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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 rowechelon 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 canespending 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
uniresponding 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“. weir1] = (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 initialvalue 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‘ . 3124:) = 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») + Aux92‘. 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—621528 so that c1 = Leg 2 —1. Therefore the solution to the given initialvalue problem is y(3) = 821 — e“: + 5382:; ...
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