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Unformatted text preview: True or False TF.1 T, by Theorem 2.4.3.
: F2 T; Let A : B in Theorem 2.4.7.
TF3 F, by Theorem 2.3.3.
TF4 T, by Theorem 2.4.8. TF5 F; Matrix AB will be 3 x 5, by Deﬁnition 2.3.11). TF6 F; Note that T 3 : (I) . A linear transformation transforms [7 into 5. 115 Chapter 2 Ch 2.TF.7 T, by Theorem 2.2.4. Ch 2.TF‘.8 T, by Theorem 2.4.6. Ch 2.TF.9 T; The matrix is [_i “i Ch 2.TF.10 F; The columns of a rotation matrix are unit vectors; see Theorem 2.2.3. Ch 2.TF.11 F; Note that det(A) : (I: i 2)2 + 9 is always positive, so that A is invertible for all values of k. Ch 2.TF .12 T; Note that the columns are unit vectors, since (—0.6)2 + (:l:0.8)2 = 1. The matrix has the form
presented in Theorem 2.2.3. Ch 2.TF.13 F; Consider A : I2 (or any other invertible 2 x 2 matrix). —1 vi
7 i 1 2 1 1 5 6 . ‘ . ‘ 7‘ .
Ch 2.TF.14 T, Note that A — [3 4] [1 1] [7 8] is the unique solution. Ch 2.TF.15 F, by Theorem 2.4.9. Note that the determinant is 0. Ch 2.TF.16 T, by Theorem 2.4.3. l—‘MEH Ch 2.TF.17 T; The shear matrix A : J works. Di—l 4y 04:1:
C112.TF.18 T, Slinphiy to see that I [y] m [_12$] —— [_12 0] Ch 2.135919 T; The equation det(A) : k2 — 6k + 10 2 (} has no real solution. Ch 2.TF.20 T; The matrix fails to be invertible for k : 5 and k 2 it, since the determinant detA : kg — 4k « 5 Z
(k — 5)(k + 1) is 0 for these values of 1:. Ch 2.TF.21 T; The product is det(A)Ig. ab
0c 0 0 Ch 2.TF.22 T; Writing an upper triangular matrix A : [ O 0 J and solving the equation A2 : [ ] we ﬁnd 0 b thatA: [O 0 J , where b is any nonzero constant. 071 Ch 2.TF.23 T; Note that the matrix [1 0 represents a rotation through 71' / 2. Thus it : 4 (or any multiple
of 4) works. 1 1 Ch 2.TF.24 F; If a matrix A is invertible, then so is A“. But [1 1 fails to be invertible. 116 True or False Ch 2.TF.25 F; If matrix A has two identical rows, then so does AB, for any matrix B. Thus AB cannot be In, so
that A fails to be invertible. Ch 2.TF.26 T, by Theorem 2.4.8. Note that A’1 : A in this case. 1 1 Ch 2.TF.27 F; For any 2 X 2 matrix A, the two columns of A [I 1 will be identical. Ch 2.TF.28 T; One solution is A : a b {Ch 2.TF.29 F; A reﬂection matrix is of the form [b in
, Where (t2 + b2 x 1. Here, e2 + b2 2 1 +1: 2. i‘iCh 2.TF.30 T; Just multiply it out. 0 0 1
'Ch 2.TF.31 F; Consider matrix 0 1 {1 ; for example.
1 0 0 2Ch 2.TF.32 T; Apply Theorem 2.4.8 to the equation (A2)’1AA 2 In, with B : (A2)“1A. h 2.TF.33 F; Consider the matrix A that represents a rotation through the angle “Zn/17. h 2.TF.34 F; Consider the reﬂection matrix A : l12.TF.35 T; We have (5A)’1 2 A71. 1
5 h 2.TF.36 T; The equation Ac}; 2 BE; means that the ith columns of A and B are identical. This observation
applies to all the columns. i1 2.TF.37 T; Note that A2B : AAB : ABA : BAA : BAZ.
i1 2.TF.38 T; Multiply both sides of the equation A2 2 A with A”.
h2.TF.39 F; Consider A 2 I2 and B : #12. h 2.TF.40 T; Since A3? is on the line onto which we project, the vector Af remains unchanged when we project
again: A(Ai") 2 AE; or Agf : ALE, for all :5. Thus A2 : A. l1 2.TF.41 T; If you reﬂect twice in a row (about the same line), you will get the original vector hack: A(A:f:') : f
01"} A23? : f 2 Igf. Thus A2 : I2 and A’1 2 A. i I 1 1 U .
2‘II1 .  ‘ : H : a :
_. F42 F, Let A [O 1] ,1; [0] ,w [1]; ior example. 1 17 Chapter ‘2 , for example. 1 0 0
Cl12.TF.43 T,LetA—[U 1 0],BM OCH
OI—IO Ch 2.TF.44 F; By Theorem 1.3.3, there is a nonzero vector such that Bf : 6, so that ABf : (1 as well. But
I35 :1? 75 0, so that AB % I3. Ch 2.TF.45 T; We can rewrite the given eqnation as A2 + 3A : i413 and wﬂA + 31'3)A : 13. By Theorem 2.4.8,
the matrix A is invertible, with A’1 — wﬂA + 313). Ch 2.TF.46 T; Note that (In + AMI” — A) : If; e A2 : In, so that (In + A)“1 : I.” — A. Ch 2.TF.47 F; A and C can be two matrices which fail to commute, and B could be In, which commutes with
anything. Ch 2.TF.48 F; Consider Th?) 2 Li’, 7? : 51, and n7 73. Ch 2_TF,49 F; Since there are only eight entries that are not 1, there will be at least two rows that contain only
ones. Having two identical rows, the matrix fails to he invertible. Ch 2.TF.50 F; Let A : B : , for example.
. ,_l U 1 , . . , . . . . ,_ . , a. 11
Ch 2.TF.51 F; We Will show that b 0 0 5 falls to be diagonal, for an arbitrary invertible matrix .5 : (i d .
 0 1 (1 Ab c d ed 612 . _
T'l ' i l — l ‘  s i . 3 r 1
New, .5 [0 0] 5 w adwa [_c a ] [0 O] — admbc [W62 Va Cd]. brnee c and 01 cannot both be aero {as 8 must be invertible), at least one of the off—diagonal entries (:3 and at?) is nonzero, proving the claim. Ch 2.TF.52 T; Consider an if such that 142:3 : 5, and lot :30 2 AE. Then Afo = A(A.’£") : A23? : , as required. Ch 2.TF.53 T; Let A : a b . Now we want A'1 : 7A, 01' i , d Mb : _a _b . This holds if
C d ed br. _C (1 HC id ad i be : 1 and d z 7e. These equations have many solutions: for example, a = d = 0, b = 1, c : ~—1. More
. 2
generally, we can choose an arbitrary e and an arbitrary nonzero 3). Then, (I. : —e and C : iHT”. e2+be cabH)?!
ac+cd obi—€12 Ch 2.TF.54 F; Consider a 2X2 matrix A x 3
o2 ml— be b(e + on] do + d) d2 + be . We make an attempt to solve the equation A2 : [ x P] J . Now the equation Ma. + d) z 0 implies that b : 0 or d 2 7a. If h : 0, then the equation al2 + be : *1 cannot be solved. If d = 70., then the tw0 diagonal entries of A2, e2 + be and d2 +be, will be equal, so that the equations (L2 + be 2 1
and (12 + be I —1 cannot be solved simultaneously. 1 U 0 _1] cannot be solved. In summary, the equation A2 2 [ 118 True or False ICh 2.TF.55 T; Recall from Deﬁnition 2.2.1 that a projection matrix has the form “:12 “1:32 , Where is
2 ,
g 2
unit vector. Thus, a2 + b2 + CZ + d2 : u‘f+(u1ug)2 + (U1U2)2 + : u? + 2(u1u2)2 + : (u? + “11%)2 z 1 :1, :Ch 2.TF.56 T; We Observe that the systems ABE : 0 and BE : U have the same solutions (multiply with A" and A, respectively, to obtain one system from the other). Then, by True or False Exercise 45 in Chapter 1,
rreﬁAB) :l‘ref(B). 119 ...
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 Spring '08
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 Linear Algebra, Algebra

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