1vektor fazo prostranstvo 2liniynaya obolochka

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1 1

(1,a,0);(0,1,0); (0,0,1) demak quyidagiga ega bulamiz x

^{Mavzu vector fazo}

^Reja
^{1vektor fazo prostranstvo}
^{2liniynaya obolochka.}
^{3(Fazolar ustida oylar globuslar)}
^4bazis.
^{1 masala Xaqiqiy sonlarmaydoni ustida 2lovchivektorlar tuplash R2=(a,b)|a,b

^{Vektornisonga kupaytmasi}

^{R2-shu ikki amalga nisbatan. vector fazo tashkil qiladmi?}

^{yechim: ma’lumki (R2,+)komutativ grupabuladi.Navbatda amalni komutativ,assos tiv,neytral,birlik element ( )distir butuvlik neytali elementlar xosil bulishini tekshirish lozim}

^{Qo`shishga nisbatan distributiblik bajariladi.}

^{Demak R2 berilgan amallarga nisbatan vector vazni tashkil qilmaydi.}

^{2 masala. xaqiqiy sonlar ustida (oa cb) a,b,c- ixtiyoriy xaqiqiy son kurinishidagi metriaer tuplami qushish va songa qulash amalga nisbatan vector fazoni tashkil qiladmi?}

^{Yechim tashkil qiladi.}

^{4 Ustanavit ,yavlytsya li podprastranstvli arifmeticheskiy prostranstva R? mnojestvo bsex vektarov suma}

^{reshinix a) pusti 41 yesti mnosejstvo vsex vektori summa koordinat katorix ravno nulyu primenim krit}^{gada podprstranstva pust a=(a1+a2++…an) 4=b1+b2++……b4=0 sledovatelno a+b+l1}^{ibo (a1+b1)+(a2+b2+)+…+(an+bn)=0 takji prilubom c}^э^{41 a.c=(a,c+a2c+++++anc)=(a1+a2+…+++an).c=0.c}^э⁴

^{samostoyatelno b)}

^{Vektor Fazo}

^{Agar skalyar kupaytma skalyar kupaytmani x.y=(x1).(y1)+|x2|.|y2| kurinishda olinib x=(x1,x2);y=(y1,y2)bulsa R2arifmetik fazo vector fazo bulishi mn x=(1,1); y=(-2,-2);(3;3)}

^{yechim: yuq chunki distribbututivlik xosasai bajarilmaydi (x+y).p=|1-2|3+|1-2(.3=b; x.p=1.3+1.3=b; y.p=|-2|.3+(-2|.3=12}

^{demak x.p+y.p=18}^≠^x+y).p=6

^{2) x=2l,+3l2-3l3; y=l5-2l3;}

^{x.y va |x|va |Y|larni xisoblang}

^{Yechish;l ortanormalangan ekanidan lk.lk=1; lk.ln=0; k}^{≠n bulsa shu sababli x.y=2l,.l5-4l,.l3+3}^l^{2.l5-6l2l3-3l2.l5+6l3.l3=6.}^{| |x|=}_+(-3)²₌

_|y|=

^{3) a=1(1,1,1); a2=(1,-1,-1,); a3=(2,1,1) lar arifmetiklik vektop fazodan olingan bulsa shu vektorlarga ortaganal vector toping}

^{Yechish;Ax tarilayoytganida |X|=1; x.a1=0; x.a2=0 x.a3=0}

^{xni x=(x1,x2,x3) deb qarab bazis ekanidan.}

^{(1,a,0);(0,1,0); (0,0,1) demak quyidagiga ega bulamiz x}1^-x2^-x₃⁼⁰ ^.x1x2+x3⁼⁰

^{x12+x22+x32=1 Birinchi urta tenglamani Tase usulida yechamiz .U xolda x1=0, x22-x3 2x2=1. x=(01 1}^2,-

^{u=(u1,u2); u12=4; u,.u2=-2 u22=9 bulsa A)skalyar kupaytma formulasini bazisda vector koordinatlari}

^{x=a1l1+a2l2 va y=b1l1+b2l2. Uxolda x.y=a1b1+a2b2}

^{Yechim. Vektorlarni qushishga nisbatan distributivlar xossasini qulaymiz x=x1u1+x2u2 va y=y1u1+y2u2 bulsin U xolda}

^{Demak a)x.y=4x1y1-2x1y2-2x2y2-2x2y1+9x2y2}

^{8 ) Bunda bazisni ortoganalib}^{narmanlashtiramiz}

|u1|= 4; ekanidan e1= ;a vector a=u1+lu2 kurinishda bulsin u1a=0 Demak bazis e,= e2=(u1+2u2)|

^-2

^{x=(1,0,-2); y=(3,-9,3);= (3,-6,0);u =(2,5,7) c1=3, c2=2 c3=3 c4=0} c1x+c2y-3 0u=0

^{5)a=(1,2,-1); b=(-2,1,3); c=(-1,3,1) chiziqni bog’lanmagan chizig’ni kombiatsiyasi y=b1x1+b2x2+….+bnxn}

^{Misol;1F(x)=10x-2x-23x+2-23x+2; 1(x)=2x+3x; l(X)=x+4x-1 F(X)=51(x0-2}
tekshirib kuramiz

2 +3

-2+3

-2-1 2 1 3 =-5 0 Demak d=0, =0,

-1 3 1

A b c chiziqli erkli

UYGA VAZIFA

Ushbu vektorlarni chiziqli bog’langanini kursating 1)f(X)=6x³_-3x2+8x- ³_+2x²_{- x+5 ‘}

_2)d= 1 2 0 -1 1 -2 0- 3 2

-3 0 1 = 2 0 7 ; = 1 0 -8
3).x=(1,2,1) y=(2,3,4) g’=(-2 ,1,2 )

Vazifa

4) Quyidagi vektorlarni chiziqli bog’lanmaganligini kursating

A)x^3—₁ x²_{-2, x-3}

_b)(4,-1,3) (-3,5,-1) (-1,-4,7) Vazifa

_{5) F(x)=2x}⁴_-4x³₊₅vektorni f,(x)=-x⁴+2x³; f₂(x)=x²+x+1; f₃(x)=-1

_{6000 400 000 000 104}

_{6) 007 vektorn 000 ; 00-3 ; 000; 351;}

_{050 000 000 020 024}

_{vektorlar orqali chiziqli ifodalang.}
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