; row vectors: 1 3 4, 1 2 5, 2 6 7

Transcription

1 column vecors: one eample: one eample: = = OR = = ; row vecors: Plug he poin ino all hree equaions. Resul: rue saemens.. Plug he poin ino all hree equaions. Resul: rue saemens.. Plug he poin ino all hree equaions. Resul: rue saemens. 4. Plug he poin ino all hree equaions. Resul: rue saemens. 5. Boh lines are parallel wih slope /. 0. A 4 7 # b A RREF 4. REF 6. RREF 8. RREF 0. one possible answer: 0 ; Rank(A) =. one possible answer: 0 0 ; Rank(A) = 4. one possible answer: 0 ; Rank(A) =

2 6. one possible answer: 0 ; Rank(A) = one possible answer: ; Rank(A) = ; Rank(A) =. ; Rank(A) = 0 4. ; Rank(A) = ; Rank(A) =.5 4. a. k = b. k = c. k ± 6. a. a = / b 0 b. a = / b = 0 c. a / 6. () 8. ( ) 40. ( 0 ).6 8. no inverible A A A (4 ) 6. ( 0 8 )

3 . 6. inverible 8. 4 inverible 0. inverible /6 4. / / All en aioms hold.. Onl A A A5 A6 A9 hold. 4. Onl A A A4 A5 A7 A8 hold es closed under addiion and scalar muliplicaion 6. no does no conain he zero vecor 8. no no closed under addiion 0. es closed under addiion and scalar muliplicaion. es closed under addiion and scalar muliplicaion 4. no does no conain he zero vecor 6. es closed under addiion and scalar muliplicaion 8. es closed under addiion and scalar muliplicaion 0. no does no conain he zero vecor 4. nullspace( A) {( r r): r } 6. nullspace( A) {( r r r) : r } 8. nullspace( A) {(8r 8 s r s r s) : r s } 4.4. es 4. no 6. es 8. no 6. { ( 0 ) (0 ) } { } 6. { ( 0) ( 0 ) }

4 8. { ( 0) ( 0 ) } 0. { (8 0) (8 0 ) }. { ( 0) } ( a b a ba b) : a b ; plane hru origin a a a a ( ) : ; line hru origin 8. es v vv 40. es p p p a b a b 4. a b a b : ab ; es B A A 4.5. linearl independen 4. linearl independen 6. linearl dependen ( 4 6) ( 69) (000) 8. linearl dependen ( ) (0 ) ( ) 5( ) = () 0. linearl independen 6. linearl dependen 5A A A 8. linearl dependen A A A 0. linearl dependen p p 0. linearl dependen 5p 7p 8p 9p one eample: { ( 5) ( ) ( ) } 8. one eample: { ( ) ( ) ( ) } 0. one eample: { }. W = never 0 linearl independen on I. W = an (sec + ) no alwas 0 over I linearl independen on I 4. W = 6 never 0 linearl independen on I 5. W = e 4 never 0 linearl independen on I 0 if 0 6. W 4 if 0 no alwas 0 over I linearl independen on I 7. linearl dependen f f f 4.6. es 4. no 6. es 0. es. no 4. es 6. es 8. no one eample: { ( 0 ) (0 ) }; dimension 6. one eample: { } 8. one eample: 0 ; dimension 0. one eample: { ( 0 ) (0 ) }; dimension

5 4. one eample: 4a. one eample: 0 0 ; dimension a. one eample: 0 0 b. one eample: A B 4 PC B PC B PB C 6 6 v ( 7) PC B (4) (4) B v v B C v (0) P B C B v B ( ) v C ( ) a. n = ; { } b. m = ; { ( 7) } 6. a. n = ; { ( ) (0 ) } b. m = ; { ( 5 9) ( 6 0) } 8. a. n = 4; { ( ) ( ) } b. m = 4; { ( 5) ( 5 7) } 0. { ( ) (5 4 ) }. { ( ) ( 4) } 4.9. nullspace(a) = span { (0 ) (6 0 0) ( ) } rank(a) = nulli(a) = # of columns = 4 4. nullspace(a) = { () } rank(a) = nulli(a) = 0 # of columns =

6 = c + p where = (4 ) and p = ( 5 0). = () + p where p = ( ) º or π/ radians. 54º or 0.95 radian. 9º or 0.68 radian 4. º or 0.56 radian. A B A B 7 or 0.54 radians. 60º 0. a. 9 b orhonormal se: ( ) ( ) (0) 6 4. no an orhogonal se w 7 ; w 5 4 ; 0 orhonormal se: () ( ) (5 4) A4 ; T( z) ( 5 z z). T( ) ( 0 ) / / / 0. / 4 / 5 / 5 / 4 7 /5 8 /5

7 6.. a. T(7 5 ) = () es b. T( 5 ) = ( ) no c. T(5 5 5) = () es. Ker(T) = span{ ( ) } dim[ker(t)] = Rng(T) = span{ ( ) } dim[rng(t)] = + = dim[ ] 4. Ker(T) = { () } dim[ker(t)] = 0 Rng(T) = dim[rng(t)] = 0 + = dim[ ] 5. Ker(T) = span{ (5 ) } dim[ker(t)] = Rng(T) = span{ ( 5) ( 8) } dim[rng(t)] = + = dim[ ] 6. Ker(T) = span{ ( 0) ( 0 ) } dim[ker(t)] = Rng(T) = span{ ( ) } dim[rng(t)] = + = dim[ ] 7. Ker(T) = span{ ( 0) } dim[ker(t)] = Rng(T) = dim[rng(t)] = + = dim[ ] 0. basis Ker S 0 dim Ker S basis Rng S 0 0 dim Rng S The book has he wrong answer on his one. Perhaps he hough he L.T. was A A T. 4. he se of marices ha commue wih B 5. Ker( T) span dim Ker T Rng( T) span dim Rng T 6. Ker(T) = span{ + + } dim[ker(t)] = Rng(T) = P ( ) dim[rng(t)] =

8 7. Ker(T) = {0} dim[ker(t)] = 0 Rng(T) = span{ + + } dim[rng(t)] = 0 8. Ker( T ) span 0 0 dim Ker T Rng( T) span dim Rng T 9. Ker( T ) span ( ) The book had he wrong answer here. dim Ker T T 0. Ker(T) = Rng( T) span 0 9 dim Rng dim[ker(t)] = 0 Rng(T) = M ( ) 4 dim[rng(t)] = Ker(T) = {0}; one-o-one Rng(T) = span{ ( ) }; no ono T DNE. Ker(T) = span{ ( ) }; no one-o-one Rng(T) = span{ ( ) }; no ono T DNE 4. Ker(T) = span{ ( ) }; no one-o-one Rng(T) = span{ (0 5 ) ( 4 4 ) }; no ono T DNE 8. T is ono bu no one-o-one T DNE 9. T is ono bu no one-o-one T DNE a b 0. T is boh one-o-one and ono T a b c b c. T is neiher T DNE a b 6. T ( a b c) 0 c a b 8. T ( a b c) b c n = 5 T( a b c d e ) ( a b c d e) / / 6. T ( ) A / / T ( ) A 5 9 8

9 C 6.5 a. T 4a. T B C B Av v 4. Av 4 4 v 6 6. Av 5 v 8 4 c 4c c 8c c 4c 4. Av c 6c c v 4 c 6c c 6 0 c 0c c 5. Av c 40c 0 4c v 4 0 c 0c c firs order 4. second order 5. ln ; ; ; 6. d LHS d d RHS d ; ; ( ) ( ) d LHS d ( ) ( ) d RHS d ( ) ( ) ( ) ( ).. nonlinear 4. nonlinear 6. nonlinear / 4. ( ) c for all 0 8. ( ) cos 40. ( ) e e ( ) e e

10 .4. an an or ( ) 4 4. ( ) cos sec e e g g 6. a g b. 5 L eac 4. eac 5. c 7.. ln ln c e c cos ln verif ha L = 0 6. verif ha L = 0 7. verif ha L = 0 0. Ker( L) ce : c D 4D 6 sin and ( ) c e c e c e ( ) c e c c e b. ( ) e cos 4cos sin ransien par: e cos sead-sae par: 4cos sin 8.7. ( ) e c c ln 4. ( ) e c c an ln 6. ( ) cos ln sec an sin 8. 5 ( ) e c c ln 4 an / e c c ( ) e c c ln 4.

11 b s b 4. s s 6 5s 8. s s 4 s s s s. s e s s s e s e s s s e cos sin

12 8. f ( ) 4 u( ) 0. f ( ) u( ) u4( ). f ( ) (6 ) u( ) ( 6) u6( ) f ( ) (sin ) u ( ) ( sin ) u ( ) 4. / / ( ) ( ) ( ) e u( ) e u( ) e 6 u( ) e 4 ( ) 44. e u( ) e e 45. The u () erm should have in he eponen insead of. 9.. ( ) ce ce and ( ) ce ce 4. ( ) ce ce and ( ) ce c 4 e 4 ( ) c e cos c e sin and ( ) c e sin c e cos ( ) 5sin and ( ) cos sin. ( ) c e c e e and ( ) c e c e e sin 4 4 e b a F( ) W[ ]( ) e 0 herefore linearl independen on ( ). W[ ]( ) 0 for some values (like = ) herefore linearl independen on ( ). W[ ]( ) ( ) e cos e sin ( ) e e ( ) e sin ( ) e cos 0 for 0 herefore linearl independen on ( ). W[ ]( ) 0 for some values (like = ) herefore linearl independen on ( ). When I = [0 ) hen herefore linearl dependen on [0 ). 4 0 herefore linearl dependen on ( ) herefore linearl dependen on ( ) herefore linearl dependen on ( ) cos sin ( ) e e cos sin cos sin 9.5. () c e c e ( ) c e e c c 4.

13 6. ( ) e c c 0 c ( ) e c c 0 c e () e e () ce ce e 8e 4e () ce ce e 8e