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- Politecnico di Milano
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- Management Engineering
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- Analisi Matematica 2
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- Exercises · By topic
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Analisi Matematica 2By topic
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103/2023 1)Siconsideriilresteraunire E 2x+ 4y+42=4 x=1+2 a)DeterminareandmatureA=M(R)- x+ 3y=2- 42 3 eunvettoreER"taliche,dettoil vetteredelle ricognite,siabbieAs= a)Determinariilnumerodisduzionidelesterior e)Caldaree soluzioniprevistealpuntoe a)2x+ 4y+42= 4 I2 (a- z= 3 8- 43][2]-x+ 3y+42=2 b)e(A)= e([AE))edèmassinio,allorailresternoammetteuna edunadasoluzione. e(A)= e([AE)mmoredelmassiniovalore erestesesteredusioniegeneredelle aree- [A1]=8444 24 1 I·alte:**]I2- 1 - 210 - 1 1s 2 I244 I 4 ITe i d e ] -0SO188 [24 Deletestess fatts- Leautenteuna I solasoluzione 444 352=- 352=- 1 (Ry - 3=1 ay= xy= e()= (i)x+4- 2=2 x=0 2)siresciailsistemalineare: E x+ y=1 alvariaredelparametrorealek.kx+ y+z=1- k y+(1- kz=1 [11 EET/a]- - -)at I 1 A I⑪1- k1 - e i· mm/jytoII 1 1- k o BI-*R 11 01 s To1 -2R-1Ie 1 27-- seR-1 I0- einiettMg+ By -41= AR) =/R-1 seRFO,1,2rEI])=e(A)=3max->ilastenabyrefore- madiR-Cammetteundedunasoloeduzione. ek=0 [ES= TO9lIT u T57 101 1/2)100x10)12-13 4z(2])= e/Ao)-2estono= soduzioni 11 [01 1--i(2)y= y= 1t(27)= [+]x+ A- t= Ax=t seR= 1 []= 1101 In·28/]/01 -12 I e(tA])= e(71)= 3maxla soluzione~(2)lastedèunico->ricadenelcaso esk= 2 Eret2]Inter
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