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1-10 CHAPTER 1 l^ Linear Equations in Linear Algebra
1.3 VECTOR EQUATIONS _______________________
Do not be deceived by the rather simple beginning of Section 1.3. The important material on Span{ v 1 ,.. ., v p} will take time to digest. Figures 8, 10, and 11 are important, along with Exercises 11–14, 17, 18, 25, and 26. Each of the exercises involves an existence question about whether a certain vector equation has a solution. (You don’t have to find the solution.) Notice how the same basic question can be asked in several different ways.
1.3 l^ Vector Equations 1-
STUDY NOTES
Develop the habit of reading the section carefully once or twice before looking at the Study Guide and before starting the exercises. (Don’t just look at the pictures and examples! Important comments lurk in between.) In nearly all of the text, a scalar is just a real number. By convention, scalars are usually written to the left of vectors, such as 5 v or c v , rather than v 5 or v c. To identify vectors in your lecture notes and homework, you can write underlined letters for vectors. (Some students write arrows above the letters, but that takes longer.) Vectors must be the same size to be added or used in a linear combination. For instance, a vector in R 3 cannot be added to a vector in R 2 .
SOLUTIONS TO EXERCISES
^ − − − + − −
u v.
Using the definitions carefully, 1 3 1 ( 2)( 3) 1 6 5 2 ( 2) 2 1 2 ( 2)( 1) 2 2 4
−^ −^ −^ −^ −^ − +
−^ −^ −^ +
u v , or, more quickly,
−^ −^ − +
−^ +
u v. The intermediate step is often not written.
7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every vector in R^2 can be written as a linear combination of u and v. To write a vector a as a linear combination of u and v , imagine walking from the origin to a along the grid “streets” and keep track of how many “blocks” you travel in the u -direction and how many in the v -direction.
1.3 l^ Vector Equations 1-
13. Denote the columns of A by a 1 , a 2 , a 3. To determine if b is a linear combination of these columns, use the boxed fact on page 34. Row reduce the augmented matrix until you reach echelon form: 1 4 2 3 1 4 2 3 0 3 5 7 0 3 5 7 2 8 4 3 0 0 0 3
− − − − − − −
:
The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns of A. 19. By inspection, v 2 = (3/2) v 1. Any linear combination of v 1 and v 2 is actually just a multiple of v 1. For instance, a v 1 + b v 2 = a v 1 + b (3/2) v 1 = ( a + 3 b /2) v 1 So Span{ v 1 , v 2 } is the set of points on the line through v 1 and 0.
Warning : Although Figures 8 and 11 provide the most common ways to view Span{ u , v },
don’t forget Exercise 19, which shows that in a special case, Span{ u , v } can be just a line through the origin. In fact, Span{ u , v } can also be just the origin itself. How?
21. Let y =
h k
. Then [ u v y ] =
h h k k h
:. This augmented matrix
corresponds to a consistent system for all h and k. So y is in Span{ u , v } for all h and k. 23. a. The alternative notation for a (column) vector is discussed after Example 1. b. Plot the points to check the assertion. Or, see the statement preceding Example 3. c. See the line displayed just before Example 4. d. See the box that discusses the matrix in (5). e. Read the geometric description of Span{ u , v } very carefully.
Study Tip : I urge my own students to work by themselves on the true/false questions and then
meet together in groups of two or three, to compare and discuss their answers.
25. a. There are only three vectors in the set { a l, a 2 , a 3 }, and b is not one of them. b. There are infinitely many vectors in W = Span{ a l, a 2 , a 3 }. To determine if b is in W , use the method of Exercise 13.
1 2 3
−^ −^ −
a a a b
1-14 CHAPTER 1 l^ Linear Equations in Linear Algebra
The system for this augmented matrix is consistent, so b is in W. c. a 1 = 1 a 1 + 0 a 2 + 0 a 3. See the discussion following the definition of Span{ v 1 ,.. ., v p }.
31. a. The center of mass is
⋅^ + ⋅^ +^ ⋅^ =
^ ^ ^ ^
b. The total mass of the new system is 9 grams. The three masses added, w 1 , w 2 , and w 3 , satisfy the equation
( 1 ) ( 2 ) ( 3 )
w w w
+^ ⋅^ +^ +^ ⋅^ +^ +^ ⋅^ =
which can be rearranged to
( 1 ) ( 2 ) ( 3 )
w w w
and
1 2 3
w w w
The condition w 1 + w 2 + w 3 = 6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown below, along with a sequence of row operations: 1 1 1 6 1 1 1 6 1 1 1 6 0 8 2 8 0 8 2 8 0 8 2 8 1 1 4 12 0 0 3 6 0 0 1 2
Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4).
