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# 2nd - EXERCISE SET 3.2 26 Let X =[33}“3 Then.‘LX =I...

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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. mes-113. In = -3. Isr=|l \$33=1 1] III . 1 32.. {a} IfA3=f,then{I—A}i=I2—2A+A2=I-2A+A=I—~A;thnsI—Aisidempetent. {b} Iii-F:l,then[2A-HE2A—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. Flt-reitnrnple1 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 fen-Ii = 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 right-hand hleelr shave. 25. The loo-efﬁcient 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 E-Liimensional 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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2nd - EXERCISE SET 3.2 26 Let X =[33}“3 Then.‘LX =I...

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