Applied Math 247 Exam#1: Summer 2008



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AnswerthequestionsinthespacesprovidedonthequestionsheetsIfyourunoutofroomforan answercontinueonthebackofthepagenonotes,books,orotheraidsmaybeusedontheexam Student Id: Answer Key 1 (1 points) 2 (1 points) 3 (1 points) 4 (1 points) 5 (1 points) 6 (1 points) 7 (1 points) 8 (1 points) 9 (1 points) 1 (1 points) Total(1 points) Page1of8

1(1 points) Determine whether each of the following statements is true or false No justification is required (a)ifavector uisalinearcombinationofvectors v and w,thenwecanwrite u=a v+b w forsome scalarsaandb Solution: True; this is the definition of a linear combination [ ] 6 8 (b) The matrix represents a rotation 8 6 [ ] a b Solution: True; the matrix has the form,wherea b a 2 +b 2 =1 (c)ifmatrixacommuteswithb,andbcommuteswithc,thenmatrixamustcommutewithc Solution: False;letAandCbetwomatricesthatfailtocommuteandletBbetheidentitymatrix (d)ifaandbaren nmatrices,andvector visinthekernelofbothaandb,then vmustbeinthe kernelofmatrixabaswell Solution: True;observethatAB v=a = (e)ifv isanythree-dimensionalsubspaceofr 5,thenV hasinfinitelymanybases Solution: True;if( v 1, v 2, v 3 )isabasisforv,then(k v 1,k v 2,k v 3 )isabasisaswell Page2of8

2(1points) Inagridofwires,thetemperatureatexteriormeshpointsismaintainedatconstantvalues asshownintheaccompanyingfigurewhenthegridisinthermalequilibrium,thetemperaturetateach interior mesh point is the average of the temperature at the four adjacent points For example, T 2 = T 3+T 1 +2+ 4 FindthetemperaturesT 1,T 2,andT 3 whenthegridisinthermalequilibrium 2 T 1 2 T 2 T 3 4 Solution: The equations for the other two mesh points are T 1 = T 2+2++ 4 and T 3 = T 2+4++ 4 Wecanrewritethethreeequationsasthesystem 4T 1 +T 2 = 2 T 1 4T 2 +T 3 = 2 T 2 4T 3 = 4 T 1 75 Reducing this system, we find T 2 = 1 T 3 125 Page3of8

3(1 points) Consider the chemical reaction ano 2 +bh 2 O chno 2 +dhno 3, wherea,b,andc,andd areunknownpositiveintegers Thereactionmustbebalanced;thatis,the numberofatomsofeachelementmustbethesamebeforeandafterthereactionforexample,because thenumberofoxygenatomsmustremainthesame, 2a+b=2c+3d Whiletherearemanypossiblevaluesfora,b,c,anddthatbalancethereaction,itiscustomarytouse the smallest possible positive integers Balance this reaction Solution: Considering hydrogen and nitrogen as well, we obtain the system 2a+b = 2c+3d 2a+b 2c 3d = 2b = c+d or 2b c d = a = c+d a c d = a 2t Reducing this system, we find that b c = t t Settingt=1,weobtain d t 2NO 2 +H 2 O HNO 2 +HNO 3 Page4of8

4(1points) SupposealineLinR 3 containstheunitvector u 1 u= u 2 u 3 FindthematrixAofthelineartransformationT( x)=proj L x GivetheentriesofAintermsofthe componentsu 1,u 2,u 3 of u Solution: u 1 x 1 u1 2+x 2u 2 u 1 +x 3 u 3 u 1 u 2 proj L x=( x u) u=(x 1 u 1 +x 2 u 2 +x 3 u 3 ) u 2 = x 1 u 1 u 2 +x 2 u2 2+x 1 u 1 u 2 u 1 u 3 3u 3 u 2 = u 1 u 2 u 2 u 3 x 1 u 1 u 3 +x 2 u 2 u 3 +x 3 u3 2 2 u 2 u 3 x u 1 u 3 u 2 u 3 u3 2 1 2 3 5(1points) Findtheinverseofthematrix 1 2 1 Solution: 1 2 3 1 1 1 1 2 1 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 1 1 1 Page5of8

6(1points) FindalllineartransformationsTfromR 2 tor 2 suchthat T [ ] 1 = 2 [ ] 2 1 and T Hint:Wearelookingforthe2 2matricesAsuchthat [ ] [ ] 1 2 A = and A 2 1 [ ] 2 = 5 [ ] 2 = 5 These two equations can be combined to form the matrix equation [ ] [ ] 1 2 2 1 A = 2 5 1 3 [ ] 1 3 [ ] 1 3 Solution: [ ] [ ] 1 2 1 1 2 A= = 1 3 2 5 [ 2 1 1 3 ] [ ] 5 2 = 2 1 [ ] 8 3 1 1 7(1 points) Consider a linear transformation T that represents the orthogonal projection onto the plane x+2y+3z=inr 3 Findbothabasisoftheimageandabasisofthekernelofthetransformation Solution: TheimageofTistheplaneitself,andthekernelofTconsistsofallvectorsorthogonaltothe plane Two linearly independent vectors that lie in the plane are 1 and 1 1 /3 2 /3 Thesetwovectorsformabasisoftheimage Tofindabasisofthekernel,weneedtofindavector orthogonaltobothofthesevectorstodoso,wefindthekernelofthematrix [ ] 1 1/3 1 2 /3 1 /3 1 Thus, 2 /3,ormoresimply 2,isabasisofthekernel 1 3 Page6of8

8(1points) AnalternatedefinitionofasubspaceW ofr n isasubsetofr n thathasthefollowingthree properties:(1) nonempty;(2) closed under addition; and(3) closed under scalar multiplication Show that this alternate definition is equivalent to the definition given in class and in the text Solution: To show equivalence, we must show that each definition implies the other First, if a subspace contains the zero vector, then it is clearly nonempty Thus, the definition given in class implies the alternativedefinitionsecond,ifasubspaceisnonempty,thenitcontainssomevector vsinceitisalso closed under scalar multiplication, v = is also in the subspace Thus, the alternative definition implies the definition given in class Hence, the two definitions are equivalent 9(1points) Findbothabasisoftheimageandabasisofthekernelofthematrix 1 5 3 5 1 1 1 5 2 1 3 Solution: We first reduce the given matrix to the matrix 1 1 Therefore,, 1 5 2 1 isabasisoftheimage,and, 1, 1 isabasisofthekernel 1 Page7of8

5 4 2 1(1points) FindthematrixBofthelineartransformationT( x)= 4 5 2 xwithrespecttothe 2 2 8 2 1 thebasis B= 2 1 1 1 2 Solution: B= [[ T( v 1 ) ] B [ T( v2 ) ] B [ T( v3 ) ] ] B = B 9 9 B 9 18 B = 9 9 Page8of8