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Unformatted text preview: EXERCISE SET 3.2
26. Let X = [33}“3. Then .‘LX =I ifsnd only if 1 U U 1111 I] 3 171;]
I] l I] :21 1‘22 1'23 =
2 [I 1 Gal— l] [l
1 El
1'31 $33 $33 {1' 1 in. Dill}? H.311 = 1.1?” = U, 1213 = 0,121= [3.333 = 1. 1‘23 = ['1 11!” +131 = 0.. 21:12 + $32 =
ﬂ, and 21:13 + :33 = 1. This is e. very simple system of equations which has the selutien $11=11312=EL $13 =lJ.1'21 =0. 322:1. mes113. In = 3. Isr=l $33=1 1]
III .
1 32.. {a} IfA3=f,then{I—A}i=I2—2A+A2=I2A+A=I—~A;thnsI—Aisidempetent.
{b} IiiF:l,then[2AHE2A—Ii=4ﬂﬂ—4A+F=4A—4A+I=I;thus2A—Iisinvert
ihle and {2A — I34 = 2.4 — r. 36. From parts Ed} and {e} sf Theerem 3.2.12, it inﬂows that Thus .4 is invertible and A“ = [ I
U
2 GHQ “(he — 3.4} = ems} — men] = e ferenytwenx n matrices Anne] B. On theother hand, trUn] =l+l++ 1 = n. Thus it is
net pussible to have AB — 3.4 = I... vs. {a} Fees. {ns}2 = {sense} = menus. It A and e commute then [.4312 = .4232. but if
EA 34 AB then this will not in general be true. a 2
(b) True. Expanding both expressions, we. have {A  3;? = A — AB  BA + B and
[B—AJ==32—se—es+e=;thus[.e—s}3=(B—n}. T I I
[1:] True. The basic fest {from Theorem 3.2.11} is that MT)“ = {1:44} , and from this it
fellows that {J's—“Jr = HAW—111": Elfin—1 = [BTW—1 =ir1 3'" 11 to _ 2 1 _
(s1 False. Fltreitnrnple1 ifA: [a 1] 5.11113: [1 1 then triﬂB}—tr[[l 1D _3, whereas triAJtrtﬁ} = [21(2) = d. 1 _ L
{e} False. For example, if B = —A then A + H = E? is mt invertible {whether A 1:: Invertlhle or not}.
EXERCISE SET 3.3
i i‘? e I] l} 1—1 h
13 {all A‘1= % % % it} A“: e 1 —s [e] .14: e t —;
e e e 0 1 4 u o i 19. If any one of the his is D, then the matrix A has a new raw and thus is neat invertible. if the iti's are all nonzero, then multiplying the 1th row nf the matrix [A I F] by lflti fenIi = 1. 2. 3.4 and then
renewing the order of the mere yields lﬂﬂﬂlﬂﬂﬂﬁ

1
elsewege
J
eeru:egue
ﬁﬂﬂHEJ[lﬂl] thus A is invertible and .44 is the matrix in the righthand hleelr shave. 25. The looefﬁcient matrix, augmented with the two columns of constants. is 1 s 5:412 I
I) s 1: tit—1 1 I] [I .' —32: 11
e 1 a E s f —s
it I] l l 3 l "1
Thus the ﬁrst system has the solution I. = —32, 39 = 8, 1:3 2 3; and the second system has the solution :1=11,33 = —2, :3 = _1‘
D6. All of these statements are true. P4, From Theorem 3.31 the system A}; = ID has only the trivial solution if and only if A is invertible.
But, since 3 is invertible. A is invertible iﬂ' BA. is invertible. Thus Ax = I] has only the trivial solution 1E [3.41.le = D has only the trivial solution. EXERCISE SET 3.4 16. The Iedueed row echequ form of the augmented matrix is l I] —4 ﬂ 1. [l
G 1 l i} E U
{l I] [I l —3 i]
E} D i} I] I] I] thus a general solution is given by {xi.mg.ea,zi.z5] = 3H, —1,1.ﬂ,ﬂ] + l{—1,21i},3, 1}. 2:2. In parts {a} and {h} the vectors are linearly independent: they do not lie in a plane. In part [c]:
the vectors are linearly dependent: they lie in a. plane hut not on a line. 26. if = span{v].v:} where 1'1 = (1,0,2. E}, —]l and v9 = (0. LI], 3,0]. This eorresponds to a plane
[Le a ELiimensional subspace] through the origin R5. ﬁll. The arguments used in Exereise 29 on easily be adapted to this more general situation. P3. First we shoe.r that W] + W2 is elosed under scalar multiplication: Suppose a = a: + 3; Where I is in
W1anel y isin W2. Then. for anysealar it. wehavekn=MX+Il =1¢x+k1nwhere kxisin W1
and Fry is in “’2 {Sines W1 and W; are subspaces}; thus it: is in H”; + W3. Finally we show that
W1+ W; is closed under addition: Suppose 21 = x1+ in and 23 = I; + yrs, where x1 and x2 are
in W1 and 3'1 311d 3': are in We. TllEIl 11 +32 = {31+ hl + (K: +f1l=lxi + x2}+[1"1+ 3'2}.
where 111 + x3 is in WI and 1n + y: is in We [since W1 and W; are ﬂubﬁpaﬂeﬂ}; thus sl + :3 is in
W1 + We. ...
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 Spring '08
 anony
 Linear Independence, Vector Space, Exercise Set

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