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· 1998 · 11.51 MB · 4,651 Downloads· English “ If you want to become full, let yourself be empty. ” ― Lao Tzu
Load more similar PDF files PDF Drive investigated dozens of problems and listed the biggest global issues facing the world today. Let's Change The World Together INTRODUCTIONTOLINEARALGEBRAFourth EditionMANUAL FOR INSTRUCTORSGilbert StrangMassachusetts Institute of Technologymath.mit/linearalgebraweb.mit/18.video lectures: ocw.mitmath.mit/gswellesleycambridgeemail:Wellesley - Cambridge PressBox 812060Wellesley, Massachusetts 02482Problem Set 1, page 81 The combinations give (a) a line in R 3 (b) a plane in R 3 (c) all of R 3. 2 vCwD; 3/andvwD;1/will be the diagonals of the parallelogram withv andwas two sides going out from; 0/. 3 This problem gives the diagonalsvCwandvwof the parallelogram and asks for the sides: The opposite of Problem 2. In this examplevD; 3/andwD;2/. 43 vCwD; 5/andcvCdwD; cC2d/. 5 uCvD.2; 3; 1/anduCvCwD; 0; 0/and 2 uC 2 vCwD first answers/D .2; 3; 1/. The vectorsu;v;ware in the same plane because a combination gives .0; 0; 0/. Stated another way:uDvwis in the plane ofvandw. 6 The components of everycvCdwadd to zero 3 anddD 9 give; 3;6/. 7 The nine combinationsc; 1/Cd; 1/withcD0; 1; 2anddD; 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 isvw(or elsewv). Adding diagonals gives 2 v(or 2 w). 9 The fourth corner can be; 4/or; 0/or.2; 2/. Three possible parallelograms! 10 ijD; 1; 0/is in the base (x-yplane).iCjCkD; 1; 1/is the opposite corner from; 0; 0/. Points in the cube have 0 x 1 , 0 y 1 , 0 z 1. 11 Four more corners; 1; 0/; .1; 0; 1/; .0; 1; 1/; .1; 1; 1/. The center point is. 12 ; 12 ; 12 /. Centers of faces are. 12 ; 12 ; 0/;. 12 ; 12 ; 1/and; 12 ; 12 /; .1; 12 ; 12 /and. 12 ; 0; 12 /;. 12 ; 1; 12 /. 12 A four-dimensional cube has 24 D 16 corners and 2 4 D 8 three-dimensional faces and 24 two-dimensional faces and 32 edges in Worked Example2 A. 13 SumDzero vector. SumD 2 : 00 vectorD 8 : 00 vector. 2 : 00 is 30 ıfrom horizontal D 6 ;sin 6 /D. p 3=2; 1=2/. 14 Moving the origin to 6 : 00 addsjD; 1/to every vector. So the sum of twelve vectors changes from 0 to 12 jD; 12/. 15 The point 34vC 14wis three-fourths of the way tovstarting fromw. The vector 1 4 vC 14wis halfway touD 12vC 12w. The vectorvCwis 2 u(the far corner of the parallelogram). 16 All combinations withcCd D 1 are on the line that passes throughvandw. The pointV DvC 2 wis on that line but it is beyondw. 17 All vectorscvCcware on the line passing through; 0/anduD 12 vC 12 w. That line continues out beyondvCwand back beyond; 0/. Withc 0 , half of this line is removed, leaving a ray that starts at; 0/. 18 The combinationscvCdwwith 0 c 1 and 0 d 1 fill the parallelogram with sidesvandw. For example, ifvD; 0/andwD; 1/thencvCdwfills the unit square. 19 Withc 0 andd 0 we get the infinite “cone” or “wedge” betweenvandw. For example, ifvD; 0/andwD; 1/, then the cone is the whole quadrantx 0 , y 0. Question : What ifwDv? The cone opens to a half-space. 5 u 1 Dv=kvk D; 1/= p 10 andu 2 Dw=kwk D; 1; 2/=3 1 D;3/= p 10 is perpendicular tou 1 (and so is.1; 3/= p 10 ).U 2 could be;2; 0/= p 5 : There is a whole plane of vectors perpendicular tou 2 , and a whole circle of unit vectors in that plane. 6 All vectorswD; 2c/are perpendicular tov. All vectors; y; z/withxCyCzD 0 lie on a plane. All vectors perpendicular to; 1; 1/and; 2; 3/lie on a line. 7 (a) cosDvw=kvkkwk D1=.2//soD 60 ıor=3radians (b) cosD 0 so D 90 ı or=2radians (c) cos D 2=.2//D 1=2so D 60 ı or= (d) cosD1= p 2 soD 135 ıor3=4. 8 (a) False:vandware any vectors in the plane perpendicular tou (b) True:u.vC 2 w/DuvC 2 uwD 0 (c) True,kuvk 2 Dv/.uv/splits into uuCvvD 2 whenuvDvuD 0. 9 Ifv 2 w 2 =v 1 w 1 D 1 thenv 2 w 2 Dv 1 w 1 orv 1 w 1 Cv 2 w 2 DvwD 0 : perpendicular! 10 Slopes2=1and1=2multiply to give 1 : thenvwD 0 and the vectors (the directions) are perpendicular. 11 vw< 0means angle> 90ı; thesew’s fill half of 3 -dimensional space. 12 .1; 1/perpendicular to; 5/c; 1/if 6 2cD 0 orcD 3 ;v.wcv/D 0 if cDvw=vv. Subtractingcvis the key to perpendicular vectors. 13 The plane perpendicular to; 0; 1/contains all vectors; d;c/. In that plane,vD .1; 0;1/andwD; 1; 0/are perpendicular. 14 One possibility among many:uD;1; 0; 0/;vD; 0; 1;1/;wD; 1;1;1/ and; 1; 1; 1/are perpendicular to each other. “We can rotate thoseu;v;win their 3 D hyperplane.” 15 12 .xCy/D/=2D 5 ; cosD 2 p 16= p 10 p 10 D8=10. 16 kvk 2 D 1 C 1 CC 1 D 9 sokvkD 3 IuDv=3D. 13 ; : : : ; 13 /is a unit vector in 9 D; wD;1; 0; : : : ; 0/= p 2 is a unit vector in the 8 D hyperplane perpendicular tov. 17 cos ̨D1= p 2 , cosˇD 0 , cos D1= p 2. For any vectorv, cos 2 ̨Ccos 2 ˇCcos 2 D 12 Cv 22 Cv 32 /=kvk 2 D 1. 18 kvk 2 D 42 C 22 D 20 andkwk 2 D.1/ 2 C 22 D 5. Pythagoras isk; 4/k 2 D 25 D 20 C 5. 19 Start from the rules/; .2/; .3/forvwDwvandu.vCw/and/w. Use rule/ for/.vCw/D/vC/w. By rule/this isv.vCw/Cw.vCw/. Rule/again givesvvCvwCwvCwwDvvC 2 vwCww. Notice vwDwv! The main point is to be free to open up parentheses. 20 We know thatw/.vw/Dvv 2 vwCww. The Law of Cosines writes kvkkwkcosforvw. When < 90ıthisvwis positive, so in this casevvCww is larger thankvwk 2. 212 vw 2 kvkkwkleads tokvCwk 2 DvvC 2 vwCwwkvk 2 C 2 kvkkwkCkwk 2. This is/ 2. Taking square roots giveskvCwkkvkCkwk. 22 v 21 w 21 C2v 1 w 1 v 2 w 2 Cv 22 w 22 v 12 w 12 Cv 21 w 22 Cv 22 w 21 Cv 22 w 22 is true (cancel 4 terms) because the difference isv 21 w 22 Cv 22 w 12 2v 1 w 1 v 2 w 2 which is 1 w 2 v 2 w 1 / 2 0. 23 cosˇDw 1 =kwkand sinˇDw 2 =kwk. Then cos.ˇa/Dcosˇcos ̨Csinˇsin ̨D v 1 w 1 =kvkkwkCv 2 w 2 =kvkkwkDvw=kvkkwk. This is cosbecauseˇ ̨D. 24 Example 6 givesju 1 jjU 1 j 12 .u 21 CU 12 /andju 2 jjU 2 j 12 .u 22 CU 22 /. The whole line becomes:96.:6/.:8/C.:8/.:6/ 12 .:6 2 C:8 2 /C 12 .:8 2 C:6 2 /D 1. True::96 < 1. 25 The cosine ofisx= p x 2 Cy 2 , near side over hypotenuse. Thenjcosj 2 is not greater than 1:x 2 = 2 Cy 2 / 1. 26 The vectorswD; y/with; 2/wDxC2yD 5 lie on a line in thexyplane. The shortestwon that line is; 2/. (The Schwarz inequalitykwkvw=kvkD p 5 is an equality when cosD 0 andwD; 2/andkwkD p 5 .) 27 The lengthkvwkis between 2 and 8 (triangle inequality whenkvkD 5 andkwkD 3 ). The dot productvwis between 15 and 15 by the Schwarz inequality. 28 Three vectors in the plane could make angles greater than 90 ıwith each other: for example; 0/; .1; 4/; .1;4/. Four vectors could not do this ( 360 ıtotal angle). How many can do this in R 3 or R n? Ben Harris and Greg Marks showed me that the answer isnC1:The vectors from the center of a regular simplex in R nto itsnC 1 vertices all have negative dot products. IfnC 2 vectors in R nhad negative dot products, project them onto the plane orthogonal to the last one. Now you havenC 1 vectors in R n 1 with negative dot products. Keep going to 4 vectors in R 2 : no way! 29 For a specific example, pickvD; 2;3/and thenwD.3; 1; 2/. In this example cos Dvw=kvkkwk D 7= p 14 p 14 D 1=2and D 120 ı. This always happens whenxCyCzD 0 : vwDxzCxyCyzD 12.xCyCz/ 2 12.x 2 Cy 2 Cz 2 / This is the same asvwD 0 12kvkkwk:Then cosD 12:30 Wikipedia gives this proof of geometric meanG D 3 p xyz arithmetic mean AD/=3. First there is equality in casexDyDz. OtherwiseAis somewhere between the three positive numbers, say for examplez < A < y. Use the known inequalitygafor the two positive numbersxandyCzA. Their meanaD 12 .xCyCzA/is 12 .3AA/ Dsame asA! Soagsays that A 3 g 2 ADxA/A. ButA/ADA/z/Cyz > yz. Substitute to findA 3 > xyzDG 3 as we wanted to prove. Not easy! There are many proofs ofGD 1 x 2 xn/1=nAD 1 Cx 2 CCxn/=n. In calculus you are maximizingGon the planex 1 Cx 2 CCxnDn. The maximum occurs when allx’s are equal. 31 The columns of the 4 by 4 “Hadamard matrix” (times 12 ) are perpendicular unit vectors: 12HD122641 1 1 11 1 1 11 1 1 11 1 1 1375 :32 The commandsVD randn .3; 30/IDD sqrt. diag .V 0 V //IUDV\DIwill give 30 random unit vectors in the columns ofU. Thenu 0 Uis a row matrix of 30 dot products whose average absolute value may be close to2=. 10 z 2 z 1 Db 1 z 3 z 2 Db 2 0 z 3 Db 3 z 1 D b 1 b 2 b 3 z 2 D b 2 b 3 z 3 D b 3 D" 1 1 10 1 10 0 1"b b 2 b 3 null D 1 b 11 The forward differences of the squares are/ 2 t 2 Dt 2 C2tC 1 t 2 D2tC 1. Differences of thenth power are/ntnDtntnCntn 1 C. The leading term is the derivativentn 1. The binomial theorem gives all the terms of/n. 12 Centered difference matrices of even size seem to be invertible. Look at eqns. 1 and 4 : 2 6 4 0 1 0 01 0 1 00 1 0 10 0 1 0375264x 1 x 2 x 3 x 4 375 D264b 1 b 2 b 3 b 4 375First solve x 2 Db 1 x 3 Db 4 264x 1 x 2 x 3 x 4 375 D264b 2 b 4 b 1 b 4 b 1 Cb 3 37513 Odd size : The five centered difference equations lead tob 1 Cb 3 Cb 5 D 0. x 2 Db 1 x 3 x 1 Db 2 x 4 x 2 Db 3 x 5 x 3 Db 4 x 4 Db 5 Add equations1; 3; 5 The left side of the sum is zero The right side isb 1 Cb 3 Cb 5 There cannot be a solution unlessb 1 Cb 3 Cb 5 D 0. 14 An example is; b/D; 6/and; d/D; 2/. The ratiosa=candb=dare equal. ThenadDbc. Then (when you divide bybd) the ratiosa=bandc=dare equal! Problem Set 2, page 401 The columns areiD; 0; 0/andjD; 1; 0/andkD; 0; 1/andbD; 3; 4/D 2 iC 3 jC 4 k. 2 The planes are the same:2xD 4 isxD 2 ,3yD 9 isyD 3 , and4zD 16 iszD 4. The solution is the same pointXDx. The columns are changed; but same combination. 3 The solution is not changed! The second plane and row 2 of the matrix and all columns of the matrix (vectors in the column picture) are changed. 4 IfzD 2 thenxCyD 0 andxyDzgive the point;1; 2/. IfzD 0 then xCyD 6 andxyD 4 produce; 1; 0/. Halfway between those is; 0; 1/. 5 Ifx; y; zsatisfy the first two equations they also satisfy the third equation. The line L of solutions containsvD; 1; 0/andwD. 12 ; 1; 12 /anduD 12 vC 12 wand all combinationscvCdwwithcCdD 1. 6 Equation 1 Cequation 2 equation 3 is now 0 D 4. Line misses plane; no solution. 7 Column 3 DColumn 1 makes the matrix singular. Solutions; y; z/D; 1; 0/or .0; 1; 1/and you can add any multiple of.1; 0; 1/;bD; 6; c/needscD 10 for solvability (thenblies in the plane of the columns). 8 Four planes in 4-dimensional space normally meet at a point. The solution toAxD .3; 3; 3; 2/isx D .0; 0; 1; 2/ifA has columns; 0; 0; 0/; .1; 1; 0; 0/; .1; 1; 1; 0/, .1; 1; 1; 1/. The equations arexCyCzCtD3; yCzCtD3; zCtD3; tD 2. 9 (a)AxD; 5; 0/and (b)AxD; 4; 5; 5/. 10 Multiplying as linear combinations of the columns gives thesameAx. By rows or by columns: 9 separate multiplications for 3 by 3. 11 Axequals; 22/and; 0/and (9; 7/. 12 Axequals; y; x/and; 0; 0/and (3; 3; 6/. 13 (a)xhasncomponents andAxhasmcomponents (b) Planes from each equation inAxDbare inn-dimensional space, but the columns are inm-dimensional space. 14 2xC3yCzC5t D 8 isAxDbwith the 1 by 4 matrixADŒ 2 3 1 5 . The solutionsxfill a 3D “plane” in 4 dimensions. It could be called a hyperplane. 15 (a)ID 1 00 1(b)PD 0 11 01690 ırotation fromRD 0 11 0, 180 ırotation fromR 2 D 1 00 1DI.17 PD"0 1 00 0 11 0 0null produces; z; x/andQD "0 0 11 0 00 1 0null recovers; y; z/ the inverse ofP. 18 ED1 01 1andED "1 0 01 1 00 0 1null subtract the first component from the second. 19 ED"1 0 00 1 01 0 1null andE 1 D "1 0 00 1 01 0 1null ,Ev D; 4; 8/andE 1 Evrecovers .3; 4; 5/.20 P 1 D1 00 0projects onto thex-axis andP 2 D 0 00 1projects onto they-axis. vD 57hasP 1 vD 50andP 2 P 1 vD 00.21 RD12p 2 p p 2 2 p 2 rotates all vectors by 45ı. The columns ofRare the results from rotating; 0/and; 1/! 22 The dot productAxDŒ 1 4 5 "x y z null D by 3/ by 1/is zero for points; y; z/ on a plane in three dimensions. The columns ofAare one-dimensional vectors. 23 ADŒ 1 2 I 3 4 andxDŒ 5 2 0 andbDŒ 1 7 0 .rDbAxprints as zero. 24 AvDŒ 3 4 5 0 andv 0 vD 50. ButvAgives an error message from 3 by 1 times 3 by 3. 25 ones; 4/ones; 1/DŒ 4 4 4 4 0 ;BwDŒ 10 10 10 10 0. 26 The row picture has two lines meeting at the solution (4; 2). The column picture will have4; 1/C2.2; 1/D 4 (column 1)C2(column 2)Dright side; 6/. 27 The row picture shows 2 planes in 3-dimensional space. The column picture is in 2-dimensional space. The solutions normally lie on aline. 8 IfkD 3 elimination must fail: no solution. Ifk D 3 , elimination gives 0 D 0 in equation 2: infinitely many solutions. IfkD 0 a row exchange is needed: one solution. 9 On the left side,6x4yis 2 times2y/. Therefore we needb 2 D2b 1 on the right side. Then there will be infinitely many solutions (twoparallel lines become one single line). 10 The equationyD 1 comes from elimination (subtractxCyD 5 fromxC2yD 6 ). ThenxD 4 and5x4yDcD 16. 11 (a) Another solution is 12 .xCX; yCY; zCZ/. (b) If 25 planes meet at two points, they meet along the whole line through those two points. 12 Elimination leads to an upper triangular system; then comes back substitution. 2xC3yC zD 8 yC3zD 4 8zD 8 gives xD 2 yD 1 If a zero is at the start of row 2 or 3, zD 1 that avoids a row operation. 13 2x3y D 3 4x5yC zD 7 2x y3zD 5 gives 2x3yD 3 yC zD 1 2yC3zD 2 and 2x3yD 3 yC zD 1 5zD 0 and xD 3 yD 1 zD 0 Subtract 2row 1 from row 2, subtract 1row 1 from row 3, subtract 2row 2 from row 3 14 Subtract 2 times row 1 from row 2 to reach20/yzD 2. Equation (3) isyzD 3. IfdD 10 exchange rows 2 and 3. IfdD 11 the system becomes singular. 15 The second pivot position will contain 2 b. IfbD 2 we exchange with row 3. If bD 1 (singular case) the second equation isyzD 0. A solution is; 1;1/. 16 (a) Example of 2 exchanges 0xC0yC2zD 4 xC2yC2zD 5 0xC3yC4zD 6 (exchange 1 and 2, then 2 and 3) (b) Exchange but then break down 0xC3yC4zD 4 xC2yC2zD 5 0xC3yC4zD 6 (rows 1 and 3 are not consistent) 17 If row 1 Drow 2, then row 2 is zero after the first step; exchange the zerorow with row 3 and there is no third pivot. If column 2 Dcolumn 1, then column 2 has no pivot. 18 Example xC2yC3zD 0 ,4xC8yC12zD 0 ,5xC10yC15zD 0 has 9 different coefficients but rows 2 and 3 become 0 D 0 : infinitely many solutions. 19 Row 2 becomes3y4zD 5 , then row 3 becomes/zDt 5. IfqD 4 the system is singular—no third pivot. Then iftD 5 the third equation is 0 D 0. Choosing zD 1 the equation3y4zD 5 givesyD 3 and equation 1 givesxD 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 rows 1 C 2 Drow 3 on the left side but not the right side:xCyCzD 0 ,x2yzD 1 ,2xyD 4. No parallel planes but still no solution. 21 (a) Pivots2; 32 ; 43 ; 54 in the equations2xCyD0; 32 yCzD0; 43 zCtD0; 54 tD 5 after elimination. Back substitution givestD4; zD 3; yD2; xD 1. (b) If the off-diagonal entries change fromC 1 to 1 , the pivots are the same. The solution is .1; 2; 3; 4/instead of.1; 2;3; 4/. 22 The fifth pivot is 65 for both matrices (1’s or 1 ’s off the diagonal). Thenth pivot is nC 1 n. 23 If ordinary elimination leads toxCyD 1 and2yD 3 , the original second equation could be2yC 24 Elimination fails on a 2 a a ifaD 2 oraD 0. 25 aD 2 (equal columns),aD 4 (equal rows),aD 0 (zero column). 26 Solvable forsD 10 (add the two pairs of equations to getaCbCcCdon the left sides, 12 and 2 Cson the right sides). The four equations fora; b; c; dare singular! Two solutions are 1 31 7and 0 42 6,AD2641 1 0 01 0 1 00 0 1 10 1 0 1375 andUD 2641 1 0 00 1 1 00 0 1 10 0 0 0375.27 Elimination leaves the diagonal matrix diag; 2; 1/in3xD3; 2yD2; zD 4. Then xD1; yD1; zD 4. 28 A;W/DA;W/ 3 A;W/subtracts 3 times row 1 from row 2. 29 The average pivots for rand(3) without row exchanges were 12 ; 5; 10in one experiment— but pivots 2 and 3 can be arbitrarily large. Their averages are actually infinite! With row exchanges in MATLAB’s lu code, the averages:75and:50and:365are much more stable (and should be predictable, also for randn with normal instead of uniform probability distribution). 30 IfA; 5/is 7 not 11 , then the last pivot will be 0 not 4. 31 RowjofUis a combination of rows1; : : : ; jofA. IfAxD 0 thenUxD 0 (not true ifbreplaces 0 ).Uis the diagonal ofAwhenAis lower triangular. 32 The question deals with 100 equationsAxD 0 whenAis singular. (a) Some linear combination of the 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:AD eye (100). A better example has 99 random rows (or the numbers 1 i; : : : ; 100iin 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 planes meeting along a common line through 0. The column picture has 100 vectors all in the same 99-dimensional hyperplane. Problem Set 2, page 631 E 21 D"1 0 05 1 00 0 1null ; E 32 D"1 0 00 1 00 7 1null ; PD"1 0 00 0 10 1 0# "0 1 01 0 00 0 1null D"0 1 00 0 11 0 0null .2 E 32 E 21 bD;5;35/butE 21 E 32 bD;5; 0/. WhenE 32 comes first, row 3 feels no effect from row 1. 3 "1 0 04 1 00 0 1null ;"1 0 00 1 02 0 1null ;"1 0 00 1 00 2 1null MDE 32 E 31 E 21 D"1 0 04 1 010 2 1null :16 (a) The ages ofXandYarexandy:x2yD 0 andxCyD 33 ;xD 22 and yD 11 (b) The lineyDmxCccontainsxD 2 ,yD 5 andxD 3 ,yD 7 when 2mCcD 5 and3mCcD 7. ThenmD 2 is the slope. 17 The parabolayDaCbxCcx 2 goes through the 3 given points when aC bC cD 4 aC2bC4cD 8 aC3bC9cD 14 .ThenaD 2 ,bD 1 , andcD 1. This matrix with columns; 1; 1/,.1; 2; 3/,.1; 4; 9/ is a “Vandermonde matrix.” 18 EFD"1 0 0a 1 0 b c 1 null ,FED"1 0 0a 1 0 bCac c 1 null ,E 2 D"1 0 02a 1 0 2b 0 1 null ,F 3 D"1 0 00 1 00 3c 1 null :19 PQD"0 1 00 0 11 0 0null . In the opposite order, two row exchanges giveQPD "0 0 11 0 00 1 0null ,IfMexchanges rows 2 and 3 thenM 2 DI(also.M/ 2 DI). There are many square roots ofI: Any matrixMD a b c a hasM 2 DIifa 2 CbcD 1. 20 (a) Each column ofEBisEtimes a column ofB (b) 1 01 11 2 41 2 4D1 2 42 4 8. All rows ofEBare multiples of 1 2 4.21 No. ED 1 01 1andFD 1 10 1giveEFD 1 11 2butFED 2 11 1.22 (a) Pa3jxj (b)a 21 a 11 (c)a 21 2a 11 (d).EAx/ 1 D/ 1 D Pa1jxj. 23 E/subtracts 4 times row 1 from row 2 (EEAdoes the row operation twice). AEsubtracts 2 times column 2 ofAfrom column 1 (multiplication byEon the right side acts on columns instead of rows). 24 A b D2 3 14 1 17!2 3 10 5 15. The triangular system is 2x 1 C3x 2 D 1 5x 2 D 15 Back substitution givesx 1 D 5 andx 2 D 3. 25 The last equation becomes 0 D 3. If the original 6 is 3, then row 1Crow 2Drow 3. 26 (a) Add two columnsbandb 1 4 1 02 7 0 1!1 4 1 00 1 2 1!xD 72andxD 41.27 (a) No solution ifdD 0 andc¤ 0 (b) Many solutions ifdD 0 Dc. No effect froma; b. 28 ADAIDA.BC/D/CDICDC. That middle equation is crucial. 29 ED2641 0 0 01 1 0 00 1 1 00 0 1 1375 subtracts each row from the next row. The result 2641 0 0 00 1 0 00 1 1 00 1 2 1375still has multipliersD 1 in a 3 by 3 Pascal matrix. The productMof all elimination matrices is 2641 0 0 01 1 0 01 2 1 01 3 3 137
30 Given positive integers withadbcD 1. Certainlyc < aandb < dwould be impossible. Alsoc > aandb > dwould be impossible with integers. This leaves row1 <row 2 OR row2 <row 1. An example isM D 3 42 3. Multiply by 1 1 0 1 to get 1 12 3, then multiply twice by 1 01 1to get 1 10 1. This shows thatMD 1 10 11 01 11 01 11 10 1.31 E 21 D266411=2 10 0 10 0 0 13775 ,E 32 D266410 10 2=3 10 0 0 13775 ,E 43 D266410 10 0 10 0 3=4 13775 ,E 43 E 32 E 21 D266411=2 11=3 2=3 11=4 2=4 3=4 13775Problem Set 2, page 751 If all entries ofA; B; C; Dare 1 , thenBAD 3 ones .5/is 5 by 5 ;ABD 5 ones .3/is 3 by 3 ;ABDD 15 ones .3; 1/is 3 by 1 .DBAandA/are not defined. 2 (a)A(column 3 ofB) (b) (Row 1 ofA)B (c) (Row 3 ofA)(column 4 ofB) (d) (Row 1 ofC)D(column 1 ofE). 3 ABCACis the same asA/D 3 86 9. ( Distributive law ). 4 A.BC/D/Cby the associative law. In this example both answers are 0 00 0from column 1 ofABand row 2 ofC(multiply columns times rows). 5 (a)A 2 D 1 2b 0 1 andAnD 1 nb 0 1 . (b) A 2 D 4 40 0andAnD 2 n 2 n 0 0 .6 .ACB/ 2 D10 46 6DA 2 CABCBACB 2. ButA 2 C2ABCB 2 D 16 23 0.7 (a) True (b) False (c) True (d) False: usually/ 2 ¤A 2 B 2. 21 ADA 2 DA 3 DD":5 ::5 :null butABD ":5 ::5 :null and/ 2 Dzero matrix! 22 AD0 11 0hasA 2 DI;BCD 1 11 11 11 1D0 00 0;DED0 11 00 11 0D1 00 1DED. You can find more examples. 23 AD"0 10 0null hasA 2 D 0. Note: Any matrixADcolumn times rowDuvTwill haveA 2 DuvTuvTD 0 ifvTuD 0 .AD 2640 1 00 0 10 0 0375 hasA 2 D 2640 0 10 0 00 0 0375butA 3 D 0 ; strictly triangular as in Problem 20. 24 .A 1 /nD 2 n 2 n 1 0 1 ,.A 2 /nD 2 n 1 1 11 1,.A 3 /nD an an 1 b 0 0 .25 24a b c d e f g h i 35241 0 00 1 00 0 135 D24a d g 351 0 0C24d e h 350 1 0C24c f i 350 0 1.26 Columns of A times rows of B "122null 3 3 0C"041null 1 2 1D"3 3 06 6 06 6 0null C"0 0 04 8 41 2 1null D" 3 3 010 14 47 8 1null DAB.27 (a) (row 3 ofA)(column 1 ofB) and (row 3 ofA)(column 2 ofB) are both zero. (b) "x x 0 null 0 x x D"0 x x 0 x x 0 0 0 null and "x x x null 0 0 x D"0 0 x 0 0 x 0 0 x null : both upper. 28 AtimesB with cuts Aˇˇˇˇˇˇˇˇˇˇˇˇ,B,ˇˇˇˇˇˇˇˇˇˇˇˇ,ˇˇˇˇˇˇˇˇ29 E 21 D"1 0 01 1 00 0 1null andE 31 D "1 0 00 1 04 0 1null produce zeros in the2; 1and3; 1entries. MultiplyE’s to getEDE 31 E 21 D "1 0 01 1 04 0 1null . ThenEAD "2 1 00 1 10 1 3null is the result of bothE’s since 31 E 21 /ADE 31 .E 21 A/. 30 In 29 ,cD 28,DD0 15 3,Dcb=aD 1 11 3in the lower corner ofEA. 31 A BB Ax y DAxBy BxCAy real part imaginary part. Complex matrix times complex vector needs 4 real times real multiplications. 32 AtimesXDŒx 1 x 2 x 3 will be the identity matrixIDŒ Ax 1 Ax 2 Ax 3 . 33 bD "358null givesxD 3 x 1 C 5 x 2 C 8 x 3 D "3816null I AD"1 0 01 1 00 1 1null will have thosex 1 D; 1; 1/;x 2 D; 1; 1/;x 3 D; 0; 1/as columns of its “inverse”A 1. 34 A ones D aCb aCb cCd cCd agrees with ones AD aCc bCb aCc bCd whenbDc andaDd ThenAD a b b a .35 AD2640 1 0 11 0 1 00 1 0 11 0 1 0375 ; A 2 D2642 0 2 00 2 0 22 0 2 00 2 0 2375 ;aba, ada cba, cda bab, bcb dab, dcb abc, adc cbc, cdc bad, bcd dad, dcd These show 16 2-step paths in the graph 36 MultiplyingABD(mbyn)(nbyp) needsmnpmultiplications. Then/Cneeds mpqmore. MultiplyBCD(nbyp)(pbyq) needsnpqand thenA/needsmnq. (a) Ifm; n; p; qare2; 4; 7; 10we compare.2/.4/.7/C.2/.7//D 196 with the larger number.2/.4/.10/C.4/.7//D 360. SoABfirst is better, so that we multiply that 7 by 10 matrix by as few rows as possible. (b) Ifu;v;wareNby 1 , then/wTneeds2Nmultiplications butuT/needs N 2 to findvwTandN 2 more to multiply by the row vectoruT. Apologies to use the transpose symbol so early. (c) We are comparingmnpCmpqwithmnqCnpq. Divide all terms bymnpq: Now we are comparingq 1 Cn 1 withp 1 Cm 1. This yields a simple important rule. If matricesAandBare multiplyingvforABv, don’t multiply the matrices first. 37 The proof of.AB/cDA/used the column rule for matrix multiplication—this rule is clearly linear, column by column. Even for nonlinear transformations,A.B//would be the “composition” ofAwithB (applyingBthenA). This compositionAıBis justABfor matrices. One of many uses for the associative law: The left-inverseB= right-inverseCfrom BDB.AC/D/CDC. Problem Set 2, page 891 A 1 D0141 3 0 andB 1 D 12 0112andC 1 D 7 45 3.2 A simple row exchange hasP 2 DIsoP 1 DP. HereP 1 D "0 0 11 0 00 1 0null . Always P 1 = “transpose” ofP, coming in Section2:7. 19 The; 1/entry requires4a3bD 1 ; the; 2/entry requires2baD 0. Then bD 15 andaD 25. For the 5 by 5 case5a4bD 1 and2bDagivebD 16 and aD 26. 20 Aones; 1/is the zero vector soAcannot be invertible. 21 Six of the sixteen 0 1 matrices are invertible, including all four with three 1’s. 22 1 3 1 02 7 0 1!1 3 1 00 1 2 1!1 0 7 30 1 2 1DI A 1;1 4 1 03 9 0 1!1 4 1 00 3 3 1!1 0 3 4=0 1 1 1=DI A 1.23 ŒA ID"2 1 0 1 0 01 2 1 0 1 00 1 2 0 0 1null !"2 1 0 1 0 00 3=2 1 1=2 1 00 1 2 0 0 1null !"2 1 0 1 0 00 3=2 1 1=2 1 00 0 4=3 1=3 2=3 1null !"2 1 0 1 0 00 3=2 0 3=4 3=2 3=0 0 4=3 1=3 2=3 1null !"2 0 0 3=2 1 1=0 3=2 0 3=4 3=2 3=0 0 4=3 1=3 2=3 1null !"1 0 0 3=4 1=2 1=0 1 0 1=2 1 1=0 0 1 1=4 1=2 3=null DŒI A 1 .24 "1 a b 1 0 0 0 1 c 0 1 0 0 0 1 0 0 1 null !"1 a 0 1 0 b 0 1 0 0 1 c 0 0 1 0 0 1 null !"1 0 0 1 a acb 0 1 0 0 1 c 0 0 1 0 0 1 null .25 "2 1 11 2 11 1 2# 1D14"3 1 11 3 11 1 3null I"2 1 11 2 11 1 2# "111null D"000null soB 1 does not exist. 26 E 21 AD 1 02 11 22 6D1 20 2.E 12 E 21 AD1 10 11 02 1AD1 00 2.Multiply byDD 1 00 1=to reachDE 12 E 21 ADI. ThenA 1 DDE 12 E 21 D 1 2 6 22 1.27 A 1 D"1 0 02 1 30 0 1null (notice the pattern);A 1 D "2 1 01 2 10 1 1null .28 0 2 1 02 2 0 1!2 2 0 10 2 1 0!2 0 1 10 2 1 0!1 0 1=2 1=0 1 1=2 0.This is I A 1: row exchanges are certainly allowed in Gauss-Jordan. 29 (a) True (IfAhas a row of zeros, then everyABhas too, andABDIis impossible) (b) False (the matrix of all ones is singular even with diagonal 1’s: ones (3) has 3 equal rows) (c) True (the inverse ofA 1 isAand the inverse ofA 2 is 1 / 2 /. 30 ThisAis not invertible forcD 7 (equal columns),cD 2 (equal rows),cD 0 (zero column). 31 Elimination produces the pivotsaandabandab 1 D 1ab/ " a 0b a a 0 0 a a null .32 A 1 D2641 1 0 00 1 1 00 0 1 10 0 0 137
A 1 is bidiagonal with 1’s on the diagonal and first superdiagonal. 33 xD; 1; : : : ; 1/hasPxDQxsoQ/xD 0. 34 I 0C Iand A 10D 1 CA 1 D 1and D II 0.35 Acan be invertible with diagonal zeros singular because each row adds to zero. 36 The equationLDLDDIsays thatLDDpascal; 1/is its own inverse. 37 hilb(6) is not the exact Hilbert matrix because fractions are rounded off. Soinv(hilb(6)) is not the exact either. 38 The three Pascal matrices havePDLUDLLTand theninv /Dinv.LT/inv/. 39 AxDbhas many solutions whenADones; 4/Dsingular matrix andbDones .4; 1/ MATLAB will pick the shortest solutionxD; 1; 1; 1/=4. This is the only solution that is combination of the rows of A (later it comes from the “pseudoinverse”ACDpinv(A) which replacesA 1 whenAis singular). Any vec- tor that solvesAxD 0 could be added to this particular solutionx. 40 The inverse ofA D 2641 a 0 0 0 1 b 0 0 0 1 c 0 0 0 1 375 isA 1 D 2641 a ab abc 0 1 b bc 0 0 1 c 0 0 0 1 37
would be a good example for the cofactor formulaA 1 DCT=detAin Section 5) 41 The product 2641a 1 b 0 1 c 0 0 1 37526410 10 d 1 0 e 0 1 375264111f 1 375 D2641a 1 b d 1 c e f 1 375that in this order the multipliers showsa; b; c; d; e; f are unchanged in the product ( important for ADLU in Section 2 ). 42 MM 1 DU V / .InCUV U / 1 V / .this is testing formula 3 / DInU VCUV U / 1 VU V UV U / 1 V .keep simplifying/ DInU VCUV U /V U / 1 VDIn 1 ; 2 ; 4 are similar/ 434 by 4 still withT 11 D 1 has pivots1; 1; 1; 1; reversing toTDULmakesT 44 D 1. 44 Add the equationsCxDbto find 0 Db 1 Cb 2 Cb 3 Cb 4. Same forFxDb. 45 The block pivots are A and S DDCA 1 B (and dcb=a is the correct second pivot of an ordinary 2 by 2 matrix). The example problem has SD 1 00 144123 3D5 66 5. |
Linear algebra and its applications solutions 4th edition
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