1.3 - 1.3 Vector Equations Vectors have important...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.3 Vector Equations Vectors have important mathematical and physical meanings, but for now, just consider them as lists of numbers. column vectors u, v ∈ R2 : u = 1 2 ,v = −3 1 8 u = v if and only if corresponding entries are equal u+v = −2 1 28 ← add corresponding entries 1 −3v = 1 1 − 24 1 ← multiply − 3 times each entry −2u + 4v =? Properties: u, v, w ∈ Rn , scalars c, d (1) u + v = v + u (2) (u + v ) + w = u + (v + w ) (3) u + 0 = u, 0 has all entries 0 (4) u + (−u) = 0, −u = (−1)u (5) c(u + v ) = cu + cv (6) (c + d)u = cu + du (7) c(du) = (cd)u (8) 1 · u = u 1 Representing vectors interpretation. EX: For w = 4 4 y a b 3 1 v by (a, b), we get a geometrical = + 1 3 = u + v , we get w u x Parallelogram Rule: 4th vertex of parallelogram is u+v x2 u+v u v 0 x1 2 Picture of vectors in R3 : x3 2u u x1 x2 Note: u is a 3 × 1 matrix · represent u in R3 by a point (x, y , z ) = (1, 2, 3) · where is 2u, 3u, −u, −2u? Pictures in Rn? u 1 u 2 · point (u1 , u2 , . . . un ) ↔ u = . . . un 3 EX: Sums of multiples of two vectors u = 2 1 and v = −3 1 3v 2v v 0 −2u + v −2u −u w 2u u 3u u−v −v −2v Describe the vertices of the intersections of the solid lines: au + bv for integers a, b (e.g., see vector −2u + 2v above) Q: By taking all combinations au + bv a, b ∈ R, can we get all vectors w ∈ R2 ? A: Yes. As an example, for the vector w labelled above, w= 3 2 4 u+v Definition: Let v1 , v2, · · · , vp ∈ Rn . We say that y ∈ Rn is a linear combination of v1 , v2, · · · , vp if y = c1 v1 + c2 v2 + · · · + cp vp for some so-called weights c1 , c2 , · · · , cp EX: example above: x2 w 2v v 0 u 2u x1 For w = 2u + 2v , the weights are 2, 2 Given another vector, say b = so that 10 −4 , find weights x1 , x2 b = x1 u + x2 v Write x1 u + x2v = b, or 5 x1 2 1 2x1 x1 + x2 −3 1 = 10 −4 10 −4 10 , or + −3x2 x2 = , or 2x1 − 3x2 x1 + x2 2x1 − 3x2 = 10 x 1 + x 2 = −4 = −4 , or Therefore, solve system by taking augmented matrix 2 −3 1 10 , 1 −4 converting to RE form, and then finding weights x2 , x1 by back substitution. Note: Can solve this for any vector b = b1 b2 Now we go directly from this special case to the general situation: Given vectors v1 , v2, · · · , vp ∈ Rn , H = Span {v1 , v2, · · · , vp} is the set of all vectors which are linear combinations of v1, v2 , · · · , vp , so any vector w ∈ Rn in this set satisfies w = c1 v1 + c2 v2 + · · · + cp vp 6 EX: previous example: u = v1 = 2 and v = v2 = −3 1 and H = Span {u, v} 1 2 is all of R . EX: 1 1 0 2 ∈ Span 1 , 1 ? Q: Is 0 1 3 1 0 1 Solution: Yes, if c1 1 + c2 1 = 2 , so solve 0 1 3 system by considering augmented matrix for =1 c2 = 3 101 R2 ← R2 − R1 A=1 1 2 −→ 013 101 R3 ← R3 − R2 0 1 1 −→ 002 101 0 1 1 013 c1 c1 +c2 = 2 inconsistent, so NO. 7 EX: 1 0 1 Q: Does Span 1 , 1 0 = R3? 0 1 1 Solution: Yes, if 1 c1 1 + c2 1 + c3 0 = y 0 1 1 z is solvable x for all y ∈ R3. z 10 1 x 101x 0 1 −1 y − x 1 1 0 y −→ 01 1 z 011z 10 1 x 0 1 −1 y−x 00 2 z−y+x 0 1 x −→ consistent for all values of x, y , z , so YES. Remark: Notice that to answer the last question, we don’t need the last column. We only need to check if every row has a leading entry (pivot position). 8 Observation about a change of notation: A vector eqn x 1 a1 + x 2 a2 + · · · + x n an = b has the same soln set as the linear system whose augmented matrix is a1 | a2 | · · · |an |b . So b ∈ Span {a1 , a2, · · · , an} iff linear system has a soln. EX: 1 1 Q: What is Span 1 , 0 0 −1 ? A: it is a plane in R3 . EX: Practical example (Problem 27 in book) In Production of goods, we have {#of days of production}×{output per day} = {total production} Problem Statement: A mining company has 2 mines. Mine # 1 produces ore with 20 metric tons of copper, 550 kg of silver per day Mine # 2 produces or with 30 metric tons of copper, 500 9 kg of silver per day so output per day is v1 = 30 500 20 550 for mine #1, v2 = for mine #2. (a) What is the physical interpretation of 5v2? A: It is 5 days production in mine #2. (b) If mine #1 operates for x1 days and #2 operates for x2 days, write a vector eqn whose soln gives # of days needed to produce 150 metric tons of copper and 2825 kg of silver. A: x1v1 + x2 v2 = 150 2825 (c) Solve for x1 , x2 . Solution: 20 30 150 → 20 30 150 , so 550 500 2825 x2 = 4, x1 = 3 (check!) 2 0 −325 −1300 10 ...
View Full Document

Ask a homework question - tutors are online