page 224 224 CHAPTER 3 Determnants e t te t e 2t 38 A = e t 2te t e 2t e t te t 2e 2t 39 If 123 A = 345, 456 compute the matrx product A adj(a) What can you conclude about det(a)? For Problems 40 43, use Cramer s rule to solve the gven lnear system 40 41 42 43 2x 1 3x 2 = 2, x 1 + 2x 2 = 4 3x 1 2x 2 + x 3 = 4, x 1 + x 2 x 3 = 2, x 1 + x 3 = 1 x 1 3x 2 + x 3 = 0, x 1 + 4x 2 x 3 = 0, 2x 1 + x 2 3x 3 = 0 x 1 2x 2 + 3x 3 x 4 = 1, 2x 1 + x 3 = 2, x 1 + x 2 x 4 = 0, x 2 2x 3 + x 4 = 3 44 Use Cramer s rule to determne x 1 and x 2 f e t x 1 + e 2t x 2 = 3snt, e t x 1 2e 2t x 2 = 4 cos t 45 Determne the value of x 2 such that x 1 + 4x 2 2x 3 + x 4 = 2, 2x 1 + 9x 2 3x 3 2x 4 = 5, x 1 + 5x 2 + x 3 x 4 = 3, 3x 1 + 14x 2 + 7x 3 2x 4 = 6 46 Fnd all solutons to the system (b + c)x 1 + a(x 2 + x 3 ) = a, (c + a)x 1 + b(x 3 + x 1 ) = b, (a + b)x 1 + c(x 1 + x 2 ) = c, where a,b,c are constants Make sure you consder all cases (that s, those when there s a unque soluton, an nfnte number of solutons, and no solutons) 47 Prove Equaton (333) 48 Let A be a randomly generated nvertble 4 4 matrx Verfy the Cofactor Expanson Theorem for expanson along row 1 49 Let A be a randomly generated 4 4 matrx Verfy Equaton (333) when = 2 and j = 4 50 Let A be a randomly generated 5 5 matrx Determne adj(a) and compute A adj(a) Use your result to determne det(a) 51 Solve the system of equatons 121x 1 + 342x 2 + 215x 3 = 325, 541x 1 + 232x 2 + 715x 3 = 461, 2163x 1 + 351x 2 + 922x 3 = 993 Round answers to two decmal places 52 Use Cramer s rule to solve the system Ax = b f 12344 68 21234 A = 32123 43212, and b = 72 87 79 44321 43 53 Verfy that BA = I n n the proof of Theorem 3316 34 Summary of Determnants The prmary am of ths secton s to serve as a stand-alone ntroducton to determnants for readers who desre only a cursory revew of the major facts pertanng to determnants It may also be used as a revew of the results derved n Sectons 31 33 Formulas for the Determnant The determnant of an n n matrx A, denoted det(a), s a scalar whose value can be obtaned n the followng manner 1 If A =[a 11 ], then det(a) = a 11
page 225 34 Summary of Determnants 225 [ ] a11 a 2 If A = 12, then det(a) = a a 21 a 11 a 22 a 12 a 21 22 3 For n>2, the determnant of A can be computed usng ether of the followng formulas: det(a) = a 1 C 1 + a 2 C 2 + +a n C n, (341) det(a) = a 1j C 1j + a 2j C 2j + +a nj C nj, (342) where C j = ( 1) +j M j, and M j s the determnant of the matrx obtaned by deletng the th row and jth column of A The formulas (341) and (342) are referred to as cofactor expanson along the th row and cofactor expanson along the j th column, respectvely The determnants M j and C j are called the mnors and cofactors of A, respectvely We also denote det(a) by a 11 a 12 a 1n a 21 a 22 a 2n a n1 a n2 a nn a 11 a 12 a 13 a 11 a 12 a 21 a 22 a 23 a 21 a 22 a 31 a 32 a 33 a 31 a 32 Fgure 341: A schematc for obtanng the determnant of a 3 3 matrx A =[a j ] As an example, consder the general 3 3 matrx a 11 a 12 a 13 A = a 21 a 22 a 23 a 31 a 32 a 33 Usng cofactor expanson along row 1, we have We next compute the requred cofactors: det(a) = a 11 C 11 + a 12 C 12 + a 13 C 13 (343) C 11 =+M 11 = a 22 a 23 a 32 a 33 = a 22 a 33 a 23 a 32, C 12 = M 12 = a 21 a 23 a 31 a 33 = (a 21 a 33 a 23 a 31 ), C 13 =+M 13 = a 21 a 22 = a a 31 a 21 a 32 a 22 a 31 32 Insertng these expressons for the cofactors nto Equaton (343) yelds det(a) = a 11 (a 22 a 33 a 23 a 32 ) a 12 (a 21 a 33 a 23 a 31 ) + a 13 (a 21 a 32 a 22 a 31 ), whch can be wrtten as det(a) = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 11 a 23 a 32 a 12 a 21 a 33 a 13 a 22 a 31 Although we chose to use cofactor expanson along the frst row to obtan the precedng formula, accordng to (341) and (342), the same result would have been obtaned f we had chosen to expand along any row or column of A A smple schematc for obtanng the terms n the determnant of a 3 3 matrx s gven n Fgure 341 By takng the product of the elements joned by each arrow and attachng the ndcated sgn to the result, we obtan the sx terms n the determnant of the 3 3 matrx A =[a j ] Note that ths technque for obtanng the terms n a 3 3 determnant does not generalze to determnants of larger matrces
page 226 226 CHAPTER 3 Determnants Example 341 Evaluate Soluton: so that 3 4 2 7 5 8 In ths case, the schematc gven n Fgure 341 s 2 1 3 4 2 3 4 7 587 5 3 4 2 = (2)(4)(8) + ( 1)(2)(7) + (1)(3)(5) (7)(4)(1) (5)(2)(2) (8)(3)( 1) 7 5 8 = 41 Propertes of Determnants Let A and B be n n matrces The determnant has the followng propertes: P1 If B s obtaned by permutng two rows (or columns) of A, then det(b) = det(a) P2 If B s obtaned by multplyng any row (or column) of A by a scalar k, then det(b) = k det(a) P3 If B s obtaned by addng a multple of any row (or column) of A to another row (or column) of A, then det(b) = det(a) P4 det(a T ) = det(a) P5 Let a 1, a 2,, a n denote the row vectors of A Iftheth row vector of A s the sum of two row vectors, say a = b + c, then where and det(a) = det(b) + det(c), B =[a 1, a 2,,a 1, b, a +1,,a n ] T C =[a 1, a 2,,a 1, c, a +1,,a n ] T The correspondng property for columns s also true P6 If A has a row (or column) of zeros, then det(a) = 0 P7 If two rows (or columns) of A are the same, then det(a) = 0 P8 det(ab) = det(a)det(b)
page 227 34 Summary of Determnants 227 The frst three propertes tell us how elementary row operatons and elementary column operatons performed on a matrx A alter the value of det(a) They can be very helpful n reducng the amount of work requred to evaluate a determnant, snce we can use elementary row operatons to put several zeros n a row or column of A and then use cofactor expanson along that row or column We llustrate wth an example Example 342 Evaluate 21 32 1 1 2 2 51 2 1 2 3 11 Soluton: Before performng a cofactor expanson, we frst use elementary row operatons to smplfy the determnant: 21 32 03 1 6 1 1 2 2 51 2 1 1 1 1 2 2 06 12 11 2 3 11 01 5 3 Accordng to P3, the determnants of the two matrces above are the same To evaluate the determnant of the matrx on the rght, we use cofactor expanson along the frst column 03 1 6 1 1 2 2 06 12 11 01 5 3 3 1 6 = ( 1) 6 12 11 1 5 3 To evaluate the determnant of the 3 3 matrx on the rght, we can use the schematc gven n Fgure 341, or, we can contnue to use elementary row operatons to ntroduce zeros nto the matrx: 3 1 6 6 12 11 1 5 3 2 = 0 16 15 0 42 29 1 5 3 = 16 15 42 29 = 166 Here, we have reduced the 3 3 determnant to a 2 2 determnant by usng cofactor expanson along the frst column of the 3 3 matrx 1 A 21 (2), A 23 (5), A 24 ( 2) 2 A 31 ( 3), A 32 ( 6) Basc Theoretcal Results The determnant s a useful theoretcal tool n lnear algebra We lst next the major results that wll be needed n the remander of the text 1 The volume of the parallelepped determned by the vectors s a = a 1 + a 2 j + a 3 k, b = b 1 + b 2 j + b 3 k, c = c 1 + c 2 j + c 3 k a 1 a 2 a 3 where A = b 1 b 2 b 3 c 1 c 2 c 3 Volume = det(a),
page 228 228 CHAPTER 3 Determnants 2 An n n matrx s nvertble f and only f det(a) 0 3 An n n lnear system Ax = b has a unque soluton f and only f det(a) 0 4 An n n homogeneous lnear system Ax = 0 has an nfnte number of solutons f and only f det(a) = 0 We see, for example, that accordng to (2), the matrces n Examples 341 and 342 are both nvertble If A s an n n matrx wth det(a) 0, then the followng two methods can be derved for obtanng the nverse of A and for fndng the unque soluton to the lnear system Ax = b, respectvely 1 Adjont Method for A 1 : If A s nvertble, then A 1 = 1 det(a) adj(a), where adj(a) denotes the transpose of the matrx obtaned by replacng each element n A by ts cofactor 2 Cramer s Rule: If det(a) 0, then the unque soluton to Ax = b s x = (x 1,x 2,,x n ), where x k = det(b k) det(a), k = 1, 2,,n, and B k denotes the matrx obtaned when the kth column vector of A s replaced by b Example 343 Use the adjont method to fnd A 1 f A = 3 4 2 7 5 8 Soluton: We have already shown n Example 341 that det(a) = 41, so that A s nvertble Replacng each element n A wth ts cofactor yelds the matrx of cofactors 22 10 13 M C = 13 9 17, 6 1 11 so that Consequently, 22 13 6 adj(a) = MC T = 10 9 1 13 17 11 A 1 = 1 det(a) adj(a) = 22 13 41 41 41 6 10 41 13 41 17 41 9 41 1 41 11 41 Example 344 Use Cramer s rule to solve the lnear system 2x 1 x 2 + x 3 = 2, 3x 1 + 4x 2 + 2x 3 = 5, 7x 1 + 5x 2 + 8x 3 = 3
page 229 34 Summary of Determnants 229 Soluton: The matrx of coeffcents s A = 3 4 2 7 5 8 We have already shown n Example 341 that det(a) = 41 Consequently, Cramer s rule can ndeed be appled In ths problem, we have det(b 1 ) = det(b 2 ) = det(b 3 ) = 5 4 2 3 5 8 221 352 738 2 1 2 3 4 5 7 5 3 It therefore follows from Cramer s rule that = 91, = 22, = 78 x 1 = det(b 1) det(a) = 91 41, x 2 = det(b 2) det(a) = 22 41, x 3 = det(b 3) det(a) = 78 41 Exercses for 34 Sklls Be able to compute the determnant of an n n matrx Know the effects that elementary row operatons and elementary column operatons have on the determnant of a matrx Be able to use the determnant to decde f a matrx s nvertble Know how the determnant s affected by matrx multplcaton and by matrx transpose Be able to compute the adjont of a matrx and use t to fnd A 1 for an nvertble matrx A Problems For Problems 1 7, evaluate the gven determnant 5 1 1 3 7 2 35 7 1 2 4 63 2 3 4 5 6 7 514 613 14 2 7 23 15 79 42 33 51 68 36 57 abc bca cab 3 5 1 2 2 1 5 2 3 2 5 7 1 1 2 1 7 1 2 3 2 2 4 6 3 1 5 4 18 9 27 54
page 230 230 CHAPTER 3 Determnants For Problems 8 12, fnd det(a) IfA s nvertble, use the adjont method to fnd A 1 [ ] 35 8 A = 27 123 9 A = 231 312 3 4 7 10 A = 2 6 1 314 1 2 5 7 11 A = 4 3 2 6 9 11 5 1 2 1 12 A = 3 1 4 5 1 1 2 1 5 9 3 2 For Problems 13 17, use Cramer s rule to determne the unque soluton to the system Ax = b for the gven matrx and vector [ ] [ ] 35 4 13 A =, b = 62 9 14 A = [ ] [ ] cos t sn t e t, b = sn t cos t 3e t 4 1 3 5 15 A = 2 1 5, b = 7 2 3 1 2 53 6 3 16 A = 24 7, b = 1 25 9 4 31 35 71 36 17 A = 22 52 63, b = 25 14 81 09 93 18 If A s an nvertble n n matrx, prove that det(a 1 ) = 1 det(a) 19 Let A and B be 3 3 matrces wth det(a) = 3 and det(b) = 4 Determne det(2a), det(a 1 ), det(a T B), det(b 5 ), det(b 1 AB) 35 Chapter Revew Ths chapter has lad out a basc ntroducton to the theory of determnants Determnants and Elementary Row Operatons For a square matrx A, one approach for computng the determnant of A, det(a), s to use elementary row operatons to reduce A to row-echelon form The effects of the varous types of elementary row operatons on det(a) are as follows: P j : permutng two rows of A alters the determnant by a factor of 1 M (k): multplyng the th row of A by k multples the determnant of the matrx by a factor of k A j (k): addng a multple of one row of A to another has no effect whatsoever on det(a) A crucal fact n ths approach s the followng: Theorem 351 If A s an n n upper (or lower) trangular matrx, ts determnant s det(a) = a 11 a 22 a nn Therefore, snce the row-echelon form of A s upper trangular, we can compute det(a) by usng Theorem 351 and by keepng track of the elementary row operatons nvolved n the row-reducton process
page 231 35 Chapter Revew 231 Cofactor Expanson Another way to compute det(a) s va the Cofactor Expanson Theorem: For n 2, the determnant of A can be computed usng ether of the followng formulas: det(a) = a 1 C 1 + a 2 C 2 + +a n C n, (351) det(a) = a 1j C 1j + a 2j C 2j + +a nj C nj, (352) where C j = ( 1) +j M j, and M j s the determnant of the matrx obtaned by deletng the th row and jth column of A The formulas (351) and (352) are referred to as cofactor expanson along the th row and cofactor expanson along the j th column, respectvely The determnants M j and C j are called the mnors and cofactors of A, respectvely Adjont Method and Cramer s Rule If A s an n n matrx wth det(a) 0, then the followng two methods can be derved for obtanng the nverse of A and for fndng the unque soluton to the lnear system Ax = b, respectvely 1 Adjont Method for A 1 : If A s nvertble, then A 1 = 1 det(a) adj(a), where adj(a) denotes the transpose of the matrx obtaned by replacng each element n A by ts cofactor 2 Cramer s Rule: If det(a) 0, then the unque soluton to Ax = b s x = (x 1,x 2,,x n ), where x k = det(b k) det(a), k = 1, 2,,n, and B k denotes the matrx obtaned when the kth column vector of A s replaced by b Addtonal Problems For Problems 1 6, evaluate the determnant of the gven matrx A by usng (a) the defnton, (b) elementary row operatons to reduce A to an upper trangular matrx, and (c) the Cofactor Expanson Theorem [ ] 7 2 1 A = 1 5 [ ] 66 2 A = 2 1 1 4 1 3 A = 02 2 22 3 2 3 5 4 A = 4 0 2 6 3 3 3 1 2 1 5 A = 0 0 1 4 0 2 1 1 0 0 0 4 0 0 0 2 6 A = 0 0 5 1 0 1 4 1 3 3 3 3
page 232 232 CHAPTER 3 Determnants For Problems 7 10, suppose that abc A = d e f, and det(a) = 4 gh Compute the determnant of each matrx below g h 7 4a 4b 4c 2d 2e 2f a 5d b 5e c 5f 8 3g 3h 3 d + 3g e + 3h f + 3 3b 3e 3h 9 c 2a f 2d 2g a d g a db ec f 10 3 2g 2h 2 d e f For Problems 11 14, suppose that A and B are 4 4nvertble matrces If det(a) = 2 and det(b) = 3, compute each determnant below 11 det(ab) 12 det(b 2 A 1 ) 13 det(((a 1 B) T )(2B 1 )) 14 det(( A) 3 (2B 2 )) 15 Let A = 16 Let [ ] 12 1, B = 21 4 Determne, f possble, 2 1 1 0 5 5 2, C = 3 1 4 4 7 2 2 6 det(a), det(b), det(c), det(c T ), det(ab), det(ba), det(b T A T ), det(bac), det(acb) A = [ ] 12, and B = 34 [ ] 54 11 Use the adjont method to fnd B 1 and then determne (A 1 B T ) 1 For Problems 17 21, use the adjont method to determne A 1 for the gven matrx A 17 A = 0 5 1 1 1 3 0 3 2 2 18 A = 0 1 1 1 1 2 3 4 1 0 0 5 0 0 0 1 19 A = 0 1 3 3 2 3 5 2 4 4 4 6 5 8 16 20 A = 4 1 8 4 4 11 266 21 A = 276 277 22 Add one row to the matrx [ ] 4 1 0 A = 5 1 4 so as to create a 3 3 matrx B wth det(b) = 10 23 True or False: Gven any real number r and any 3 3 matrx A whose entres are all nonzero, t s always possble to change at most one entry of A to get a matrx B wth det(b) = r 124 24 Let A = 316 k 32 (a) Fnd all value(s) of k for whch the matrx A fals to be nvertble (b) In terms of k, determne the volume of the parallelepped determned by the row vectors of the matrx A Is that the same as the volume of the parallelepped determned by the column vectors of the matrx A? Explan how you know ths wthout any calculaton 25 Repeat the precedng problem for the matrx k + 121 A = 0 3 k 1 1 1 26 Repeat the precedng problem for the matrx A = 2 k 3 k2 2 1 4 1 k 0
page 233 27 Let A and B be n n matrces such that AB = BA Use determnants to prove that f n s odd, then A and B cannot both be nvertble 28 A real n n matrx A s called orthogonal f AA T = A T A = I n IfA s an orthogonal matrx, prove that det(a) =±1 For Problems 29 31, use Cramer s rule to solve the gven lnear system 29 30 31 3x 1 + x 2 = 3, x 1 + 2x 2 = 1 2x 1 x 2 + x 3 = 2, 4x 1 + 5x 2 + 3x 3 = 0, 4x 1 3x 2 + 3x 3 = 2 3x 1 + x 2 + 2x 3 = 1, 2x 1 x 2 + x 3 = 1, 5x 2 + 5x 3 = 5 35 Chapter Revew 233 Project: Volume of a Tetrahedron In ths project, we use determnants and vectors to derve the formula for the volume of a tetrahedron wth vertces A = (x 1,y 1,z 1 ), B = (x 2,y 2,z 2 ), C = (x 3,y 3,z 3 ), and D = (x 4,y 4,z 4 ) Let h denote the dstance from A to the plane determned by B,C, and D From geometry, the volume of the tetrahedron s gven by Volume = 1 3h(area of trangle BCD) (353) (a) Express the area of trangle BCD n terms of a cross product of vectors (b) Use trgonometry to express h n terms of the dstance from A to B and the angle between the vector AB and the segment connectng A to the base BCD at a rght angle (c) Combnng (a) and (b) wth the volume of the tetrahedron gven above, express the volume of the tetrahedron n terms of dot products and cross products of vectors (d) Followng the proof of part 2 of Theorem 3111, express the volume of the tetrahedron n terms of a determnant wth entres n terms of the x,y, and z for 1 4 (e) Show that the expresson n part (d) s the same as Volume = 1 6 x 1 y 1 z 1 1 x 2 y 2 z 2 1 (354) x 3 y 3 z 3 1 x 4 y 4 z 4 1 (f) For each set of four ponts below, determne the volume of the tetrahedron wth those ponts as vertces by usng (353) and by usng (354) Both formulas should yeld the same answer () (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1) () ( 1, 1, 2), (0, 3, 3), (1, 1, 2), (0, 0, 1)
page 234 CHAPTER 4 Vector Spaces To crtcze mathematcs for ts abstracton s to mss the pont entrely Abstracton s what makes mathematcs work Ian Stewart The man am of ths text s to study lnear mathematcs In Chapter 2 we studed systems of lnear equatons, and the theory underlyng the soluton of a system of lnear equatons can be consdered as a specal case of a general mathematcal framework for lnear problems To llustrate ths framework, we dscuss an example Consder the homogeneous lnear system Ax = 0, where 1 1 2 A = 2 2 4 3 3 6 It s straghtforward to show that ths system has soluton set S = {(r 2s, r, s) : r, s R} Geometrcally we can nterpret each soluton as definng the coordnates of a pont n space or, equvalently, as the geometrc vector wth components v = (r 2s, r, s) Usng the standard operatons of vector addton and multplcaton of a vector by a real number, t follows that v can be wrtten n the form v = r(1, 1, 0) + s( 2, 0, 1) We see that every soluton to the gven lnear problem can be expressed as a lnear combnaton of the two basc solutons (see Fgure 401): v1 = (1, 1, 0) and v2 = ( 2, 0, 1) 234