Mathematics > QUESTION PAPER (QP) > University of TorontoBMAT 301BMidterm solutions 2015 summer (All)
University of Toronto - St. George MAT301 - Midterm solutions - Wednesday, June 17, 2015 110 minutes Make sure you have 7 pages. Student’s Information (please PRINT) Surname: First Name: Stud ... ent Number: PUT INSTRUCTIONS HERE. Q Mark Question 1 /20 Question 2 /10 Question 3 /10 Question 4 /20 Question 5 /20 Question 6 /20 Total /100 2 Question 1. (20 marks) Let G = 8<: 0@ a b c d 1A : a + d = b + c; a; b; c; d 2 R 9=; Show that G is a group under (the usual) matrix addition. Solution: Notice that matrix addition is associative (2nd year linear algebra). Given A; B 2 G, then A = 0 @x y z w1 A ; B = 0 @xz11 wy111 A for some x; y; z; w; x1; y1; z1; w1 2 R with x + w = y + z; x1 + w1 = y1 + z1 ) (x + x1) + (w + w1) = (y + y1) + (z + z1). That is, A + B = 0 @x y z w1 A + 0 @xz11 wy111 A = 0 @xz ++ xz11 wy ++ wy111 A 2 G This proves closure. Notice that E = 0 @0 0 0 01 A 2 G Moreover, A + E = A = E + A for all A 2 G. Hence, E is the identity element. Given A 2 G, then A = 0 @x y z w1 A for some x; y; z; w 2 R with x + w = y + z ) -x - w = -y - z. That is, -A 2 G. Moreover, A-1 = -A (since we are under addition and A + (-A) = E = (-A) + A). 3 Question 2. (10 marks) Suppose x 2 G (G a group) and jxj = 7. Show that x is the cube of some element in G. That is, there exists a 2 G such that x = a3. Solution: Since jxj = 7 ) x7 = e ) x7k = e for any integer k. Let a = x5. Then a3 = (x5)3 = x15 = x14x = ex = x [Show More]
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