Page 839 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 819

⎛ ⎞ ⎛ ⎞
1 0 0 −8 −2 38
1 ⎝ 37
7. A R = ⎝ 0 1 0 ⎠ ; = 270 43 −7 ⎠
0 0 1 19 −29 11

1 0 0 0 0 1
9. A R = ; =
0 1 3/2 1/2 1/2 1/2
⎛ ⎞ ⎛ ⎞
1 0 0 0 1/2 −1
11. A R = ⎝ 0 1 0 ⎠ ; = ⎝ 0 0 1 ⎠
0 0 1 −1/7 2/7 −3/7
Section 7.4 Row and Column Spaces
1. (a)

1 0 −3/5
A R = ,
0 1 3/5
so A has rank 2. (b) < −4,1,3 > and < 2,2,0 > form a basis for the row space. (c) The column vectors

−4 and 1
2 2
form a basis for the column space of A.
3. (a)
⎛ ⎞
1 0
A R = ⎝ 0 1 ⎠ ,
0 0
so rank(A) = 2. (b) The row vectors < −3,1 > and < 2,2 > form a basis for the row space. The column space has
basis consisting of
⎛ ⎞ ⎛ ⎞
−3 1
⎝ 2 ⎠ and ⎝ 2 ⎠ .
4 −3
5. (a)

1 0 −1/4 1/2
A R = ,
0 1 −5/4 1/2
so rank(A) = 2. (b) The row vectors < 8,−4,3,2 > and < 1,−1,1,0 > form a basis for the row space. (c) The
column space has basis consisting of

8 and −4 .
1 −1
7. (a)
1 0 0
⎛ ⎞
⎜0 1 0⎟
0 0 1 ⎠ ,
A R = ⎝
0 0 0
so A has rank 3. (b) The row vectors < 2,2,1 >, < 1,−1,3 >,and < 0,0,1 > form a basis for the row space. (c) The
column space has a basis consisting of
2 2 1
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎠ , and ⎝ ⎠ .
⎜1⎟ ⎜−1⎟ ⎜3⎟
0 0 1
⎝ ⎠ , ⎝
4 0 7
9. (a)
⎛ ⎞
1 0 0
R = ⎝ 0 1 0 ⎠ ,
0 0 1

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October 14, 2010 17:50 THM/NEIL Page-819 27410_25_Ans_p801-866