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Unformatted text preview: ill/inter 2010 MTH 141 Test 3 Student Number: Last Name (Print): Ryerson University Department of Mathematics Test 3
MTH 141 . Last Name: Signature: Date: April 16, 2010, ll: 30 am Duration: 1hr. 30 min. Instructions: . Have your student card available on your desk. . This is a closed—book test. Notes, calculators and other aids are not permitted. Verify that your test has pages 18. Do not separate the pages of this test booklet. . The point value of each question is indicated by the question number. . Include all signiﬁcant steps in your solutions to the questions. presented in the correct order. Unjustiﬁed
answers will be given little or no credit. Cross out or erase all rough work not relevant to your solution. . Present your solutions neatly and legibly in the space provided. Messy or illegible solutions will receive no credit. If you need more space, use the back of the previous page. Indicate this fact on the original page. Linear Algebra . First Name (Print): Professor (circle one) C. Grandisou B. Task? Section : For Instructor’s use only. Question Winter 2010 MTH 141 Test 3 Last Name (Print): ‘2 1. [(5+3+2) marks] Let T : R2 ——> R3 be a. linear transformation such that
"HI, 1) = (1.0, ——1) and T(1,~1) = (3,—2, —5). (a) Determine T(1:,y).
TCI,:)+T(I,—l) =2 (x,o,)+—L3,—2)5) « (4,—2,—6) vb
T(<:,a>+u,—u)= (4,4,4) =§ T(2,o)=<4,—2,6) g. QTCI,0)=(4,2,c;) =>
T010) = (2,—I,—a) 0.}
T Ch!) —T(I,: ) a C,o,~J)—(3,—2,—5)=C2 ,2,4) =9
“C ’2') ‘C'NU  02,2,4) t> T(0,1)=(—2,2,4) =3 2T(o,) =C2;2;4) =>
Tco,l) a(—I,t.2) . (u)
Hen” fem Standard matrix of T L‘J [Tl =£T(€,)TC€3U=[_ZI 7'] .. ’5 2
and T (12,?) =[3‘ “][;] .__‘ (lac—3, *lffj/ ?>:z_+?j)
‘5 .2 (b) Find the kernel ker(T). yer(T){C1.%)6Rz (T095) = (0,0,o)}. gupposc Téx«y)=CU/0;O) We” (1X‘9lXi'y ) =C0,0JO).
61mg 9334‘9011 2.13 =0 ) “1‘15 30 / ﬂax+2j=o Has may
ﬂu; hivial, Wham x =5 =1 0 1+7 JCQHO'HJS M44 (,T') ={(0,o)_}. (c) 13 T one—t0 one? 5mm; KMLT)={(0,O)II 5:, 71%... 6.3.11 H: RiloquaL T ‘3
W~+U—me. Winter 2010 MTH 141 Test 3 Last Name (Print): 3 2. ((64%) marks} Let T : R3 ——> R3 be a linear operator that ﬁrst rotates a. vector about the xaxis through an angle 9, then projects the resulting vector orthogonale onto the arty—plane. (3.) Find the standard matrix for the linear operator T. T can Be. wmsScA as 4AA: cmpoﬁttc‘n
T = TB oTA 6A where A ,B are. the, standma matrices *0?“ rotaizim W‘LA/
projectimn respectively. These, MatrLCc/s one: 1 O o i ‘9’ a
A : O we ‘5‘” ’ B u D l O
o sme cose ° 0 °
anal. karma the shon—LA maJS'WX 41m" T 1':
l O 0
BA ‘: 0 00195!
0 o (D (b) Is the linear opeartor T onto ? Justify your answer. I O O 7C
T (323.2) = 0 Lose ‘S’ y = (3C } j 0, o o 2 If aaéo is an arbitrme rad, number) Jthen (0,0,a)¢rc1n(T)
becausgkhere, is 410 (x,3,z)éR3 sudﬂ that T0953): (0,0,CL).
56/ T is «not mtd’. Winter 2010 MTH 141 Test 3 Last Name (Print): 3. [(4+6) marks} Let t be a real parameter and T: : R3 —) R3 be a linear upcrator defined by Tt(r,y‘z)=(;vay+23, r+2y+tz, ~y+z) (3.) Determine all values of t for which T is invertible Tt is: mvevlnllole iﬁiF its sfcmclcucl mairéx [7;] "LTCCJTLEQTCGQJ
it anve/btlloLe 3H clef. (ETJ) =75 0. I at 9.
detLCTJ): l 1 t =t+1¢o sff £¢—l O «l l (b) Find Tt"1(;r,y,z) fort: 1. 1 I ’5 ‘l “'5 On: finals =5 l I I
«I l ’5 ol 1) “l ‘5 ’XI
T1 (1"1319‘3)‘ ‘Ii —1 I r 1:. C51 Winter 2010 MTH 111 Test 3 Last Name (Print): 4. [(4+2+6) marks] Let u1 = (1,1,1), u2 = (~1,l,0) and us 2 (1,2,1) be vectors in RJ and let 8 = {111, 02.03} (a) Show that B = {111, u2,u3} is a basis for R3. (Hint: Use determinants) Solution: Since we have 3 vectors in R“ in order to show that B is a basis it sufﬁces to show that B is
linearly independent set. The set 8 is going to be linearly independent if and only if the only solution of the system: ru1+yu2 + zu3 =0 is trivial, i.e., .L‘ = y = z = 0. The above system is homogeneous and will have only the trivial solution if the determinant of the system is not zero. Computing the determinant of the above system one gets: 1 —l l
1 1 2 =—l#0.So.BisabasisforlR3.
l 0 l (b) Express the vector w = (1, 2, 3) as a linear combination of the vectors in the basis B = (m, L12, 113}. We are looKl’nn‘j M scalars 36,3,2 Sit Th‘s f5 lagHa: s‘jsrl‘zrﬂ Minoye wgmanlecl Matrix {S 1—; (:1 l0 1(254 _.)~..—>°'°§z
l O §2 0013—} M Solw'l'l'm is x=J=Z=l 0410'. Linnet. w =(9U , +2vz~ 31.23 , Winter 2010 MTH 141 Test 3 Last Name (Print): 6 (c) Let m = (1,1,1), 11; = (—1_.l.0) and 113 = (1,2,1) be vectors in R3 and let B = {ul,u2,u3}. Use
the GramSchmidt orthogonalization process to transform this basis into an orthogonal basis. and then normalize the orthogonal basis vectors to obtain an orthonormal basis for R3. Solution: Let {V1, V2. V3} denote the orthogonal basis produced by GramSchmidt orthogonalization process. and let {wh W2. W3} denote the orthonormal basis that results from normalizing V1, V2, V3.
Step 1: Let V1: u1 = (1,1,1). Step 2: Let V2 = “2 — projvluz = u;  ﬁn = (—1,1,0)~ 30.1.1) =(—1.1.0) Step 3: Let V3 = u;  projvhnug = 113 — V1 — sz = (1,2,1)  $0, 1.1) — %(——l.l,0) =
(in tipll Thus the vectors: v1 = (1.1.1). V2 = (—1.1,0) and V3 = (é. é,3‘;) form an orthogonal basis for R3.
The norms of these vectors are: “V1” = J5, “V2” = ﬂ, and “V3” = @. So, an orthonormal basis for R3 is given by: W1: 11%?” = %9 w? = 11%?" =(_7159 715,0)y W3 = “’2'” = Last Name (Print): 7
5 . 1; (5+3) marKS] 3 —l l
LetA: 7 45 1
6 —6 2 (8.) Given that pA(/\) = (A — 2)2(A + 4) is the characteristic polynomial of the matrix A ﬁnd the algebraic
and geometric multiplicities of the eigenvalues of A. Solution: The algebraic multiplicity of eigenvalue 2 is 2 and the algebraic multiplicity of eigenvalue 4 is 1. To determine geometric multiplicities of eigenvalues we need to ﬁnd eigenspaces. Eigenspace corresponding to the eigenvalue A = 2 is the solution set of the system of uations Ax = 2x
1 —— 1 1 0
or equivalently (A— 213)): = O. The augmented matrix of the system (A—213)X = 0 is 7 _7 1 0 6 —6 0 O
l ——l I 0
Reducing it to the RREF‘we get 0 0 1 0V
0 0 0 0
The corresponding equivalent system is: 1'1 — 1:2 +13 = 0 and 3:3 = 0. Hence we get that the eigenspace
s 1
is the set of all vectors x = 3 = s 1 , where s 6 R. This implies that the geometric multiplicity
0 0
of eigenvalue 2 is 1.
Eigenspace corresponding to the eigenvalue A = —4 is the solution set of the system of equations Ax = —4x or equivalently (A + 41:3)x == 0. The augmented matrix of the system (A + 413)x = 0 is
7 —l 1 0 l —l 1 0 7 ‘1 1 0 .Reducingit totheRREFweget 0 1 1 0 6 —6 6 0 0 0 0 0
The corresponding equivalent system is: 1:1 — 1:2 + .173 = O and 12 — $3 = O . Hence we get that the
0 0
eigenspace is the set of all vectors x = 3 = s 1 , where s 6 ER. This implies that the geometric
.9 l
l 0
multiplicity of eigenvalue 4 is 1. Let us denote by: p1 = 1 .pg 2 1 O 1 8 Last Name (Print): (b) Determine whether A is diagunniizable. If 50. find a matrix P that diagonalizes the matrix A. and determine P“ 1 AP. Solution: Since geometric multiplicity of eigenvalue 2 (which is 1) is not the same as algebraic multi plicity of eigenvalue 2 (which is 2) the matrix A is not diagonalizable by Thm 8.2.10. Alternatively one can use: Thm 8.2.9 and explain that A is not diagonalizable since the sum of the geometric multiplicities of its eigenvalues is 2. or Thin 8.2.6 and explain that A is not diagonalizable since A has only 2 linearly independent eigenvectors p1 and p2. ...
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 Fall '10
 poliakov
 Linear Algebra, Algebra

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