Linear Algebra 2270-2Due in Week 3The third week finishes the first half of the work from chapter 2. Here’s the list of problems, followed byproblem notes and a few answers.Section 2.1. Exercises 9, 10, 21, 29, 34Section 2.2. Exercises 5, 9, 12, 13, 32Section 2.3. Exercises 3, 13, 19, 25, 27, 28Problem NotesResolution of issues for Strang’s problems will appear here in response to communications. If there is adifficulty, like a missing definition or incomplete information, then please write email, call 581-6879, or visitJWB 113.2.1-29. A matrix A is Markov if each column sum is one and the entries are positive. It is known that A~x haspositive components that add to one, provided the same is true for ~x. An important example is the marketshare matrix A for three telephone companies ATT, MCI, SPRINT. It helps to solve the problem by mapleassist.Some Answers2.1. Exercises 29, 34 have answers in the textbook. Reproduced here.2.1-9. (a) Ax = (18, 5, 0) and (b) Ax = (3, 4, 5, 5).2.1-10. Multiplying as linear combinations of the columns gives the same Ax. By rows or by columns: 9separate multiplications for 3 by 3.2.1-21. R =1√2 1 −11 1!. The entries are sines and cosines of 45 degrees.2.1-29. u2= .7.3!and u3= .65.35!. The components add to 1. They are always positive. u7, v7, w7areall close to .6.4!. Their components still add to 1.2.1-34. A =2 −1 0 0−1 2 −1 00 −1 2 −10 0 −1 2and A~x =1234has solution ~x =4786.2.2. Exercises 5, 9, 12, 13, 32 do not have textbook answers. Below for answers.2.2-5. 6x + 4y is 2 times 3x + 2y. There is no solution unless the right side is 2 × 10 = 20. Then all thepoints on the line 3x + 2y = 10 are solutions, including (0, 5) and (4, −1). (The two lines in the row pictureare the same line, containing all solutions).2.2-9. On the left side, 6x − 4y is 2 times 3x − 2y. Therefore we need b2= 2b1on the right side. Then therewill be infinitely many solutions (two parallel lines become one single line).2.2-12. Elimination first gives an upper triangular system2x + 3y + z = 8y + 3z = 48z = 8Back-substitution sorts out the answer.2.2-13. Whatever the elimination details, x = 3, y = 1, z = 0.2.2-32. The question deals with 100 equations Ax = 0 when A is singular. (a) Some linear combination ofthe 100 rows is the row of 100 zeros.(b) Some linear combination of the 100 columns is the column of zeros.(c) A very singular matrix has all ones: A = eye(100). A better example has 99 random rows (or the numbers1i, . . . , 100iin those rows). The 100th row could be the sum of the first 99 rows (or any other combination ofthose rows with no zeros).(d) The row picture has 100 planes meeting along a common line through 0. The column picture has 100vectors all in the same 99-dimensional hyperplane.2.3. Exercises 3, 25, 27, 28 have a textbook answer.2.3-13. (a) E times the third column of B is the third column of EB. A column that starts at zero will stayat zero. (b) E could add row 2 to row 3 to change a zero row to a nonzero row.2.3-19. Any swap elementary matrix S satisfies S2=