Math 214 — Midterm 1 Blue Version Answers by Michael Cap Khoury Problem 1: Let d, s, m be the numbers of sacks of Dragon, Serenity, and Miracle Tea that Datura buys, respectively. d+s+m=9 2d+3s+5m=25 25d+30s+20m=230 This system has a unique solution (d, s, m) = (6, 1, 2). Datura bought six sacks of Dragon Tea, one sack of Serenity Tea, and two sacks of Miracle Tea. Problem 2: We use the projection formula. Let L be the line spanned by 3 4 12 . projL −1 1 1 = −1 1 1 · 3 4 12 3 4 12 · 3 4 12 3 4 12 = 13169 3 4 12 = 3/13 4/13 12/13 Problem 3: There are infinitely many solutions. What all solutions have in common is that each column consists of cnumbers in arithmetic progression and that the five columns are not all a multiple of a common vector. An example follows. 1 1 1 1 1 1 2 3 4 5 1 3 5 7 9 1 4 7 10 13 Problem 4: Since T is compatible with scalar multiples, T parenleftbigg 1 0 parenrightbigg = 13 21 15 12 = 7 5 4 T parenleftbigg 0 1 parenrightbigg = 12 10 12 14 = 5 6 7 . This gives the matrix representing T, column-by-column: 7 5 5 6 −4 7 . 1 Problem 5: So many possibilities... B1= 1 0 0 , 0 1 0 , 0 0 1 B2= 1 1 2 , 1 2 1 , 2 1 1 B3= 91 423√ , 019√ 0 , pi 6 1 Problem 6: First, we need the rref of A. rref(A)= 1 0 −2 5 0 0 1 1 −1 0 0 0 0 0 1 . a) The solutions of AxM =0M have the form 2r−5s −r+s r s 0 ; a basis for kerA is 2 −1 1 0 0 , −5 1 0 1 0 . b) The third and fourth columns are redundant. A basis of im A is braceleftBigg 1 0 1 , 4 1 5 , 1 0 0 bracerightBigg . (Indeed the image is all of R3, so any basis of R3 will do.) c) The rank is 3. d) The nullity is 2. Problem 7: a) C =AB= −10 10 3 16 −5 3 −5 2 1 . b) A−1= 1 −4 −15 0 1 3 0 0 1 . c) B−1= 1 0 0 −1 1 0 7 −2 1 . d) C−1=B−1A−1= 1 0 0 −1 1 0 7 −2 1 1 −4 −15 0 1 3 0 0 1 = 1 −4 −15 −1 5 18 7 −30 −110 Problem 8: a) False b) True c) True d) False e) True f) True g) True h) True i) False j) True 2 /home/cap/.TeXmacs/system/tmp/tmp_1087425880.ps