By Matthews K.R.
Read or Download Elementary linear algebra: solutions to problems PDF
Similar study guides books
Capture this new ACT! and construct higher relationshipsOrganize client info, agenda actions, create stories, and moreIt's time to behave! in your goal to enhance consumer courting administration! ACT! 2005 bargains nice new instruments, and with this booklet in hand, you've got a best specialist displaying you the way to take advantage of them.
This new version might help you pinpoint your analyzing strengths and weaknesses, gauge your development and skim quicker.
This can be a significant new textbook protecting the most modules of the OCR A2 syllabus inside of a unmarried source. The publication is in particular designed to assist scholars of all studying kinds comprehend the topic. the color layout and pedagogical units corresponding to key case packing containers, instance packing containers, diagrams and tables carry the topic to existence and make research attention-grabbing.
- Basic Math and Pre-Algebra. Cliffs Quick Review
- Fiske WordPower
- MCDST Microsoft Certified Desktop Support Technician Study Guide [Exams 70-271, 70-272]
- Language and the Worship of the Church
Additional info for Elementary linear algebra: solutions to problems
Example text
X r ⊆ Y 1 , . . , Y s , (1) if each of X1 , . . , Xr is a linear combination of Y1 , . . , Ys . 1) Suppose that each of X1 , . . , Xr is a linear combination of Y1 , . . , Ys and that each of Y1 , . . , Ys is a linear combination of X1 , . . , Xr . Then by (a) equation (1) above X1 , . . , X r ⊆ Y 1 , . . , Y s and Y1 , . . , Y s ⊆ X 1 , . . , X r . Hence X1 , . . , X r = Y 1 , . . , Y s . 1) Suppose that each of Z1 , . . , Zt is a linear combination of X1 , . . , Xr . Then each of X1 , .
Also Arg (4 + i) = α, where tan α = 1/4 and 0 < α < π/2. Hence α = tan −1 (1/4). 1 = 2 (−3)2 + (−1)2 y ✻ ✛ 1√ 9+1= = 2 √ 10 . 2 Also Arg (−1 + 2i) = π − α, where tan α = 2 and 0 < α < π/2. Hence α = tan −1 2. 60 ✲x α ✏ ✏✏ ✮ −3−i 2 ❄ Also Arg ( −3−i ) = −π + α, where 2 1 3 tan α = 2 / 2 = 1/3 and 0 < α < π/2. Hence α = tan −1 (1/3). (iii) The number lies in the second quadrant of the complex plane. √ | − 1 + 2i| = (−1)2 + 22 = 5. 4+i ✲x ❄ (ii) The number lies in the third quadrant of the complex plane.
12 12 (d) z5 w2 = 25 (cos ( 5π − 2π ) + i sin ( 5π 32 4 6 4 32 11π 11π (cos 12 + i sin 12 ). 9 = − 2π )) 6 (a) (1 + i)2 = 2i, so (1 + i)12 = (2i)6 = 26 i6 = 64(i2 )3 = 64(−1)3 = −64. √ )2 = −i, so (b) ( 1−i 2 1−i √ 2 −6 = 1−i √ 2 = (−i)−3 = 2 −3 −1 1 −1 = = = −i. 3 i −i i √ √ 8. (i) To solve the equation z 2 = 1 + 3i, we write 1 + 3i in modulus– argument form: √ π π 1 + 3i = 2(cos + i sin ). 3 3 Then the solutions are π π √ + 2kπ + 2kπ 3 zk = 2 cos + i sin 3 , k = 0, 1. 2 2 62 Now k = 0 gives the solution z0 = √ √ √ π π 2(cos + i sin ) = 2 6 6 3 i + 2 2 √ 3 i =√ +√ .