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# Calculus II MATH 1502

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This 0 page Class Notes was uploaded by Chelsea Nolan MD on Monday November 2, 2015. The Class Notes belongs to MATH 1502 at Georgia Institute of Technology - Main Campus taught by Doron Lubinsky in Fall. Since its upload, it has received 12 views. For similar materials see /class/233950/math-1502-georgia-institute-of-technology-main-campus in Mathematics (M) at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

MATH1502 Homework 3 due Thursday April 2 2009 Question 1 Consider the following four matrices A B C D 0126 A 201517 0 1 1 B 1 4 0 2 71 2 C 5 i1 0 i4 71 72 D 7 0 3 1 Which of the 16 possible products AAABACAD make sense Com pute the products that do make sense Solution First let us consider multiplication in which A is the matrix on the left Ais2by4 Bis3by3 Cis1by4 Dis2by2 i AA is not de ned 2 X 4 2 X 4 ii AB is not de ned 2 X 4 3 X 3 iii AC is not de ned 2 X 4 1 X 4 iv AD is not de ned 2 X 4 2 X 2 Second let us consider multiplication in which B is the matrix on the left i BA is not de ned 3 X 3 2 X 4 ii BB is de ned 3 X 3 3 X 3 331115113151 12712112712 1i 13i1 137461 iii BC is not de ned 3 X 3 1 X 4 iv BD is not de ned 3 X 3 2 X 2 Third let us consider multiplication in Which C is the matrix on the left i CA is not de ned 1 X 4 2 X 4 ii CB is not de ned 1 X 4 3 X 3 iii CC is not de ned 1 X 4 1 X 4 iv CD is not de ned 1 X 4 2 X 2 Fourth let us consider multiplication in Which D is the matrix on the left i DA is de ned 2 X 2 2 X 4 71 72 0 1 2 6 0 3 2 0 1 5 i4 71 74 716 6 0 3 15 ii DB is not de ned 2 X 2 3 X 3 iii DC is not de ned 2 X 2 1 X 4 iv DD is de ned 2 X 2 2 X 2 DD H m 3 H 3 Question 2 Let 11 be real numbers and 3 1 Ho alkl a Compute AB 10 Show that A is always invertible and nd A71 1 cube w 1 E 0 7 Solution a 3 a l b 7 3 AB 107311031 1 3b 7 g 7 0 1 b We want a matrix A 1 such that 1 7 1 0 7 1 AA 7 0 1 7 A A We see that B almost does half the job Indeed if we choose I so that 31 7 3 0 a b 9 3 9 then AB I So let us set I Thus 1 2 B 1 a t 1 0 3 We hope that BA I also Now BA Question 3 Let 7 sin2 0 cos2 0 50 1 1 Use the formula for the inverse of a 2 by 2 matrix to compute S 071 Solution Wesee a b 869 C d where 11 b1 c sin20 dcos2 Then adibccos20sin0210 so the inverse does exist and is given by SW47 1 d 7bcos2 71 7 ad 7 b0 0 a Sin2 9 1 Question 4 Consider the vectors l 1 l l 2 l v1 4 v2 2 39v l 2 l l 1 a Compute ViVj for ij 17 27 3 What are the lengths of these vectors c What are the angles between the 3 pairs of vectors ls any pair orthogo d Draw a diagram showing the vectors as arrows Do the lengths angles 7 look right Solution M M V1V1 4 4 21 M M 1 2 Vl39Vg Vg39Vl 4 39 2 2 1 V139V3 V339V1 4 39 M l2 0 So V1 and V3 are orthogonal So V2 and V3 are orthogonal lvll 1242222l v2l x2222123 M 0271222 E c The angle between V1 and V2 is The angle between V1 and V3 is V139V3 7 0 7 7139 0 1 7 1 7 1 0 7 cos ltlV1l lVal COS wkE COS 2 So V1 and V3 are orthogonal The angle between V2 and V3 is 0 7r 0 1 V2 V3 1 7 1 0 7 cos V2 V3 cos 3 cos 2 So V2 and V3 are orthogonal Question 5 10 points Let C be an invertible 2 by 2 matrix such that CHHHamorww Find 2 by 2 matrices A and B so that CAB and then solve for C Solution We see from our rst and second equations that 2 i 1 i 1 1 lllwl 4H l 1l and hence multiplying by C 2 i 1 1 7 1 all l 1H ll We then have the two equations 1 2 2 1 Ol4ll1land0l1ll1l We can rewrite this as 1 2 2 1 Cl 41H1 1 or CA B where 712 271 A 41andB1 1 Let us assume that A71 exists Then CAA 1 BA l gt OBA 1 By our formula for inverses of 2 by 2 matrices 2 i 1 7 1 1 1 A 711724 74 SO 1 79 Question 6 10 points Solve if possible the system of linear equations 1 2 23 9 15283 3 21 7 22 73 3 Identify pivotal columns Say which variables are pivotal and which are non pivotal Say if the solution is unique Solution The augmented matrix is 1 12 91 1 58 3 l2 72 7 i3 Step1Row2Row1andRow3 2 X Rowl 1 12 9 0 4 6 76 3 l0 74 i21l 7 1 1 2 9 l 0 4 6 76 l 0 0 9 727 l We now have a row reduced matrix and see the 1st 2nd 3rd columns are pivotal so all of x1 x2 x3 are pivotal There are no non pivotal variables Step 3 Back Substitution Step 2 Row 3 Row 2 1223 9 4263 6 We see that Then gt23 Then 197272x3973612 XJlii Hi In particular the solution is unique So the solution set is 10 Question 7 15 points Solve if possible the system of linear equations 12790329047905 3 41 62 7 43 ll4 7 25 21 7 23 34 7 45 Identify pivotal columns Say which variables are pivotal and which are non pivotal Say if the solution is unique Solution The augmented matrix is 11 i1 271 3 4674117213 l2072 374 7l Step1RDW24XROW1ROW32XROW 1 171 2713 0 2 0 3 21 lo 72 071721l Step2ROW3Row2 11712713 02 03 21 loo 02 02l This is in row reduced form The pivotal columns are the 1st 2nd and 4th So x1 x2 x4 are pivotal while x3 x5 are non pivotal We set 3 t1 and 5 t2 Step 3 Back substitution 12790329047905 3 223425 l 7 2 We solve for 904 Then 22l3425 gt2lit2 Next 1 37237245 37717t2t172t2 2t12t2 Then a parametric description of the solution set is 2t12t2 2 1 2 71712 i1 0 71 x t1 0 11 1 12 0 1 1 0 0 12 0 0 1 So the solution set contains in nitely many solutions and is given by Ethtg E HOOHM 2 l 1 0 X 0 t1 1 232 1 0 0 0 12 Question 8 Consider the system 21 7 22 3 2 2x1 7 31mg 2903 7a 21 7 32 41 7 a For which 11 does this system have a unique solution Give the solution for such 11 10 For which 11 does this system have no solution c For which 11 are there in nitely many solutions Give the solutions for such 11 Solution The augmented matrix is 2 72 1 2 2 73b 2 7a 2 73 4 3 Step1Row2Row1Row3Row1 2 72 1 2 0 2731 1 7a72 0 71 3 1 Step 2 Swop rows 2 and 3 to prevent no pivot When 2 7 3b 0 2 i2 1 21 0 71 3 1 l0 2 73b 1 7a7 2 Step 3 Row 3 2731 gtlt Row2 2 i2 1 21 0 71 3 1 l0 0 7 7 9b 7a 7 31 Now we have a row reduced matrix 1f 1 y 3 then 7 7 9b y 0 and there is a pivot in the 3rd column If I then 7 7 9b 0 and there is no pivot in 13 the 3rd column We consider the two cases separately Step 4 Back Substitution Case I b y g Then we substitute back 21i223 2 i233 1 7791903 71731 Then 771731 903 7791f Then i2 li33 71731 173 71 i 779b3a9b 7791 7 73a 7791 i 73a 902 77 Then 1 1 Elt22gi3 1 73a 71731 7 22 7 2 7791 7791 14718b71476aa3b 2779b 75a715b7 5a3b 2779b 277 The solution is N H l I 1 m 3 9v 14 In this case there is a unique solution Step 4 Back Substitution Case 11 b g Then we substitute back 21 22 3 2 902 3903 0 7a 7 31 l H Then there is no solution unless iai3b0 gta 73b In this case we obtain 217223 2 i233 1 Here x1 x2 are pivotal and x3 is not pivotal so we set 903 t Then i2 Then 5 gt 1 it 2 and the solution set is 15 SUMMARY I If b 3 there is a unique solution 7 n3 l 2 a l i 7 a X i 7791 7 n3 7791 H H b there is no solution unless a 731 When a 731 there are 00 many solutions and the solution set is Flil lihel lol lll 16 Question 9 10 points Give a one to one parametrization of KerA for 71 1 2 Solution Recall that KerA is the solution set of the equation AX 0 So we solve this by applying row reduction to AMJZE 1 1f 81 76 6 31 OJ Step1Row24gtltRow1Row36gtltRow1 711 2 01 0 0 19 0 L 0 0 19 OJ Step 2 Row3 Row2 71 1 2 0 0 0 19 0 0 0 0 0 This is in row reduced form The rst and third columns are pivotal So the rst and third variables 901903 are pivotal the second 902 is not Set 902 t Step 3 Back Substitution i1223 0 19903 0 0 0 We see that 903 0 Then i1 i2 7 23 it gt 1 17 R t E t t t 0 1 1 0 t Ix 2 3 x x m a x K 18 Quest ion 1 0 Let AJZ i ii 425 Compute the rank of A and decide if A71 exists If so nd A71 Solution Since the algorithm for nding A71 will tell us what the rank is anyway it will give the row reduced form of A we might as well try nd A71 So form the augmented matrix 24 0 54 LE 0 04545 1 Step 2 Multiply rows 2 and 3 by 3 to get rid of fractions 3 2 4 1 0 0 l 0 5 4 4 73 0 L 0 2 1 4 0 73 J Step3Row3 ngow2 At this stage we see that we have 3 pivots in a 3 by 3 matrix7 so we have rank 3 and then the matrix is invertible Step 4 Divide by diagonal entries to obtain 1 s along diagonal H LO Q HW M Hmmww rheumle w o O O 4 L0 Step 5 Clear out entries above the 1 in the 3rd column ROW2g gtltRow3Row1 gtltRow3 2 17 8 20 130 5 3 0104174 0 2 L 0 Step 6 Clear out entries above the 1 in the 2rd column Row1 gtltRow2 H i 3 i i ii 0 1 74 72 SJ SO 3 2 74 717 7 A 11 if Not required of students Check A l i i 1 742 5 liiilli fiil L425JL74 i2 5 HTSLI 001 The other side is similar 20 Question 11 10 points 1 1 71 1 i2 1 A 1 5 1 3 J 9 5 3 1 Compute the rank of A and decide if A71 exists It so nd A71 Solution Again we apply the algorithm for nding an inverse It will tell us the rank and if the inverse exists Form the augmented matrix Let 1 71 1 72 1 0 0 0 71 3 71 0 0 1 0 0 Ami 2 72 i2 3 0 0 1 0 9 75 i3 71 0 0 0 1 Step1Row2Row1Row32 gtlt Row1Row49 gtlt Rowl 1 71 1 i2 1 0 0 0 0 2 0 i2 1 1 0 0 0 0 74 7 i2 0 1 0 0 4 712 17 79 0 0 1 Step2RDW42XROW2 1 71 1 i2 1 0 0 0 l I 0 2 0 i2 1 1 0 0 I 0 0 i4 7 i2 0 1 0 L 0 0 712 21 711 i2 0 1 J Step3Row43 gtlt Row3 1 71 1 72 1 0 0 0 0 2 0 72 1 1 0 0 0 0 i4 7 i2 0 1 0 L 0 0 0 0 75 72 73 1 J We see that there are 3 pivots in the row reduced form of A so the matrix has rank 3 which is less than the number of columns namely 4 and is NOT invertible 21 Question 12 15 points Let 52621 A 0100 4252 0201i Compute the rank of A and decide if A71 exists It so nd A71 Solution Again we apply the algorithm for nding an inverse It will tell us the rank and if the inverse exists Form the augmented matrix 5 2 6 2 1 0 0 0 0 1 0 0 0 1 0 0 Ami 4 2 5 2 0 0 1 0 0 2 0 1 0 0 0 1 Step1 Row 3 g X Row 1 5 2 6 2 1 0 0 0 0 1 0 0 0 1 0 0 2 1 2 4 0 5 5 3 3 0 1 0 L 0 2 0 1 0 0 0 1 J Step 2 Multiply Row 3 by 5 5 2 6 2 1 0 0 0 0 1 0 0 0 1 0 0 0 2 1 2 74 0 5 0 0 2 0 1 0 0 0 1 Step3Row32gtltRow2Row42gtltRow2 52621000 01000100 0012747250 0001072011 We see the rank is 4 and so the matrix is invertible Step 4 Divide through by diagonal entries 22 2621 1gggg0001 01000100 001247250 000107201 Step 5 Clear out entries above 44 position Row 3 2 X Row 4 Rowlg gtltRow4 26 14 2 13303307 01000100 0010742572 10001072011 Step 5 Clear out entries above 33 position Row 1 g X Row 3 1 1 0 0 5 0 0 0 1 0 74 0 1 0 OOOH OOHOWHQ 1000 57276 2 0100 010 0 001074 2 572 0001072 01 So 57276 2 010 0 71 A7 4 2 572 072 01 23 Not required of students Let s check 76 0 5 0 1 I 24 72 1 2 2 7 2021 0001 040 6050 0010 7 5 2122 0100 5040 1000 A Question 13 30 points Let 2 72 i3 33 i 4 6 2 66 A ll l 8 732 264 a Find a lower triangular matrix R and the row reduced form U of A such that RA U 10 points b Find a unit lower triangular matrix L such that A LU 5 points c Let 0 b 2 11 and calculate c Rb Hence use c and U to solve the equation AH l El 7 points l 11 l for x d Let 73 b 74 710 Solve the equation Ly b for y and hence solve the equation l 3 l AX b 4 8 points l 710 l 25 for x Solution a 10 points We apply row reduction to 2 42 43 33 1 0 0 A11 4 76 42 66 0 1 0 14 3 432 264 0 0 1 Step1 Row22 gtlt Row1Row37 XROW 2 i2 73 33 1 0 0 0 i2 4 0 i2 1 0 0 22 711 33 i7 0 1 Step 2 Row 3 11 gtlt Row 2 2 42 i3 33 1 0 0 0 42 4 0 i2 1 0 U1R 0 0 33 33 729 11 1 1 2 i2 i3 33 1 1 1 0 0 1 7 7 U 0 2 4 0 d R 2 1 0 10 0 33 331 17291111 10 5 points To nd L R71 we apply row reduction to 100100 R11 4210 010 429111001 Step1Row22gtltRow1Row329gtltRow1 1 0 0 10 0 0 10 210 0111 29 01 Step 2 Row 3 11 X Row 2 10 0 1 0 01 0 1 0 2 10 0 01 7 7111 Then 7 11 001 1 L R 13111 Checking our answers are correct not required from students 1 1 0 0 1 R 2 1 0 L 7 711 1 J 1 1 0 0 1 1 1 0 0 1 72 1 0 2 1 0 729111JL77111J 1 1 0 0 1 0 1 0 I L 0 0 1 J Also not required from students 1 0011272 73 331 1 LU 13411118 333031 1 2 72 73 33 1 4 6 2 66 A 1 14 8 732 2641 c 7 points Recall that Ax b We rst form 27 Now we solve Uxc 22 73 33 i0 S Sg jx ig y We can solve this directly by back substitution 21 722 73 0 722 43 2 33903 33 33 Then x4 is non pivotal and we set 904 t so Then and 33mg 33733433733t gt 31t 722243i24tgt212t 02x23x3733x4274t373t733t5740t 5 1 E i So the solution is Ax BEN WHO Checking our answers are correct not required from students Vii HEW HM 5 5 572m 5 17 1 1 l 2 i2 i3 33 32 0 4 76 i2 66 1172 2 L14 8 732 264 t in 28 d 8 points Recall the idea Ax b a LUX b If we set y UK we then have Ly b So we rst solve Ly b or H mum i7 7111 710 We can use forward substitution 3J1 3 231 92 4 731 11372 93 10 Then yl 73 y242912 and so 343 710 i 7341 11342 710 21 22 33 So Next we solve UX y or 272 73 33 73 0 72 4 0 X 2 i 0 0 33 33 J i which we can do by back substitution 21 722 73 3 722 43 2 33903 33 33 Then set 904 t so so 29 2x1 732233733473274t373t733t2740t 9011720t So the solution is 901 17201 1 720 7 x2 7 1721 7 1 72 X 903 lit 1 H 71 4 t 0 1 Checking our answers are correct not required from students 1720t 141 3 2211111111 14 8 732 264JL liijiquj 30 Quest ion 1 4 Let 1 1 i1 71 l 0 1 1 A i 2 4 1 2 0 3 i Find a basis for KerA What is the dimension of KerA ii Find a basis for 1mgA What is the dimension of 1mgA ii Find a least squares solution of the equation Is it also a solution to Ax b If not nd the error Solution i 8 points We solve Ax 0 Form the augmented matrix 1 1 1 0 0 1 1 0 2 4 1 0 2 0 3 0 and apply row reduction Step1 Row32 gtlt Row1Row42 gtlt Rowl 171710 071710 1072101 10 27101 Step2 Row32gtltRow2Row42gtltRow2 1 71 71 0 0 71 71 0 0 0 3 0 0 0 i3 0 Step 3 Row 4 Row 3 1 71 71 0 0 71 71 0 0 0 3 0 0 0 0 0 This is in row reduced form We see that there are pivots in the rst second third columns so every column is pivotal Since the number of pivots equals the number of columns in A one of our theorems tells us that KerA Alternatively we can back substitute to obtain KerA Then the dimension of KerA is 0 ii 2 points Since in the row reduced form of A all of its columns are pivotal one of our theorems tells us that the columns of the original matrix A form a basis for lmgA Thus a basis for lmgA is 1 71 71 0 71 71 2 74 71 2 0 73 and its dimension is 3 iii 10 points To solve this least squares we use the normal equations ATAx ATb We could also use QR but I use the rst method Here Audi 7 7i El 1 1 1 3 and T 1 0 2 2 15 Abtiii gimt t Now we solve this Form the augmented matrix 1 9 79 79 15 79 18 6 721 79 6 12 712 and apply row reduction Step 1 Row 2 Row 1 Row 3 Row 1 7 9 79 79 15 l 0 9 73 76 to 73 3 3 Step 2 Swap Rows 2 and 3 to simplify arithmetic 9 79 7915 073 3 3 to 9 73 76 Step 3 Row3 3 X Row2 9 79 7915 0 73 3 3 to 0 63 Step 4 Back Substitution 91 ig2 79 73902 3903 3 3 Then SO SO 2 91 9293 1 Then the least squares solution is AFR ilpgillgl l sill a lisl Then L N l U39 H l I H gtJgt H mimwm own I l mlemw own l and the error is 2732 2 12 121 le blilt3 0 6 6 72 Since this is positive X is not a solution to AX b and is only a least squares solution The error is 2 34 Question 15 Let 7 1 71 2 A l 3 i 2 j 39 a Find an orthonormal basis for my A b Are the columns of A linearly independent ls A invertible Give a reason for your answer 0 Compute the QR factorisation of A d Which of the following is true UQWQE ii QQT I Brie y explain your answer e Find the least squares solution of l Axb 71 l ls this also a solution to Ag b If so give a reason If not nd the error Mri z Solutions a We write AV1V2V3ll and apply the Gram Schmidt process Step 1 form 111 Now memwa 1 1M WUM Q So u17v17 1 1 w w no Step 2 form u2 Now W2 V27 V2111 111 35 i1 1 1 1 2 73 7 72 2 1 71 1 2 0 1 wZ111 1 1 1 3 2 0 8 2 000 1W39936 3 36 1 l 31 JjUj Sl W Wiai am on Then the orthonormal basis for ImgA is 1 1 1 1172713 11117114171141 lol 31 ll 61 b Yes we obtained three different vectors in our orthonormal basis and there are three columns So the columns of A are linearly independent A has rank 3 and A is invertible c Next we factorize A QR Firstly WE WE um Q 1N5 iw 4 l 0 1N3 iw l Next 11139V1 11139V2 11139V3 0 llg39Vg 11239V3 0 11339V3 7 11139V1 7 0 llg39Vg l 0 0 11339V3 J Recall that 11ng V2 111 and this has already been computed etc Here 111 1111 1 110i l l 2 0 Then 72 W R 0 M3 72 l0 0 3 l So PM 1N3 1N6 72 1 i1 2 QB L1N 71 il JL 2M3 A 0 0 1 3 72 6 d QTQ I is true because the columns of Q are always orthonormal Because Q is square and of full rank it is invertible and its left inverse is also its right inverse So QQT I is also true e We must solve Rx QTb Here QTb W 4N3 W5 1 1M 1M 0 1 we 4N6 2N l l 1l 0 Sowesolve 72 W l 0 0 ME 72 x ME lo 0 ml l ol 01 5901 72 x2 9c3 0 V3902 2 3 WE 0 38

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