# (MT103) Linear Algebra And Its Applications 4th Edition

- Description
- Full Document

Linear Algebra And Its Applications 4th Edition Textbook Solutions 1-

- UniversityNational University of Computer and Emerging Sciences
- CourseLinear Algebra (MT103)

15. a. False. See the definition of AB.

b. False. The roles of A and B should be reversed in the second half of the statement. See the box

after Example 3.

c. True. See Theorem 2(b), read right to left.

d. True. See Theorem 3(b), read right to left.

e. False. The phrase “in the same order” should be “in the reverse order.” See the box after Theorem

3.

16. a. True. See the box after Example 4.

b. False. AB must be a 3×3 matrix, but the formula given here implies that it is a 3×1 matrix. The

plus signs should just be spaces (between columns). This is a common mistake.

c. True. Apply Theorem 3(d) to A2=AA

d. False. The left-to-right order of (ABC)T, is CTBTAT. The order cannot be changed in general.

e. True. This general statement follows from Theorem 3(b).

17. Since [ 1 2 ]

3 11

Note: An alternative solution of Exercise 17 is to row reduce [A Ab1 Ab2] with one sequence of row

operations. This observation can prepare the way for the inversion algorithm in Section 2.2.

18. The third column of AB is also all zeros because Ab3 = A0 = 0

19. (A solution is in the text). Write B = [b1 b2 b3]. By definition, the third column of AB is Ab3. By

hypothesis, b3 = b1 + b2. So Ab3 = A(b1 + b2) = Ab1 + Ab2, by a property of matrix-vector

multiplication. Thus, the third column of AB is the sum of the first two columns of AB.

20. The first two columns of AB are Ab1 and Ab2. They are equal since b1 and b2 are equal.

21. Let bp be the last column of B. By hypothesis, the last column of AB is zero. Thus, Abp = 0.

However, bp is not the zero vector, because B has no column of zeros. Thus, the equation Abp = 0 is a

linear dependence relation among the columns of A, and so the columns of A are linearly dependent.

Note: The text answer for Exercise 21 is, “The columns of A are linearly dependent. Why?” The Study

Guide supplies the argument above, in case a student needs help.

22. If the columns of B are linearly dependent, then there exists a nonzero vector x such that Bx = 0.

From this, A(Bx) = A0 and (AB)x = 0 (by associativity). Since x is nonzero, the columns of AB must

be linearly dependent.

23. If x satisfies Ax = 0, then CAx = C0 = 0 and so Inx = 0 and x = 0. This shows that the equation Ax = 0

has no free variables. So every variable is a basic variable and every column of A is a pivot column.

(A variation of this argument could be made using linear independence and Exercise 30 in Section

1.7.) Since each pivot is in a different row, A must have at least as many rows as columns.

2.1 • Solutions 91

Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.

24. Write I3 =[e1 e2 e3] and D = [d1 d2 d3]. By definition of AD, the equation AD = I3 is equivalent

|to the three equations Ad1 = e1, Ad2 = e2, and Ad3 = e3. Each of these equations has at least one

solution because the columns of A span R3. (See Theorem 4 in Section 1.4.) Select one solution of

each equation and use them for the columns of D. Then AD = I3.

25. By Exercise 23, the equation CA = In implies that (number of rows in A) > (number of columns), that

is, m > n. By Exercise 24, the equation AD = Im implies that (number of rows in A) < (number of

columns), that is, m < n. Thus m = n. To prove the second statement, observe that CAD = (CA)D =

InD = D, and also CAD = C(AD) = CIm = C. Thus C = D. A shorter calculation is

C =C In = C(AD) = (CA)D = In D= D

26. Take any b in Rm. By hypothesis, ADb = Imb = b. Rewrite this equation as A(Db) = b. Thus, the

vector x = Db satisfies Ax = b. This proves that the equation Ax = b has a solution for each b in Rm.

By Theorem 4 in Section 1.4, A has a pivot position in each row. Since each pivot is in a different

column, A must have at least as many columns as rows.

27. The product uTv is a 1×1 matrix, which usually is identified with a real number and is written

without the matrix brackets.

**PREVIEW**

**NOTE: Please check the details before purchasing the document.**