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# 5th - EXEHGISE SETIIJ ﬁt{a 6[b I{c E{d —12 15 MIA =-2...

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Unformatted text preview: EXEHGISE SETIIJ ﬁt {a} 6 [b] I] {c} E {d} —12 15. MIA] = -2 22. det{A} = {e — Hat-:1. + 35) as. {a} Wehnwdetﬂ :]=5—-4=1anddet[: ﬂ=3—ﬁ—-3;thusdat{M}=[l][—3}=-3. 1 2 ﬂ 1 2 3- U I] {It} Wehm 323:9”: 5|n=2{ﬁ-4]=2m11 2 l ={a}(1}[—4)=—12;tmtedet{m= _ —3 ﬂ ._ {21912} - —24. D8. {a} True. HA is invertible, than dat(A] #- 0. Since dEt{ABA} = dat{A)det[E}det[A] it Fallen: that if A te invertible and detL-IBA] = 0. than detw) = n. {It} True. IfA = A-l. theeeteeetletm-l) = 33h?- tt tellewethetmetm}? : 1 endeedet{.4} = i1. (:1 True. cheredumdmechelmformulﬂ hunrwufzaermJL-endiamtinmﬂhle. {a} True. Since det[.4T} - time}, It Follows that dat(AﬂT} n dqu] detmf} = (deem)? :3 u. {a} Thle. It'detIA] it‘- I] then A is Emu-tibia, and an invertible matrih: can always be mitten an a. product of elementary matrim. 3! 1n 3: In P1. Ifx- : andy—chmmaingmﬁcmrexpmaiomﬂnngﬂmﬁhwlummmm (1313(0) ={31 +m}01; + [12 +yeJC'g, + ...+ {an +ynJC'ﬂj = [310:5 +32%,- + - - - + tecnﬂ + {#1011 + 320:: + "' + HnC-tj} = M3” +d3ti3} EIEFIﬁIEE BET 4.3 14. Usingtheideutityl+Izan3a=anc2mwhmredetﬁd}=1forttaéi+mn 'I'hus.forthaaevalueaof tmam’n Dania D a, the matrix A is inwrtible and A" = adjfd} = -DDI2IEI tenuous“:- I} I} 0 sec”: i j I: 34. {a} Exﬁn —1 2 2 n-ﬂ+j—3k eteeMBC=g||EeEH=£F 1 1 —1 {h} maaac= e? HEN}: - gt; thee t. = gig—1 = [email protected] = tutu—ill”:- '|='1‘!J'5‘.t']:lm5 511. {a} Wehuwvxw (lwzw' wlwtmw]. 111 1:1 1:3 11': 1’: u-[vxw}=u1 11': ”3—413: §+ﬂaw I: #1 tr: 1!; ”I w: a 3 m 102 W3 {h} |u~(trxw”inequalmthewlumaofthspualklpipedhaﬁngthsmtnmu,muraaadjmnt edges. Aproufcmbeiuundinmstandudnnkuluatm. m. LetA—l ‘ ‘ﬂc‘ﬂ.mammu'+(1-c}“=2c3-2c+1;tafouum1ueaofc.nus. l-c formqthamhasamiquemluﬁongimby 3 -[1—c} c 3| 4 c _ Tc—4 {1+6} -4 -c—3 ”1: 2:22-2:+1 —ij-2c+l 3*: i3-2c+1 ‘2 —2c+l P1. We Emu-1r: lullllvllmaﬂ and luxvﬂ = ||u||||v||5inﬂ; thus my: ltﬁll ...
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