answer(s) submitted: Answer(s) submitted: -3 -4 (correct) (correct) 6. (1 pt) Find a linearly independent set of vectors that spans the same of R3as that spanned by the vectors subspace 3 -6 0 1 , 2 , 2 . 2 -2 1 2. (1 pt) Find a basis of the subspace of R4 spanned by the following vectors: 3 3 -6 -3 1 2 0 1 8 , 16 , -6 , -4 . 3 3 -6 -3 , Linearly independent set: . 3 (correct) -3 7. (1 pt) Find a linearly independent set of vectors that spans 4 the same subspace of R asthat spanned by the vectors -4 -1 7 2 4 3 -10 -3 -2 , 0 , 4 , 2 . -3 -2 8 3 (correct) 3. (1 pt) Find a basis of the subspace of R4 defined by the equation 7x1 7x2 + 8x 4x4 = 0. 3 , , . Answer(s) submitted: Answer(s) submitted: , . Linearly independent set: Answer(s) submitted: Answer(s) submitted: 1 -4 (correct) (correct) 1 , , . 8. (1 pt) Determine if the set of vectors is a basis of R4 . If not, determine spanned by the subspace dimension of the the -7 3 -5 -6 0 1 -4 2 vectors. 5 -1 -1 -6 -8 5 -3 -2 The dimension of the subspace spanned by the vectors is 12. (1 pt) Let A= (correct) 9. (1 pt) Let 12 6 2 A= 3 0 4 16 4 0 3 1 3 7 3 1 . 4 0 9 1 Find a basis for the column space of A. Answer(s) submitted: 4 2 5 2 2 5 9 3 . 6 ( , ) . Answer(s) submitted: 2 Find a non-zero vector in the column space of A. (correct) 4 2 13. (1 pt) Let A = 6 3 0 0 Find a basis of nullspace(A). Answer(s) submitted: (incorrect) 10. (1 pt) Let 5 A= 5 4 2 5 2 5 4 4 3 1 1 (correct) -1 -6 1 3 14. (1 pt) Let A = 0 3 -2 -9 Find a basis of nullspace(A). . Answer(s) submitted: (incorrect) , -2 2 0 -4 4 -2 . -2 6 . Answer(s) submitted: -2 11. (1 pt) Let W be the set of all vectors of the form 5r + 3s + 5t 3r + 4s 3t 2r 4s t 4r + 2s 2t (correct) 15. (1 pt) Find a basis of the column space of the matrix 1 2 4 0 1 4 4 1 A= 2 8 8 2 . with r, s and t real. Find a matrix A such that W = Col(A). A= . Answer(s) submitted: 1/2 4 3 . 4 Give a non-zero vector ~x in the null space of A. ~x = , 8 12 0 . , . Answer(s) submitted: 5 Answer(s) submitted: 1 (correct) (correct) 2 Answer(s) submitted: 16. (0 pts) Find bases for the column space, the row space, and the null space of matrix A. You should verify that the RankNullity Theorem holds. 1 -3 3 A = -3 11 -11 3 -13 13 Basis for the column space of A = Basis for the row space of A = Basis for the null space of A = (incorrect) 3 3 19. (1 pt)If T :R R is alineartransformation such that 1 -2 0 3 T 0 = -3 ,T 1 = 0 , 0 1 -3 0 0 0 and T 0 = -4 , 1 1 -3 . then T -5 = -4 Answer(s) submitted: Answer(s) submitted: (incorrect) \u0014 \u0015 \u0014 \u0015 3 -1 20. (1 pt) Let v1 = and v2 = . 4 -1 Let T : R\u00142 R\u00152 be the linear \u0014transformation satisfying \u0015 -38 10 T (v1 ) = and T (v2 ) = . -15 6 \u0014 \u0015 x Find the image of an arbitrary vector . y \u0014 \u0015 \u0014 \u0015 x T = . y (incorrect) 17. (1 pt) Find bases for the column space, the row space, and the null space of matrix A. You should verify that the RankNullity Theorem holds. 1 5 -1 1 A = 3 18 0 6 4 26 2 10 Basis for the column space of A = Answer(s) submitted: Basis for the row space of A = Basis for the null space of A = (incorrect) 21. (1 pt) Let 6 A= 0 0 0 0 9 0 . 0 1 3 3 Define the linear transformation T :R R by T (~x) = A~x. 4 a Find the images of ~u = 4 and ~v = b under T . 4 c Answer(s) submitted: 1 1 6 (correct) T (~u) = T (~v) = 18. (1 pt) Let ~e1 = (1, 0), ~e2 = (0, 1), ~x1 = (7, 9) and ~x2 = (1, 1). Let T : R2 R2 be a linear transformation that sends ~e1 to ~x1 and ~e2 to ~x2 . If T maps (2, 6) to the vector ~y, then Answer(s) submitted: ~y = . (Enter your answer as an ordered pair, such as (1,2), including the parentheses.) (incorrect) 3 22. (1 pt) Let 0 0 A= 0 9 9 0 25. (1 pt) Let T : R3 R3 be the linear transformation defined by 6 0 . 0 T (x1 , x2 , x3 ) = (x1 x2 , x2 x3 , x3 x1 ). Find a vector ~w R3 that is not in the image of T . 3 3 Define the linear transformation T :R R by T (~x) = A~x. 2 a Find the images of ~u = 4 and ~v = b under T . 4 c T (~u) = T (~v) = ~w = Answer(s) submitted: (incorrect) \u0014 \u0015 8 6 2 26. (1 pt) Let A = 4 3 1 Find bases of the kernel and image of A (or the linear transformation T (x) = Ax). Answer(s) submitted: , Kernel: (incorrect) 23. (1 pt) Let \u0014 A= 6 0 0 7 \u0014 Image: \u0015 . R2 . \u0015 . Answer(s) submitted: R2 Define the linear transformation T : \u0014 \u0015 by T (~x) = A~x. \u0014 \u0015 4 a Find the images of ~u = and ~v = under T . 1 b \u0015 \u0014 T (~u) = \u0014 \u0015 T (~v) = -3/4 8 (correct) 6 27. (1 pt) Let A = 2 4 Find a basis of the image T (x) = Ax). Answer(s) submitted: , (incorrect) 2 1 3 of A 4 3 1 (or the linear transformation . Answer(s) submitted: 24. (1 pt) Let 1 A= 0 3 3 1 8 6 1 1 2 and ~b = 3 . 1 0 (correct) \u0014 \u0015 6 8 . 3 4 Find bases of the kernel and image of A (or the linear transformation \u0014T (x) = Ax). \u0015 28. (1 pt) Let A = Define the linear transformation T : R3 R3 by T (~x) = A~x. Find a vector ~x whose image under T is ~b. ~x = . Kernel: \u0014 Image: . Is the vector ~x unique? ? . \u0015 . Answer(s) submitted: Answer(s) submitted: 4/3 -6 (correct) (incorrect) 4 3 1 0 1 29. (1 pt) Let A = 1 2 0 5 6 2 0 2 Find dimensions of the kernel and image of A (or the linear transformation T (x) = Ax). , dim(Ker(A)) = . dim(Im(A)) = \u0014 \u0015 Kernel: Image: . . Answer(s) submitted: 1/2 -4 (correct) Answer(s) submitted: 4 2 31. (1 pt) Let A = 0 0 0 0 Find dimensions of the kernel and image of A (or the linear transformation T (x) = Ax). , dim(Ker(A)) = dim(Im(A)) = . 2 2 (correct) 4 2 30. (1 pt) Let A = 8 4 8 4 Find bases of the kernel and image of A (or the linear transformation T (x) = Ax). Answer(s) submitted: 1 1 (correct) c Generated by WeBWorK, http://webwork.maa.org, Mathematical Association of America 5