{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

key-test3-fall09

# key-test3-fall09 - MATH 2090 NAME KEY TEST 2 SHOW ALL WORK...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 2090 NAME: KEY_ TEST 2 SHOW ALL WORK 1. Let T : V —> V be the linear transformation satisfying T(v1)= 3121—122 F“ d T av +bv T022): 121+2v2 m ( l 2) T(av1 + bvz) = aT(v1) + .bThwg) = o(3v1 -— V2) + b(v1 + 21:2) 33v: — mm + bv1 + 25v; 2 (3:: + 5)»: + {23? — (1)1’2. 7—0“; +£VA>SGQ+B> V, 7:24:58 2- Let T : R3 —> R2 be the linear transformation deﬁned? T(x],x2,x3)=(x1—x2 + 2.7c3,,—33c1 + 31c2 — 6x2) Find a basis for the Ker(T) and its dimension; 5. Ker(T} = {X E R3 : T(x) = O} = {x E R3 : A}: 2 0}. The augmented matrix of the system Ax 2 0 is: [ _; w; _§ | 3 J, With reduced row-echelon form of [ {1) —{1] g 3 1.311115 Ker(T) = {x e R3 : x : r(1.,1,0)+s(—2,o,1)_. as e R}. Geometrically, this dmcribes the plane through the origin in R3, which is spanned by the linearly indepen— dent set {(1, 1,0). (—2.0, 1)}. d5miKer(T)] = 2. —4 3 0 3.LetA=—6 5 0 3 —3 —1 (a) Find the eigenvalues of A and for each eigenvalue indicate its algebraic multiplicity. (b) For each eigenvalue of a ﬁnd a basis of the caneSponding eigenSpace. (c) Is the matrix A defective? Explain your answer. 20. To compute the eigemvalues, we ﬁnd the characteristic equation —-1- — A 3 0 det(A~AI}=det —6 5—/\ 0 ={—1—A){(—4—A}{5—)i)+18} 3 —3 -1 — A = (—1 — no” — A — 2.} = —[A +1}'°'{A — 2). so the eigenvalues are A: = —1 and A2 = Eigenvalue A1 = —1: To get eigeaimtors, we consider --3 3 0 1 —1 nullspaee(A + I) = nullspnee —6 6 0 ~ 0 0 3 —3 D 0 0 ODD I—I Them are two free variables : = it Md: 3; ._ s From the ﬁrst equation 32— — a This, _£w0 linearly independent eigenwactors am he obtained canes-pending to )4: {i} and /El [\ﬂjﬂinsﬁa{(ll 0)/(/0/[ 0 ]( “LIME“:aéa —6 3 U 1 —1 —1 nullspaoe(A—2I}=null.space —6 3 0 .. O 1 9 . Eigemlue {\2 = 2: To get an eigenvector. we consider 3w3 -3 0 0 0 we let 2 z t. Then y = —2t from the middle line, and u: = —t. from the top line. Thus, an eigenvector uni-responding to A2 2 2 may be. chosen as 4. Determine two LI solutions to the DE 2x2y”+ 51y'+ y = 0, x > 0 of the form y(x) = x’ , and thereby determine the general solution. 32. Given 22:23:" + Ery' + y z 0, the trial solution gives 21211:" — Dam—2 + ﬁzrmr‘l + :r' = 0, or z’[2r(r — .1) + 5r +1] 2 0. Therefore. 2'!“2 + 31" + I = O. which factors as (21' + 1)(r +1) 2 0. Therefore, Hence, we obtain the solutions 191(1) : 2:712 and ygtt) = I“. weir-1] = (ﬁx—g — (grew—1) = ﬁx“? aé 0, so that, {\$_%,z_l} is a linearly independent set of solutions to the given differential equation on (O, 00}. Consequently, from Theorem 6.1.3, the general solution is given by y(3) = c1:r_% + 621—1- 5. Find the general solution to the homogeneous linear differential equation, on w we : a PM) :Cra +r~ 42..)Cr+-1)3:: :3 6/ . z/ y"—y'—~2y=1582", 6. Solve the following initial-value problem- . y(0) = O, y (0) = 8. 46. In operator form the given differential equation is (02 — D n 2);; = 15.223. that is (D — 2}(D + My = 15.223. Therefore the complementenr function is 2‘ ‘22:E + 035‘ . 31-24:) = C} The annihilator of F(af) = 1522" is AU?) = D — 2. Operating on (00.44} with D — 2 yields (D w 2WD +1)y = 0 which has general solution le) = 9dr») + Aux-92‘. Consequently, an appropriate trial solution for (0.0.44) is yp(:z:) = Anrez". Diﬂ'm‘entiating this trial solution with respect to 2: yields yum} = Aocgtﬂa‘ + 1), ygm : A0621(‘1-:r + 4). Inserting these expressions into the given diﬂ'erential equation yields Agegﬂn: + 4 — (21 +1)— 2.1!] = 153”, that is, 3.40 = 15. Therefore, .49 = 5, so that 319(3) 2 533323? and the general solution to the given differential equation is y[:::} = cleh + (326—: + 53:22:. The initial conditions y(0) = D, y’(0) = 8 require, respectively, c1+02=0._ 281—62-1-528 so that c1 = Leg 2 —1. Therefore the solution to the given initial-value problem is y(3) = 821 — e“: + 5382:; ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern