Given: The vectors [123] and [369]. Since, [369]=3[123] That is., the given vector is some scalar...Given: Three sides of a box meet at solutions x=(x,y,z)=(2,3,4)⇒[ 1 0 0 0 1 0 0 0 1][ x y z]=[ 2 3...Given information: (1) x+y=y+x(2) x+(y+z)=(x+y)+z(3) There is a unique "zero vector" such that x+0=x...Given: Let the matrix A of order 2 by 3 be A=[123246]is of rank 1, as 1ndrow is obtained by...Given: detA=12, where A is 4×4 matrix Calculation :- Since detA=12 and we know that if A is n×n...Given: Matrices are A=[.8.3.2.7], A2=[.70.45.30.55] and A∞=[.6.6.4.4] Concept Used: If T is a linear...Given information: A=12 3 4 24 6 8 3 4 6 8 9 12 12 16 B=23 4 5 34 5 6 4 5 5 6 6 7 7 8 Not requiring...Given: "A linear transformation must leave the zero vector fixed; T(0)=0 prove this from...Given: z1=2+i,z2=2−i Concept Used: If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then z1 + z2... Show
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Sorry, something went wrong. Reload? Sorry, we cannot display this file. Sorry, this file is invalid so it cannot be displayed. INTRODUCTIONTOLINEARALGEBRAFifth EditionMANUAL FOR INSTRUCTORSGilbert StrangMassachusetts Institute of Technologymath.mit/linearalgebraweb.mit/18.video lectures: ocw.mitmath.mit/∼gswellesleycambridgeemail: linearalgebrabook@gmailWellesley - Cambridge PressBox 812060Wellesley, Massachusetts 02482Problem Set 1, page 81 The combinations give (a) a line inR 3 (b) a plane inR 3 (c) all ofR 3 . 2 v+w= (2,3)andv−w= (6,−1)will be the diagonals of the parallelogram with vandwas two sides going out from(0,0). 3 This problem gives the diagonalsv+wandv−wof the parallelogram and asks for the sides: The opposite of Problem 2. In this examplev= (3,3)andw= (2,−2). 43 v+w= (7,5)andcv+dw= (2c+d, c+ 2d). 5 u+v= (− 2 , 3 ,1)andu+v+w= (0, 0 ,0)and 2 u+2v+w= (add first answers) = (− 2 , 3 ,1). The vectorsu,v,ware in the same plane because a combination gives (0, 0 ,0). Stated another way:u=−v−wis in the plane ofvandw. 6 The components of everycv+dwadd to zero because the components ofvand ofw add to zero= 3andd= 9give(3, 3 ,−6). There is no solution tocv+dw= (3, 3 ,6) because3 + 3 + 6is not zero. 7 The nine combinationsc(2,1) +d(0,1)withc= 0, 1 , 2 andd= (0, 1 ,2)will lie on a lattice. If we took all whole numberscandd, the lattice would lie over the whole plane. 8 The other diagonal isv−w(or elsew−v). Adding diagonals gives 2 v(or 2 w). 9 The fourth corner can be(4,4)or(4,0)or(− 2 ,2). Three possible parallelograms! 10 i−j= (1, 1 ,0)is in the base (x-yplane).i+j+k= (1, 1 ,1)is the opposite corner from(0, 0 ,0). Points in the cube have 0 ≤x≤ 1 , 0 ≤y≤ 1 , 0 ≤z≤ 1. 11 Four more corners(1, 1 ,0),(1, 0 ,1),(0, 1 ,1),(1, 1 ,1). The center point is( 1 2 ,1 2 ,1 2 ).Centers of faces are( 1 2 ,1 2 ,0),(1 2 ,1 2 ,1)and(0, 1 2 ,1 2 ),(1,1 2 ,1 2 )and( 1 2 , 0 ,1 2 ),(1 2 , 1 ,1 2 ).12 The combinations ofi= (1, 0 ,0)andi+j= (1, 1 ,0)fill thexyplane inxyzspace. 13 Sum=zero vector. Sum=− 2 : 00 vector= 8: 00 vector. 2 : 00 is 30 ◦ from horizontal = (cos π 6 ,sin π 6 ) = (√3 / 2 , 1 /2).14 Moving the origin to 6 : 00 addsj= (0,1)to every vector. So the sum of twelve vectors changes from 0 to 12 j= (0,12). 26 Two equations come from the two components:c+ 3d= 14and 2 c+d= 8 solution isc= 2andd= 4. Then2(1,2) + 4(3,1) = (14,8). 27 A four-dimensional cube has 2 4 = 16corners and 2 ·4 = 8three-dimensional faces and 24 two-dimensional faces and 32 edges in Worked Example2 A. 28 There are 6 unknown numbersv 1 , v 2 , v 3 , w 1 , w 2 , w 3. The six equations come from the components ofv+w= (4, 5 ,6)andv−w= (2, 5 ,8). Add to find 2 v= (6, 10 ,14) sov= (3, 5 ,7)andw= (1, 0 ,−1). 29 Fact : For any three vectorsu,v,win the plane, some combinationcu+dv+ewis the zero vector (beyond the obviousc=d=e= 0). So if there is one combination Cu+Dv+Ewthat producesb, there will be many more—just addc, d, eor 2 c, 2 d, 2 e to the particular solutionC, D, E. The example has 3 u− 2 v+w= 3(1,3)−2(2,7) + 1(1,5) = (0,0). It also has − 2 u+ 1v+ 0w=b= (0,1). Adding givesu−v+w= (0,1). In this casec, d, e equal 3 ,− 2 , 1 andC, D, E=− 2 , 1 , 0. Could another example haveu,v,wthat could NOT combine to produceb? Yes. The vectors(1,1),(2,2),(3,3)are on a line and no combination producesb. We can easily solvecu+dv+ew= 0but notCu+Dv+Ew=b. 30 The combinations ofvandwfill the planeunlessvandwlie on the same line through (0,0). Four vectors whose combinations fill 4 -dimensional space: one example is the “standard basis”(1, 0 , 0 ,0),(0, 1 , 0 ,0),(0, 0 , 1 ,0), and(0, 0 , 0 ,1). 31 The equationscu+dv+ew=bare 2 c −d = 1 −c+2d −e= 0 −d+2e= 0 Sod= 2e thenc= 3e then 4 e= 1 c= 3/ 4 d= 2/ 4 e= 1/ 4 Problem Set 1, page 181 u·v=− 2 .4 + 2 = 0,u·w=−.6 + 1 = 1,u·(v+w) =u·v+u·w= 0 + 1,w·v= 4 + 6 = 10 =v·w. 2 ‖u‖= 1and‖v‖= 5and‖w‖= √
5√5 , confirming the Schwarz inequality. 3 Unit vectorsv/‖v‖= ( 4 5 ,3 5 ) = (0. 8 , 0 .6). The vectorsw,(2,−1), and−wmake 0◦ , 90 ◦ , 180 ◦ angles withwandw/‖w‖= (1/ √5 , 2 /√5). The cosine ofθis v ‖v‖ ·w ‖w‖ = 10/ 5√5.4 (a) v·(−v) =− 1 (b)(v+w)·(v−w) =v·v+w·v−v·w−w·w= 1+( )−( )−1 = 0 soθ= 90 ◦ (noticev·w=w·v) (c) (v− 2 w)·(v+2w) = v·v− 4 w·w= 1−4 =− 3. 5 u 1 =v/‖v‖= (1,3)/ √10 andu 2 =w/‖w‖= (2, 1 ,2)/ 3 .U 1 = (3,−1)/ √10 is perpendicular tou 1 (and so is(− 3 ,1)/ √10 ).U 2 could be(1,− 2 ,0)/ √5 : There is a whole plane of vectors perpendicular tou 2 , and a whole circle of unit vectors in that plane. 6 All vectorsw= (c, 2 c)are perpendicular tov. They lie on a line. All vectors(x, y, z) withx+y+z= 0lie on aplane. All vectors perpendicular to(1, 1 ,1)and(1, 2 ,3) lie on alinein 3 -dimensional space. 7 (a) cosθ=v·w/‖v‖‖w‖= 1/(2)(1)soθ= 60 ◦ orπ/ 3 radians (b)cosθ= 0 soθ = 90 ◦ orπ/ 2 radians (c) cosθ = 2/(2)(2) = 1/ 2 soθ = 60 ◦ orπ/ 3 (d)cosθ=− 1 / √2 soθ= 135 ◦ or 3 π/ 4. 8 (a) False:vandware any vectors in the plane perpendicular tou (b) True:u· (v+ 2w) =u·v+ 2u·w= 0 (c) True,‖u−v‖ 2 = (u−v)·(u−v)splits into u·u+v·v= 2 whenu·v=v·u= 0. 9 Ifv 2 w 2 /v 1 w 1 =− 1 thenv 2 w 2 =−v 1 w 1 orv 1 w 1 +v 2 w 2 =v·w= 0: perpendicular! The vectors(1,4)and(1,− 1 4 )are perpendicular. 212 v·w≤ 2 ‖v‖‖w‖leads to‖v+w‖ 2 =v·v+2v·w+w·w≤ ‖v‖ 2 +2‖v‖‖w‖+ ‖w‖ 2 . This is(‖v‖+‖w‖) 2 . Taking square roots gives‖v+w‖ ≤ ‖v‖+‖w‖. 22 v 2 1 w 2 1
because the difference isv 2 1 w 2 2 +v 2 2 w 2 1 − 2 v 1 w 1 v 2 w 2 which is(v 1 w 2 −v 2 w 1 ) 2 ≥ 0. 23 cosβ=w 1 /‖w‖andsinβ=w 2 /‖w‖. Thencos(β−a) = cosβcosα+sinβsinα= v 1 w 1 /‖v‖‖w‖+v 2 w 2 /‖v‖‖w‖=v·w/‖v‖‖w‖. This iscosθbecauseβ−α=θ. 24 Example 6 gives|u 1 ||U 1 | ≤ 1 2 (u 2 1 +U2 1 )and|u 2 ||U 2 | ≤ 1 2 (u 2 2 +U2 2 ). The whole line becomes. 96 ≤(.6)(.8) + (.8)(.6)≤ 1 2 (. 62 +. 8 2 ) + 1 2 (. 82 +. 6 2 ) = 1. True:. 96 < 1. 25 The cosine ofθisx/ √x 2 +y 2 , near side over hypotenuse. Then|cosθ| 2 is not greater than 1:x 2 /(x 2 +y 2 )≤ 1. 26–27(with apologies for that typo !) These two lines add to 2 ||v|| 2 + 2||w|| 2 : ||v+w|| 2 = (v+w)·(v+w) =v·v+v·w+w·v+w·w ||v−w|| 2 = (v−w)·(v−w) =v·v−v·w−w·v+w·w 28 The vectorsw= (x, y)with(1,2)·w=x+ 2y= 5lie on a line in thexyplane. The shortestwon that line is(1,2). (The Schwarz inequality‖w‖ ≥v·w/‖v‖= √5 is an equality whencosθ= 0andw= (1,2)and‖w‖= √5 .)29 The length‖v−w‖is between 2 and 8 (triangle inequality when‖v‖= 5and‖w‖= 3 ). The dot productv·wis between− 15 and 15 by the Schwarz inequality. 30 Three vectors in the plane could make angles greater than 90 ◦ with each other: for example(1,0),(− 1 ,4),(− 1 ,−4). Four vectors couldnotdo this ( 360 ◦ total angle). How many can do this inR 3 orR n ? Ben Harris and Greg Marks showed me that the answer isn+ 1 vectors from the center of a regular simplex inR n to itsn+ 1 vertices all have negative dot products. Ifn+2vectors inR n had negative dot products, project them onto the plane orthogonal to the last one. Now you haven+ 1vectors in Rn− 1 with negative dot products. Keep going to 4 vectors inR 2 : no way! 31 For a specific example, pickv= (1, 2 ,−3)and thenw= (− 3 , 1 ,2). In this example cosθ= v·w/‖v‖‖w‖ =− 7 / √14√14 =− 1 / 2 andθ = 120 ◦ . This always happens whenx+y+z= 0: v·w=xz+xy+yz= 12(x+y+z) 2 − 12(x 2 +y 2 +z 2 ) This is the same asv·w= 0− 12‖v‖‖w‖.Thencosθ= 12.32 Wikipedia gives this proof of geometric mean G = 3 √xyz ≤arithmetic mean A= (x+y+z)/ 3. First there is equality in casex=y =z. OtherwiseAis somewhere between the three positive numbers, say for examplez < A < y. Use the known inequalityg≤afor thetwopositive numbersxandy+z−A. Their meana = 1 2 (x+y+z−A)is 1 2 (3A−A) =same asA! Soa ≥gsays that A3 ≥g 2 A=x(y+z−A)A. But(y+z−A)A= (y−A)(A−z) +yz > yz. Substitute to findA 3 > xyz=G 3 as we wanted to prove. Not easy! There are many proofs ofG= (x 1 x 2 · · ·xn) 1 /n ≤A= (x 1 +x 2 +· · ·+xn)/n. In calculus you are maximizingGon the planex 1 +x 2 +· · ·+xn=n. The maximum occurs when allx’s are equal. 33 The columns of the 4 by 4 “Hadamard matrix” (times 1 2 ) are perpendicular unit vectors: 12H=12 1 1 1 11 −1 1 − 11 1 − 1 − 11 − 1 −1 1 .34 The commandsV=randn(3,30);D=sqrt(diag(V ′ ∗V));U=V\D;will give 30 random unit vectors in the columns ofU. Thenu ′ ∗Uis a row matrix of 30 dot products whose average absolute value should be close to 2 /π. Problem Set 1, page 2913 s 1 + 4s 2 + 5s 3 = (3, 7 ,12). The same vectorbcomes fromStimesx= (3, 4 ,5): 7 All three rows are perpendicular to the solutionx(the three equationsr 1 ·x= 0and r 2 ·x= 0andr 3 ·x= 0tell us this). Then the whole plane of the rows is perpendicular tox(the plane is also perpendicular to all multiplescx). 8 x 1 −0 =b 1 x 2 −x 1 =b 2 x 3 −x 2 =b 3 x 4 −x 3 =b 4 x 1 =b 1 x 2 =b 1 +b 2 x 3 =b 1 +b 2 +b 3 x 4 =b 1 +b 2 +b 3 +b 4 = 1 0 0 01 1 0 01 1 1 01 1 1 1 b 1 b 2 b 3 b 4 =A− 1 b 9 The cyclic difference matrixChas a line of solutions (in 4 dimensions) toCx= 0 : 1 0 0 − 1−1 1 0 00 −1 1 00 0 −1 1 x 1 x 2 x 3 x 4 = 0000 whenx= c c c c = any constant vector. 10 z 2 −z 1 =b 1 z 3 −z 2 =b 2 0 −z 3 =b 3 z 1 =−b 1 −b 2 −b 3 z 2 = −b 2 −b 3 z 3 = −b 3 =− 1 − 1 − 10 − 1 − 10 0 − 1b 1 b 2 b 3 = ∆− 1 b 11 The forward differences of the squares are(t+ 1) 2 −t 2 =t 2 + 2t+ 1−t 2 = 2t+ 1. Differences of thenth power are(t+ 1) n −t n =t n −t n +nt n− 1 +· · ·. The leading term is the derivativent n− 1 . The binomial theorem gives all the terms of(t+ 1) n . 12 Centered difference matrices ofeven sizeseem to be invertible. Look at eqns. 1 and 4 : 0 1 0 0−1 0 1 00 −1 0 10 0 −1 0 x 1 x 2 x 3 x 4 = b 1 b 2 b 3 b 4 First solve x 2 =b 1 −x 3 =b 4 x 1 x 2 x 3 x 4 = −b 2 −b 4 b 1 −b 4 b 1 +b 3 13 Odd size: The five centered difference equations lead tob 1 +b 3 +b 5 = 0. x 2 =b 1 x 3 −x 1 =b 2 x 4 −x 2 =b 3 x 5 −x 3 =b 4 −x 4 =b 5 Add equations 1 , 3 , 5 The left side of the sum is zero The right side isb 1 +b 3 +b 5 There cannot be a solution unlessb 1 +b 3 +b 5 = 0. 14 An example is(a, b) = (3,6)and(c, d) = (1,2). We are given that the ratiosa/cand b/dare equal. Thenad=bc. Then (when you divide bybd) the ratiosa/bandc/d must also be equal! 9 (a) Ax= (18, 5 ,0)and (b)Ax= (3, 4 , 5 ,5). 10 Multiplying as linear combinations of the columns gives the sameAx= (18, 5 ,0)and (3, 4 , 5 ,5). By rows or by columns: 9 separate multiplications whenAis 3 by 3. 11 Axequals(14,22)and(0,0)and ( 9 ,7). 12 Axequals(z, y, x)and(0, 0 ,0)and ( 3 , 3 ,6). 13 (a) xhasncomponents andAxhasmcomponents (b) Planes from each equation inAx=bare inn-dimensional space. The columns ofAare inm-dimensional space. 142 x+3y+z+5t= 8isAx=bwith the 1 by 4 matrixA= [ 2 3 1 5 ]:one row. The solutions(x, y, z, t)fill a 3D “plane” in 4 dimensions. It could be called ahyperplane. 15 (a) I= 1 00 1=“identity” (b) P= 0 11 0=“permutation” 1690◦ rotation fromR= 0 1−1 0, 180 ◦rotation fromR 2 = −1 00 − 1=−I.17 P =0 1 00 0 11 0 0produces y z x andQ= 0 0 11 0 00 1 0recovers x y z . Qis the inverseofP. Later we writeQP=IandQ=P − 1 . 18 E=1 0−1 1andE= 1 0 0−1 1 00 0 1subtract the first component from the second. 19 E=1 0 00 1 01 0 1andE − 1 = 1 0 00 1 0−1 0 1,Ev= 348andE − 1 Evrecovers 345.20 P 1 =1 00 0projects onto thex-axis andP 2 = 0 00 1projects onto they-axis. The vectorv= 57projects toP 1 v= 50andP 2 P 1 v= 00.21 R=12√2 −√2√2√2rotates all vectors by 45◦. The columns ofRare the results from rotating(1,0)and(0,1)! 22 The dot productAx= [ 1 4 5 ] x y z = (1 by 3)(3 by 1)is zero for points(x, y, z) on a plane in three dimensions. The 3 columns ofAare one-dimensional vectors. 23 A= [ 1 2 ; 3 4 ]andx= [ 5 −2 ] ′ or[ 5 ; −2 ]andb= [ 1 7 ] ′ or[ 1 ; 7 ]. r=b−A∗xprints as two zeros. 24 A∗v= [ 3 4 5 ] ′ andv ′ ∗v= 50. Butv∗Agives an error message from 3 by 1 times 3 by 3. 25 ones(4,4)∗ones(4,1) =column vector[ 4 4 4 4 ] ′ ;B∗w= [ 10 10 10 10 ] ′ . 26 The row picture has two lines meeting at the solution ( 4 , 2 ). The column picture will have4(1,1) + 2(− 2 ,1) = 4 (column 1)+ 2 (column 2)=right side(0,6). 27 The row picture shows2 planesin3-dimensional space. The column picture is in 2-dimensional space. The solutions normally fill aline in 3 -dimensional space. 28 The row picture shows fourlinesin the 2D plane. The column picture is infour- dimensional space. No solution unless the right side is a combinationofthe two columns. 29 u 2 = . 7. 3andu 3 = . 65. 35.The components add to 1. They are always positive. Their components still add to 1. 30 u 7 andv 7 have components adding to 1 ; they are close tos= (. 6 , .4). . 8. 3. 2. 7. 6. 4=. 6. 4=steady states. No change when multiplied by . 8. 3. 2. 7.31 M=8 3 41 5 96 7 2=5 +u 5 −u+v 5 −v 5 −u−v 5 5 +u+v 5 +v 5 +u−v 5 −u ;M 3 (1, 1 ,1) = (15, 15 ,15);M 4 (1, 1 , 1 ,1) = (34, 34 , 34 ,34)because1 + 2 +· · ·+ 16 = 136which is4(34). 56 x+ 4yis 2 times 3 x+ 2y. There is no solution unless the right side is 2 ·10 = 20. Then all the points on the line 3 x+2y= 10are solutions, including(0,5)and(4,−1). The two lines in the row picture are the same line, containing all solutions. 6 Singular system ifb= 4, because 4 x+ 8yis 2 times 2 x+ 4y. Theng= 32makes the lines 2 x+ 4y= 16and 4 x+ 8y= 32become thesame: infinitely many solutions like (8,0)and(0,4). 7 Ifa= 2elimination must fail (two parallel lines in the row picture). The equations have no solution. Witha= 0, elimination will stop for a row exchange. Then 3 y=− 3 givesy=− 1 and 4 x+ 6y= 6givesx= 3. 8 Ifk= 3elimination must fail: no solution. Ifk=− 3 , elimination gives0 = 0in equation 2: infinitely many solutions. Ifk= 0a row exchange is needed: one solution. 9 On the left side, 6 x− 4 yis 2 times(3x− 2 y). Therefore we needb 2 = 2b 1 on the right side. Then there will be infinitely many solutions (two parallel lines become one single line in the row picture). The column picture has both columns along the same line. 10 The equationy= 1comes from elimination (subtractx+y= 5fromx+ 2y= 6). Thenx= 4and 5 x− 4 y= 20−4 =c= 16. 11 (a) Another solution is 1 2 (x+X, y+Y, z+Z). (b) If 25 planes meet at two points, they meet along the whole line through those two points. 12 Elimination leads to this upper triangular system; then comes back substitution. 2 x+ 3y+ z= 8 y+ 3z= 4 8 z= 8 gives x= 2 y= 1 If a zero is at the start of row 2 or row 3, z= 1 that avoids a row operation. 13 2 x− 3 y = 3 4 x− 5 y+ z= 7 2 x− y− 3 z= 5 gives 2 x− 3 y= 3 y+ z= 1 2 y+ 3z= 2 and 2 x− 3 y= 3 y+ z= 1 − 5 z= 0 and x= 3 y= 1 z= 0 Here are steps 1 , 2 , 3 : Subtract 2×row 1 from row 2, subtract 1×row 1 from row 3, subtract 2×row 2 from row 3 14 Subtract 2 times row 1 from row 2 to reach(d−10)y−z= 2. Equation (3) isy−z= 3. Ifd= 10exchange rows 2 and 3. Ifd= 11the system becomes singular. 15 The second pivot position will contain− 2 −b. Ifb=− 2 we exchange with row 3. Ifb=− 1 (singular case) the second equation is−y−z= 0. But equation(3)is the same so there is aline of solutions(x, y, z) = (1, 1 ,−1). 16 (a) Example of 2 exchanges 0 x+ 0y+ 2z= 4 x+ 2y+ 2z= 5 0 x+ 3y+ 4z= 6 (exchange 1 and 2, then 2 and 3) (b) Exchange but then breakdown 0 x+ 3y+ 4z= 4 x+ 2y+ 2z= 5 0 x+ 3y+ 4z= 6 (rows 1 and 3 are not consistent) 17 If row1 =row 2 , then row 2 is zero after the first step; exchange the zero row with row 3 and row 3 has no pivot. If column2 =column 1 , then column 2 has no pivot. 18 Examplex+ 2y+ 3z= 0, 4 x+ 8y+ 12z= 0, 5 x+ 10y+ 15z= 0has 9 different coefficients but rows 2 and 3 become0 = 0: infinitely many solutions toAx= 0 but almost surely no solution toAx=bfor a randomb. 19 Row 2 becomes 3 y− 4 z= 5, then row 3 becomes(q+ 4)z=t− 5. Ifq=− 4 the system is singular—no third pivot. Then ift= 5the third equation is0 = 0which allows infinitely many solutions. Choosingz= 1the equation 3 y− 4 z= 5givesy= 3 and equation 1 givesx=− 9. 20 Singular if row 3 is a combination of rows 1 and 2. From the end view, the three planes form a triangle. This happens if rows1+2=row 3 on the left side but not the right side: x+y+z= 0,x− 2 y−z= 1, 2 x−y= 4. No parallel planes but still no solution. The three planes in the row picture form a triangular tunnel. 21 (a) Pivots 2 , 3 2 ,4 3 ,5 4 in the equations 2 x+y= 0, 3 2 y+z= 0, 4 3 z+t= 0, 5 4 t= 5 after elimination. Back substitution givest= 4, z =− 3 , y= 2, x=− 1. (b) If the off-diagonal entries change from+1to− 1 , the pivots are the same. The solution is (1, 2 , 3 ,4)instead of(− 1 , 2 ,− 3 ,4). 22 The fifth pivot is 6 5 for both matrices (1’s or− 1 ’s off the diagonal). Thenth pivot is n+ n .(a) Some linear combination of the 100 rows isthe row of 100 zeros. (b) Some linear combination of the 100columnsisthe column of zeros. (c) A very singular matrix has all ones:A=ones(100). A better example has 99 random rows (or the numbers 1 i ,... , 100 i in those rows). The 100th row could be the sum of the first 99 rows (or any other combination of those rows with no zeros). (d) The row picture has 100 planesmeeting along a common line through 0. The column picture has 100 vectors all in the same 99-dimensional hyperplane. Problem Set 2, page 661 E 21 =1 0 0−5 1 00 0 1, E 32 =1 0 00 1 00 7 1, P=1 0 00 0 10 1 00 1 01 0 00 0 1=0 1 00 0 11 0 0.2 E 32 E 21 b= (1,− 5 ,−35)butE 21 E 32 b= (1,− 5 ,0). WhenE 32 comes first, row 3 feels no effect from row 1. 3 1 0 0−4 1 00 0 1,1 0 00 1 02 0 1,1 0 00 1 00 −2 1M=E 32 E 31 E 21 =1 0 0−4 1 010 −2 1.ThoseE’s are in the right order to giveM A=U. 4 Elimination on column 4: b = 100 E 21→ 1− 40 E 31→ 1− 42 E 32→ 1− 410 . The originalAx=bhas becomeUx=c= (1,− 4 ,10). Then back substitution gives z=− 5 , y= 1 2 , x= 1 2 .This solvesAx= (1, 0 ,0). 5 Changinga 33 from 7 to 11 will change the third pivot from 5 to 9. Changinga 33 from 7 to 2 will change the pivot from 5 tono pivot. 6 Example: 2 3 72 3 72 3 7 13− 1 = 444 . If all columns are multiples of column 1 , there is no second pivot. 7 To reverseE 31 ,add 7 times row 1 to row 3. The inverse of the elimination matrix E= 1 0 00 1 0− 7 0 1 isE − 1 = 1 0 00 1 07 0 1 . Multiplication confirmsEE − 1 =I. 8 M =a b c d andM*= a b c−ℓa d−ℓb . detM*=a(d−ℓb)−b(c−ℓa) reduces toad−bc! Subtracting row 1 from row 2 doesn’t changedetM. 9 M=1 0 00 0 1−1 1 0. After the exchange, we needE 31 (notE 21 ) to act on the new row 3. 10 E 13 =1 0 10 1 00 0 1;1 0 10 1 01 0 1;E 31 E 13 =2 0 10 1 01 0 1.Test on the identity matrix! 11 An example with two negative pivots isA= 1 2 21 1 21 2 1. The diagonal entries can change sign during elimination. 12 The first product is 9 8 76 5 43 2 1rows and also columns reversed. The second product is 1 2 30 1 − 20 2 − 3.Is Linear Algebra harder than calculus?The pure mechanics of Linear algebra are very basic, being far easier than anything of substance in Calculus. The difficulty is that linear algebra is mostly about understanding terms and definitions and determining the type of calculation and analysis needed to get the required result.
Is Linear Algebra hard?Many students regard linear algebra as a difficult study. It is more challenging than discrete mathematics which is usually a first-year program taught in most STEM majors. Linear algebra is taught in its second year and demands robust reasoning and analytical skills.
How do you introduce in linear algebra?Introduction to Linear Algebra
It is the study of vector spaces, lines and planes, and some mappings that are required to perform the linear transformations. It includes vectors, matrices and linear functions. It is the study of linear sets of equations and its transformation properties.
Who discovered linear algebra?Finding the eigenvectors and eigenvalues for a linear transformation is often done using matrix algebra, first developed in the mid-19th century by the English mathematician Arthur Cayley. His work formed the foundation for modern linear algebra.
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