MATH 218 Problems Solution Manual Chapter 1.8, Fall 2011

32 Chapter 1 Systems of Linear Equations and Matrices ; 100 214 -1 27. Using the LU factorization A= LU=j} 1 1 0 0 1 -1 |, we have that oid 00 38 1 1 1 2 1 9 1 0 0 1 -4 0 2 2 2 Atsutpt=|]0 1 3 -1 1 Oj;=]-1 § 3 4 1 J 0 0 4 0 -1 1 | 0 -$ 3 28. 1 0 OJ [-3 2 47\7* F-d 2 -1]7 710 0 At =(LUy b= -1 1 0 0 12 =| 0 1 -2 1 1,0 1 -1 1 0 O01 0 0 1 011 i-} =| 1 -1l -2 0 1 1 29, Suppose (eel[s s]eLi al: This gives the system of equations ad = 0,ae = 1,bd = 1,be + cf = 0. The first two equations are satisfied ‘only when a 4 0 and d = 0. But this incompatible with the third equation. 30. Since A is row equivalent to B there are elementary matrices such that B = E,...,A and since B is row equivalent to C' there are elementary. matrices such that C = Dy...D,B. Then C = Dy... DiB = Dn ....D Em... EA and hence, A is row equivalent to C. 31. If A is invertible, there are elementary matrices E1,...,#, such that I = E,-+-E,A. Similarly, there are elementary matrices D,,...,.D¢ such that J = De---D,B. Then A= By! tee Ey" De «+ DB, so A is row equivalent to B. : 32. a. Since L is invertible, the diagonal entries are all nonzero. b. The determinant of A is the product of the diagonal entries of L and U, that is det(A) = @11+++€nntii''*Unn. ¢ Since L is. lower triangular and invertible it is row equivalent to the identity matrix and can be reduced to J using only replacement operations. Exercise Set 1.8 1, We need to find positive whole numbers 21, v2, @3, and xq such that a, Al3 + v2CuO0 — agAlQ3 + eqCu is. balanced, That is, we need to solve the linear system 8a, = ug 9 1 %q = 3x3 , which has infinitely many solutions given by #1 = ge 03 = ter = 02,02 ER, BQ = 4 ; . A particular solution that balances the equation is given by v1 = 2,%2 = 9,23 = 3,04 = 9. 2. To balance the equation vIg + vgNagSg03 —> asNal + xq4Na254O¢, we solve the linear system 20, = 23 Qu, = ug + 2x4 : : : ow An , 80 that v1 = 24,@2 = 24,03 = 204,24 € R. For a particular solution that balances 2 = 4h4 ‘ 8x2 = 624 the equation, let 24 = 1, so v1 = 1, aq = 2, and ay = 2, 1.8 Applications of Systems of Linear Equations 33 3. We need to find positive whole numbers 21, %2,%3,24 and zs such that av1NaHCOs + aeCgHgO7 —> w3NagCgHsO7 + r4HeO + vgCOg. The augmented matrix for the resulting homogeneous linear system and the reduced row echelon form are 10-3 0 0 |0 10 0 0-1/0 18 -5 -2 0/0] ~/0 10 0 ~g/0 16 -6 0 -1]0 0 0 10 -t]0 37 -7 -1 ~-2]0 00 0 1,-1/0 particular solution that balances the equation is v1 = %q = vs = 3, a2 = 23 = | 4. To balance the equation w,MnS + ag AsoCr19O35 + @3H,SO4 — awaHMnO4 + #5 AsH3 a aeCr$3012 + @7H20, we solve the linear system wy = 04 @1 + = 326 2x2, = 25 10x = %6 . 35aq +403 = 404 + 1206 + v7 223 = 04 + 8x5 + 2x7 The augmented matrix for the equivalent homogeneous linear system 1000 0 0 ofo 100 -1 0 0 1021 0 0 -38 0 jo 01 0 0 0 0200 -1 0 olfo]. 001.00 0 0 100 0 0 -1 0 Jo | etucests} oo 0 1 Oo 0 0 3 4 —4 0 -12 -1]0 00001 0 002-1 -3 0 2/0 000 00 0 A particular solution that balances the equation is v7 = 327,01 = 16,44 = 138,49 = 374,24 = 16,25 = 26, and ag = 130. . 7 : 5. Let 21,%2,...,@7 be defined as in the figure. The total flows in and out of the entire network and in and out each intersection are given in the table. 300 " | Flow In . Flow Out 800+ 2 4500 700+300+500+300 a1 + 800+ a4 + 27 tq + 2X3 21 + 800 ash Yt tp +700 U3 + u4 25 xe + 300 Ly +27 me ‘ “m . 500-800 Lg + 26 700 Vite Equating the total flows in and out gives a linear system with solution x = 1000-—24—~27, 29 = 800~26, 23 = 1000 — a4 +g ~ &7, "5 = 800 + 4g — @7, with v4,26, and «7 free variables. Since the network consists of one way streets, the individual flows are nonnegative. As a sample solution let, rq = 200,2¢ = 300,27 = 100, then 7 = 700, v2 = 500, v3 = 1000, vs = 500. 34 Chapter 1 Systems of Linear Equations and Matrices 6, Let #1,%2,...,%8 be defined as in the figure. . ‘00 a ts 400+ + 4500 4300 Yu $00» > > 300 . C4 Vos Lz 400 x ug + 4 + 4200 ry + “or 500 Balancing all in and out flows generates the linear system 21 +600 + ag + 200+ 500 = 100+ 400 + xg + 500 + 300 az + 500 = Do + Bs as +300 = 100 + 400 a7 +500 = xq + 300 ae +400 = U7 + Xe ag3+200 . = 400.+ 500 2+ @4 . = v3 +800. The set of solutions is given by 21 = 500 — a7,a = 800 — 27,03 = 700,24 = 200+ 27,23 = 200,26 = 400 ++ 27 + ag, %7,2g € R. In order to-have all positive flows, for example, let x7 = 300 and xg = 200, so a1 == 200, % = 500,23 = 700,24 = 500, vy = 200, 2g = 100. 7, Equating total flows in and out gives the linear system xy +24 = 150 21 — 2g -2, =100 @2 +43 = 100 . —@3 +a4+a_ = —-50 with solution x1 = 150 ~ x4, 9 = 50 — a4 — a5, and 23 = 50+24 +45. Letting, for example, v4 = 5 = 20 gives the particular solution 7, = 130, %2 = 10,73 = 90. 8. The set, of solutions is given by v1 = 200+ %g, vq = —100+ 7g, 03 = 100+, 24 = %g, U5 = 150+ 08,26 = —150 + vg, v7 = 100 + ag, 2g € R. If wg > 150, then all the flows will remain positive. 9. If 21,2,23, and xq denote the number of grams required from each of the four food groups, then the specifications yleld the linear system 2021 + 802g + 40an3 +1004 = 250 4021 + 20a + 35a3-+20r4 = 300 50x + 40x29 + 1023+ 3024 = 400° 5a, +529+l0a3+524 | =70. The solution is v1 = 1.4, 29 = 3.2, 03 = 1.6, 04 = 6.2, 10. If x1,%2, and vg denote the number of grams required from each of the three food groups, then the specifications yield the linear system 20021 + 400a2 + 30023 = 2400 30021 + 500%2 + 400%3 = 3500 40a + 50ag + 2023 = 200 5ay.+ 38uq + 2x3 = 25 which is inconsistent, and hence it is not possible to prepare the required.diet.