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TTU MATH 2360 - MATH 2360 Exam II

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Math 2360 Exam II 28 October 2009Make-up Form BSection IFor each of the following problems at least one of the choices is correct. For many of the problems more thanone of the choices is correct. Record your answers for each problem on the answer sheet (last page). For theissection, turn in (only) the answer sheet (with your answers on it) at the end of the exam. Do your own work. You may keep the exam for your own records.1. (12 pts) Consider the V set of polynomials with degree less than 3. Define “addition” for V by if with , then is defined by,pqV∈22() and ()pxabxcx qx x xαβγ=+ + = + +pq⊕. (We will use symbol for “addition” since333 33 33233()pqx a b x c xαβ γ⊕=+++++⊕this is not the usual addition for polynomials.) Define scalar multiplication for V by ifwith then is defined bypV∈2()px a bx cx andα=+ + ∈\pα . (This is usual scalar multiplication for polynomials.)2()pxabxcxαααα=+ +Recall the vector space axioms for addition for a vector space V. i. Addition is commutative, i.e. for every p, q in V we have ⊕=⊕pqq pii. Addition is associative, i.e., for every p, q, r in V we have () ()⊕⊕= ⊕ ⊕pq rp qriii. Existence of an additive identity in V, i.e. there exists an additive identity (called 0) in V so thatfor every p in V we have ⊕=0ppiv. Existence of additive inverses in V, i.e. for each p in V there exists an additive inverse (called -p)in V so that ⊕− =0ppWhich of the above axioms are not satisfied by V with its defined “addition” and scalar multiplication?2. (9 pts) Consider the following subsets of :4\i. ii.1234 13 24 12 34{( , , , ) | }Txxxx xx xx xx xx+=+1234 1 3 2 4{( , , , ) | 2 3 4 }Txxxx x x x x+=+iii. 12341234{( , , , ) | 0 or 0 or 0 or 0}Txxxx x x x x====Which of the above subsets of are subspaces of ?4\4\3. (9 pts) Consider the following subsets of :4Pi. ii. 4{ | is an even function and (-1) -1}pPp p∈=4{|(0)(1)}pPp p∈=iii. 4{|(1)0}pPp′∈=Which of the above subsets of are subspaces of ?4P4P4. (12 pts) Consider the following subsets of :3\i. ii.102,101⎧⎫⎡⎤⎡⎤⎪⎪⎢⎥⎢⎥−⎨⎬⎢⎥⎢⎥⎪⎪⎢⎥⎢⎥⎣⎦⎣⎦⎩⎭0111, 0 , 1211⎧⎫−⎡⎤⎡ ⎤⎡ ⎤⎪⎪⎢⎥⎢ ⎥⎢ ⎥−⎨⎬⎢⎥⎢ ⎥⎢ ⎥⎪⎪⎢⎥⎢ ⎥⎢ ⎥−−⎣⎦⎣ ⎦⎣ ⎦⎩⎭iii. iv. 12 01,1, 1021⎧⎫−⎡⎤⎡⎤⎡⎤⎪⎪⎢⎥⎢⎥⎢⎥−⎨⎬⎢⎥⎢⎥⎢⎥⎪⎪⎢⎥⎢⎥⎢⎥−⎣⎦⎣⎦⎣⎦⎩⎭11111, 1 ,0, 1111 1⎧⎫−⎡⎤⎡ ⎤⎡⎤⎡ ⎤⎪⎪⎢⎥⎢ ⎥⎢⎥⎢ ⎥−−⎨⎬⎢⎥⎢ ⎥⎢⎥⎢ ⎥⎪⎪⎢⎥⎢ ⎥⎢⎥⎢ ⎥−−⎣⎦⎣ ⎦⎣⎦⎣ ⎦⎩⎭Which of the above subsets of are spanning sets for ?3\3\5. (12 pts) Consider the following subsets of :3\i. ii.112, 211⎧⎫−−⎡⎤⎡⎤⎪⎪⎢⎥⎢⎥−⎨⎬⎢⎥⎢⎥⎪⎪⎢⎥⎢⎥⎣⎦⎣⎦⎩⎭12 01,1, 1021⎧⎫−⎡⎤⎡ ⎤⎡ ⎤⎪⎪⎢⎥⎢ ⎥⎢ ⎥−⎨⎬⎢⎥⎢ ⎥⎢ ⎥⎪⎪⎢⎥⎢ ⎥⎢ ⎥−⎣⎦⎣ ⎦⎣ ⎦⎩⎭iii. iv. 0111, 0 , 1211⎧⎫−⎡⎤⎡⎤⎡⎤⎪⎪⎢⎥⎢⎥⎢⎥−⎨⎬⎢⎥⎢⎥⎢⎥⎪⎪⎢⎥⎢⎥⎢⎥−−⎣⎦⎣⎦⎣⎦⎩⎭12210,1, 1,020 11⎧⎫−−⎡⎤⎡ ⎤⎡ ⎤⎡⎤⎪⎪⎢⎥⎢ ⎥⎢ ⎥⎢⎥−⎨⎬⎢⎥⎢ ⎥⎢ ⎥⎢⎥⎪⎪⎢⎥⎢ ⎥⎢ ⎥⎢⎥−⎣⎦⎣ ⎦⎣ ⎦⎣⎦⎩⎭Which of the above subsets of are linearly independent?3\6. (6 pts) Consider the following subsets . 3Pi. ii.22{1 , ,1 }Sxxxx=+ + +22 2{1 2 , 1 , 1 3 }Sxxxxxx=+ − +− + −Which of the above subsets of are linearly independent?3P7. (9 pts) Consider the following subsets of :2\i. ii. 64,12 8⎧⎫−⎡⎤⎡⎤⎨⎬⎢⎥⎢⎥−⎣⎦⎣⎦⎩⎭21,11⎧⎫−⎡⎤⎡ ⎤⎨⎬⎢⎥⎢ ⎥−⎣⎦⎣ ⎦⎩⎭iii. 123,,342⎧⎫−⎡⎤⎡⎤⎡⎤⎨⎬⎢⎥⎢⎥⎢⎥−−⎣⎦⎣⎦⎣⎦⎩⎭Which of the above subsets of form a basis for ?2\2\8. (12 pts) Consider the following subsets of :3\i. ii.121,001⎧⎫⎡⎤⎡⎤⎪⎪⎢⎥⎢⎥−⎨⎬⎢⎥⎢⎥⎪⎪⎢⎥⎢⎥⎣⎦⎣⎦⎩⎭1121, 2,1021⎧⎫−⎡⎤⎡ ⎤⎡⎤⎪⎪⎢⎥⎢ ⎥⎢⎥−−⎨⎬⎢⎥⎢ ⎥⎢⎥⎪⎪⎢⎥⎢ ⎥⎢⎥−⎣⎦⎣ ⎦⎣⎦⎩⎭iii. iv. 1130, 1 , 2131⎧⎫−−⎡⎤⎡ ⎤⎡ ⎤⎪⎪⎢⎥⎢ ⎥⎢ ⎥−⎨⎬⎢⎥⎢ ⎥⎢ ⎥⎪⎪⎢⎥⎢ ⎥⎢ ⎥−⎣⎦⎣ ⎦⎣ ⎦⎩⎭11121, 1 , 3,11011⎧⎫−−⎡⎤⎡ ⎤⎡ ⎤⎡⎤⎪⎪⎢⎥⎢ ⎥⎢ ⎥⎢⎥−−⎨⎬⎢⎥⎢ ⎥⎢ ⎥⎢⎥⎪⎪⎢⎥⎢ ⎥⎢ ⎥⎢⎥⎣⎦⎣ ⎦⎣ ⎦⎣⎦⎩⎭Which of the above subsets of forms a basis for ?3\3\9. (3 pts) Consider the following statement:Theorem. Let and let . The following are equivalent:{}123,,,,nn⊂"\aaa a[]123 nA = "aaa ai. A is invertibleii. has a unique solutionA = 0xiii. A is row equivalent to Iiv. has a solution for every A =xbn∈ \bv. det( ) 0A ≠vi. diag( ) 0A ≠vii. is a basis for {}123,,,,n"aaa an\viii. 123dim( ({ , , , , }))nspan n="aaa aix. () {}NA= 0Which one of the above statements needs to be deleted to make the theorem true?10. (3 pts) Let A be a matrix. If the dimension of the null space of A is 11, then the rank of A is:816×a. 3 b. 4 c. 5 d. 8e. Undeterminable from the given informationSection II Answer the problem in this section on back of the answer sheet. You do not need to rewrite theproblem statement. Work carefully. Do your own work. Show all relevant supporting steps!11. (15 pts) Consider the matrix A given by . A straightforward reduction by12 1 3 111 119 41131234 5 1 5A−−⎡⎤⎢⎥−−⎢⎥=⎢⎥−−−⎢⎥−⎣⎦elimination shows that A is row equivalent to U where 10 7 5 301 44 100 00 00000 0U−−⎡⎤⎢⎥−−⎢⎥=⎢⎥⎢⎥⎣⎦a. Find a basis for the row space of A.b. Find a basis for the column space of A.c. Find a basis for the null space of A.Answer Sheet - Form BName _________________________1. i. Is Not Satisfied Is Satisfied ii. Is Not Satisfied Is Satisfiediii. Is Not Satisfied Is Satisfied iv. Is Not Satisfied Is Satisfied2. i. Yes No ii. Yes No iii. Yes No3.i. Yes No ii. Yes No iii. Yes No4. i. Yes No ii. Yes No iii. Yes No iv. Yes No5. i. Yes No ii. Yes No iii. Yes No iv. Yes No6. i. Yes No ii. Yes No7. i. Yes No ii. Yes No iii. Yes No8.


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TTU MATH 2360 - MATH 2360 Exam II

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