Solution Manual for Elementary Linear Algebra: Applications Version 12th Edition by Howard Anton, Chris Rorres , Anton Kaul

Elementary
Linear Algebra
Solution

Howard Anton
12th edition

, Table of Contents.

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, 1.1 Introduction to Systems of Linear Equations 1




1.1 Introduction to Systems of Linear Equations

1. (a) This is a linear equation in x1 , x2 , and x3 .

(b) This is not a linear equation in x1 , x2 , and x3 because of the term x1 x3 .

(c) We can rewrite this equation in the form x1  7 x2  3 x3  0 therefore it is a linear equation in x1 , x2 , and x3 .

(d) This is not a linear equation in x1 , x2 , and x3 because of the term x12 .

(e) This is not a linear equation in x1 , x2 , and x3 because of the term x13/5 .

(f) This is a linear equation in x1 , x2 , and x3 .

2. (a) This is a linear equation in x and y .

(b) This is not a linear equation in x and y because of the terms 2x1/3 and 3 y .

(c) This is a linear equation in x and y .

(d) This is not a linear equation in x and y because of the term 7
cos x .

(e) This is not a linear equation in x and y because of the term xy .

(f) We can rewrite this equation in the form  x  y  7 thus it is a linear equation in x and y .

3. (a) a11 x1  a12 x2  b1
a21 x1  a22 x2  b2

(b) a11 x1  a12 x2  a13 x3  b1
a21 x1  a22 x2  a23 x3  b2
a31 x1  a32 x2  a33 x3  b3

(c) a11 x1  a12 x2  a13 x3  a14 x4  b1
a21 x1  a22 x2  a23 x3  a24 x4  b2



4. (a) (b) (c)
 a11 a12 b1   a11 a12 a13 b1   a11 a12 a13 a14 b1 
a b2  a
 21 a22
 21 a22 a23 b2  a
 21 a22 a23 a24 b2 
 a31 a32 a33 b3 

, 1.1 Introduction to Systems of Linear Equations 2


5. (a) (b)
2 x1  0 3 x1  2 x3  5
3 x1  4 x2  0 7 x1  x2  4 x3  3
x2  1  2 x2  x3  7


6. (a) (b)
3 x2  x3  x4  1 3 x1  x3  4 x4  3
5 x1  2 x2  3 x4  6 4 x1  4 x3  x4  3
 x1  3 x2  2 x4  9
 x4  2


7. (a) (b) (c)
 2 6  6 1 3 4   0 2 0 3 1 0
 3 8 0  3 1 1 0 0 1
   5 1 1   
 9 3   6 2 1 2 3 6 



8. (a) (b) (c)
 3 2 1  2 0 2 1 1 0 0 1 
4  3 1 4 7  0 1 0 2 
 5 3    
 7 3 2   6 1 1 0  0 0 1 3 



9. The values in (a), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (b) and (c) are not solutions of the system.
10. The values in (b), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (a) and (c) are not solutions of the system.
11. (a) We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system

3x  2 y  4
0  1

The second equation is contradictory, so the original system has no solutions. The lines represented by the
equations in that system have no points of intersection (the lines are parallel and distinct).
(b) We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system

2x  4y  1
0  0