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2 Math 205A Final Exam page 1 April 10, 2008 NAME uy Solas -1 4 -6 ‘3 1A. Let A=] 0 3 0 | and verify that |3] is an eigenvector of A. 232 0 0 What is the corresponding eigenvalue? ving Se ft 4 3 03 © }] = [i]=3]}] > es) 0 fa) oO “i. 2, © 1B. Fact: Another eigenvalue of A is 4 = —4. Find a basis for the eigenspace of A = —4. (a--4t)= > ss ~ : . es the. sdbtiins 4 (9--11)x=8 a x, 2 2 ea 4 oO Oe [*]- [+ % | Le x; I He eae ‘ : 7 } 1C. Fact: There are only two distinct eigenvalues for this matrix (so one of them has multiplicity 2). Show that A zs diagonalizable: exhibit appropriate P and D that demonstrate this. Since te ejerpued N21 6 Dy mat be Ma , te dijeratce f Ned'e 9p “cis ee et ae dieganalicalle, ° I~ lo a o | hx, whe 2 21 73 o 9 oO x, a fee Gnk alee fr the digenepace be Nae ELE) PF] [we 2 2 9 alr oe be hie P= : a O-|73 ‘| “aranemont" l go °6 On; esol =Y ar That is a : a 5 ot ake P 1D. What is the characteristic polynomial of A, in factored form? 6 f Can cil, ff (-3) (+4) pstop) MH) en Hr, Math 205A Final Exam page 2 April 10, 2008 NAME. Seayested solde —22 25 36 2A. Fact: The matrix B= | —1 4 2 | has exactly the same characteristic polynomial as matrix —12 12 20 Ain problem 1, yet B is not diagonalizable. What must “go wrong”? at least one eiyenepace mmx rave olinension strchy vhanlln. Poon hy. mobipliiby | covcapording eigenvaloc. Sree, the milfplits of A*3 aut \=-Y ane 2. Aral L , rapectively, Mon imple A. epengiice thr A3 dat be (only Zt 2B. Support your conclusion: your answer will be in terms of one of the eigenspaces of BR. loth ad ok: about A=2: (B-3I) = me % | “tz iP Shee. ae olay one ‘roe tiaialele ty Ha Miley the elneainf He eigngpter ti tela nly 2. , bot cya te mitplil, fA-3o 2. 4 Ped wud, Har A=PDP" (hee Di digonee & te. ide F fn LT egimvedys. corypadly & A Ns om he grok f O ) cyt 66 Ne ( son.’ 3 Ps Ae prlem) Math 2054 Final Exam page 4 April 10, 2008 NAME. 44. In terms of x (the amount produced), C (the consumption matrix) and d (the final demand), what is the equation used to model the Leontief Input-Output Model for an economy? ya Chad 4B. Suppose a certain economy has three producing sectors, Math, Music and Food, and an open sector of People who just consume Math, Music and Food while producing nothing. As it turns out, the silly People demand no Math, but lots of Food, 708 units worth, and an amount of Music, $, which we will determine by the end of the problem. Making one unit of Math consumes 4/10 units of Math, 0.15 of one unit of Music, and an as yet unknown amount f of a unit of Food, whereas Music requires 1/10 of a unit each of Music and Food, but an unknown amount m. of a unit of Math, and Food requires 0.05, 0.25 and 0.2 units of Math, Music and Food, respectively. The production levels of Math, Music and Food are 360, 800, and 1120 units. Explicitly write out the Leontief Input-Output Model for this system in matrix notation. There will be three unkowns: f, m and S and lots of numbers in the matrix equation you write. 5@0 | m 0,057 )| $60 oO §00 }=|015 Ol 0.25 |[ 800 }4 1 S {120 § fy pete 708 40. Find f,m and S. the above matrix equation tls Kets thear gestions? 360-M1-Sb 60 360 = OF K 360 + m* 800 t 0.08 #I!Z09M=—— = BOD GOO = GIS xo + 0,1 # BQO + OLS KNIOFS2S = 80D-S1~GO -2 = 2 = 20 - & -224-F0F 20 = £ 8M%O + Ole far 402 H+ OR f= Mo~@aw1-Re = 108-03 360 4D. Explicitly find the intermediate demands for Math, Music, and Food. Th 1 gliten by CR alkenebieh , CX X~al on [rd .2 05 ][300) | 200 & th pia #2 |e Plas af 2S |] oo] =)44 | Me u20} [og al ae dee. 1! Revol ; 2 an Ie ~/ one as thi) Math 205A Final Exam page 5 April 10, 2008 NAME 25 58 11 36 5. Suppose T is a linear transformation with standard matrixC =| 1 4 1 2 | (the same C as 1220 8 AT in problem 3). 5A. Find any conditions on b = (6,, 6, b3) (written sideways to save space) such that b is in the image of T. Explain your answer. we Kaew be See fT info cheb om thi Oram Bf os e three some Re cama fT fost a Th) elt] [eo C= 8 he Fhe, uid. Sys O= b-b, ~2b, , Het &, X oni > b,~b, 2, 6 O, chermae 5B. Give an explicit example of a vector not in the image of T. he ae iy GH sibe i Mconsd tate, fn ig ple bs |i) 5C. Find T(w) where (written sideways) w = (2,0,0, 1). ES enreaa In 2 ae Je} | 50) (% T(@) = Cw = ete, 2 =2 i gobkp2 pS zie ak 4 ip i !z /? 24 oO 4 5D. Recall the kernel of a linear transformation T is the set of all vectors in the domain which T maps to the zero-vector. Find a basis for this subspace of T. We Know (8) =o = CR =O; ths ee Ker(1)=Mil(c), Zn G8) f he Exam, Al \z We. ford a Lacie 4 Mill(c) ble =) -Y, Math 205A Final Exam page 7 April 10, 2008 NAME 7. Let S be a set of vectors in RB”. : : het» well: these Gre. ROME or 7A. Define what it means for $ to be linearly independent. at. best ryhe: let f= J Mo ~-9% J, ay. *vtee bac en te tial os a i gs Me OMLY whole ble £8 @ SEF not « Inadtix oe es OARS has onl He tial cl * ts the Hritieh Solution y X,a --- =%e=O aa whol Unless gee ot He columns f A are Gyn % 7B. Define what it means for S to be orthegonal. fo) te no fea wealies fig di: “ Dike pis d mee Sa A y 07, beter a reult bt at aif hes, Yikey Ye yi Lee O nme f Hates 6a Lp hehen 7c. wena btu) set of nonzero vectors. Prave S is linea Sintec! “i ie Speicher, Suppote. = [Ms ae Yp} blatant = A's e 5, Syppise Hat XU, 4. X= 0. Flan ( ‘dt' each oil, vith U i v get...) xi-Wr-- i ca oe as All dot preclich, VV, 9 Wee¥ 9° a) are © becuse YAY y yor : So the LHS os simply 7 ¥), The RHE a fae OO" B abt aayKhiy 6 O sO x, (G- Vj=o Ab, %A4e ott FO. x,=O Sirilady Ki2Oyy Heo, Hence S 6 LE sees € RewsitEeD —— afer obit iy ye 2944 -1é9y +O.19x* [ES =x lated & A ieddh necklice.. ¢. reel slota hoe hy ole yoo Sap he the me : pa ink I ae - =a Behe! vie Peas HS as, yee necklace ce 20 40 50 60 70 80 90 | Fig. 2 Necklace Calan and Best-Fit Parabola Math 205A Final Exam page 8 April 10, 2008 NAME. 8. What is the dimension of each of the following vector spaces? BA. P*® 8B. Col(Js) (4) © 8C. F, our space of continuous functions. 8D. {0} 8E. The eigenspace of A = —4 in problem 1. Math 2054. Final Exam page INFO April 10, 2008 is row equivalent to 10 -1/8 2/30 +2 £ 01 1/3 1/3/0 4/7 —i/a 020° 0 Oe 8 [26.4 0 58 4 1 Heats: 0 | 36 2 0 is row equivalent to 0 0 0 1 [10 1/2 |o 0 -1/7 i/f14 0 1 -1/2]0 Bu 00 0 Al 1 —l [oo o {0 —2 =1 NAME 
