Math 54. Selected Solutions for Week 2 Section 1.4 (Page 42)

Math 54. Selected Solutions for Week 2

Section 1.4 (Page 42)

0

3 -5

13. Let u = 4 and A = -2 6 . Is u in the plane in R3 spanned by the columns

4

11

of A ? (See the figure [omitted].) Why or why not?

3 -5

First of all, the plane in R3 is just the set Span -2 , 6 , so the question

1

1

is asking whether or not u lies in that set.

As in previous exercises, this leads to a linear system whose augmented matrix is

reduced to echelon form as:

3 -5 0 3 -5 0 3 -5 0

-2

6

4 0

8 3

4 0

8 3

4 .

1 14

0

8 3

4

000

The system is consistent (since the last column is not a pivot column), so u does lie in the given plane.

33. Suppose A is a 4 ? 3 matrix and b is a vector in R4 with the property that Ax = b has a unique solution. What can you say about the reduced echelon form of A ? Justify your answer.

Since Ax = b has a unique solution, the associated linear system has no free

variables, and therefore all columns of A are pivot columns. So the reduced echelon

form of A must be

1 0 0

0 0

1 0

0

1

.

000

36. Suppose A is a 4 ? 4 matrix and b is a vector in R4 with the property that Ax = b has a unique solution. Explain why the columns of A must span R4 .

As in the previous exercise, since Ax = b has a unique solution, all columns of A must be pivot columns. Therefore there are four pivot columns, hence four pivot elements. These must all lie in different rows, so since there are four rows, all rows must contain a pivot element. By Theorem 4 on page 39, it follows that the columns of A span R4 .

1

2

Section 1.5 (Page 49)

21. Let p =

3 -3

and q =

4 1

. Find a parametric equation of the line M through p

and q . [Hint: M is parallel to the vector q - p . See the figure below [omitted].]

We have q - p =

1 4

.

The line containing this vector is Span{q - p} , and is

given in parametric form as

x=t

1 4

( t in R ) .

Therefore (as on page 47) the line through p and q is obtained by translating that line by p ; it is given in parametric form as

x=

3 -3

+t

1 4

( t in R ) .

(You could also use q in place of p .)

37. Construct a 2 ? 2 matrix A such that the solution set of the equation Ax = 0 is the line in R2 through (4, 1) and the origin. Then, find a vector b in R2 such that the solution set of Ax = b is not a line in R2 parallel to the solution set of Ax = 0 . Why does this not contradict Theorem 6?

We can find a homogeneous linear equation in (x1, x2) that has solution x1 = 4 , x2 = 1 ; it is x1 - 4x2 = 0 (or any nonzero scalar multiple of this equation). We need a linear system with two such equations, so we can just use this equation twice. The coefficient matrix of this linear system is our matrix A :

A=

1 1

-4 -4

.

For any vector x in R2 , the two entries of the product Ax must be the same. So, let

b=

0 1

.

Then the matrix equation Ax = b is inconsistent, because when you row reduce the matrix A b you find that the last column is a pivot column. The solution set of this matrix equation is empty, so it is not a line in R2 parallel to the solution set of Ax = 0 .

This does not contradict Theorem 6, because Theorem 6 applies only to consistent equations, and this system is not consistent.

3

38. Let A be an m ? n matrix and let w be a vector in Rn that satisfies the equation Ax = 0 . Show that for any scalar c , the vector cw also satisfies Ax = 0 . [That is, show that A(cw) = 0 .]

By Theorem 5(b) (page 41) and the fact that w satisfies Ax = 0 ,

A(cw) = c(Aw) = c0 = 0 .

Therefore cw satisfies the equation.

Section 1.7 (Page 62)

8. Determine if the columns of the matrix

1 -2 3 2 -2 4 -6 2

0 1 -1 3

form a linearly independent set. Justify your answer.

They are linearly dependent, by Theorem 8 on page 61 (the matrix has more columns than rows).

13. Find the value(s) of h for which the vectors

1 5 ,

-3

-2 -9 ,

6

3 h

-9

are linearly dependent. Justify your answer. Let us row reduce the matrix whose columns are these vectors:

1 -2 3 1 -2 3

5 -9 h 0 1 h - 15

-3 6 -9

00 0

The vectors are linearly dependent for all values of h , because the third column of the above matrix is never a pivot column.

(This is similar to Example 2 on page 59, except that we left out the last column of the augmented matrix, since it is always zero and therefore does not affect the process of row reduction (it is never a pivot column).)

4

36. The following statement is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If the statement is true, give a justification. (One specific example cannot explain why a statement is always true.) If v1 , v2 , v3 are in R3 and v3 is not a linear combination of v1 and v2 , then {v1, v2, v3} is linearly independent.

0

1

Take v1 = 0 , v2 = 1 , and v3 = 0 . This (indexed) set is linearly dependent

0

0

(because the first vector is zero), but one cannot write v3 as a linear combination of v1

and v2 , because the first coordinates of v1 and v2 are zero, but the first coordinate of v3 is not zero.

40. Suppose an m ? n matrix A has n pivot columns. Explain why for each b in Rm the equation Ax = b has at most one solution. [Hint: Explain why Ax = b cannot have infinitely many solutions.

The matrix A has n pivot columns, which is equal to its number of columns. Therefore every matrix of A is a pivot column. Therefore, in an augmented matrix

A b , all columns except for possibly the last one will be pivot columns since the pivots of the " A part" of this matrix are the same as the pivots of A . So the equation Ax = b cannot have infinitely many solutions (regardless of what b is).

Section 1.8 (Page 70)

10. Find all x in R4 that are mapped into the zero vector by the transformation x Ax

for the matrix

3 2 10 -6

A

=

1 0

0 1

2 2

-4

3

.

1 4 10 8

The question amounts to solving the matrix equation Ax = 0 , so we row reduce its augmented matrix to reduced echelon form:

3 2 10 -6 0 1 0 2 -4 0 1 0 2 -4 0

10 0 1

2 2

-4 3

0 0

3 0

2 1

10 2

-6 3

0 0

0 0

2 1

4 2

6 3

0

0

1 4 10 8 0

1 4 10 8 0

0 4 8 12 0

1 0 2 -4 0 1 0 2 -4 0

0 0

2 0

4 0

6 0

0 0

0 0

1 0

2 0

3 0

0

0

.

000 0 0

000 0 0

5

In parametric vector form, the solutions are therefore

-2 4

x

=

x3

-2 1

+

x4

-3 0

.

0

1

-1

12.

Let

b=

3 -1

,

and

let

A

be

the

matrix

in

Exercise

10.

Is

b

in the

range

of

the

linear

4

transformation x Ax ? Why or why not?

This leads to a linear system whose augmented matrix is (partially) row reduced as follows:

3 2 10 -6 -1 1 0 2 -4 3

10 0 1

2 2

-4 3

3 -1

3 0

2 1

10 2

-6 3

-1

-1

1 4 10 8 4

1 4 10 8 4

1 0 2 -4 3 1 0 2 -4 3

0 0

2 1

4 2

6 3

-10 -1

0 0

2 0

4 0

6 0

-10

4

.

0 4 8 12 1

0 0 0 0 21

At this point we can stop, because it is clear that the last column is a pivot column, so

the linear system is inconsistent. Therefore b is not in the range of x Ax .

(Another way to see this is to notice that if b is to equal the expression in the

answer to Exercise 10, then x3 must be -1 and x4 must be 4 . But using those values 18

gives

x

=

-10 -1

,

which

is

not

b .)

4

20.

Let x =

x1 x2

, v1 =

-3 5

, and v2 =

7 -2

, and let T : R2 R2

be a linear

transformation that maps x into x1v1 + x2v2 . Find a matrix A such that T (x) is

Ax for each x .

We have

A = [ T (e1)

T (e2) ] = [ v1

v2 ] =

-3 5

7 -2

.

32. Show that the transformation T defined by T (x1, x2) = (x1 - 2|x2|, x1 - 4x2) is not linear. [In this exercise, column vectors are written as rows; for example, x = (x1, x2) and T (x) is written as T (x1, x2) .]

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