TEST BANK FOR Introduction to Linear Algebra with Applications By James DeFranza, Daniel Gagliardi

,1.1 Systems of Linear Equations 1




Solutions to All Exercises

Systems of Linear Equations and
1 Matrices

Exercise Set 1.1

In Section 1.1 of the text, Gaussian Elimination is used to solve a linear system. This procedure utilizes
three operations that when applied to a linear system result in a new system that is equivalent to the original.
Equivalent means that the linear systems have the same solutions. The three operations are:
• Interchange two equations.
• Multiply any equation by a nonzero constant.
• Add a multiple of one equation to another.
When used judiciously these three operations allow us to reduce a linear system to a triangular linear system,
which can be solved. A linear system is consistent if there is at least one solution and is inconsistent if there
are no solutions. Every linear system has either a unique solution, infinitely many solutions or no solutions.
For example, the triangular linear systems
  
x1 − x2 + x3
 =2 x1 − 2x2 + x3
 =2 2x1
 + x3 = 1
x2 − 2x3 = −1 , −x2 + 2x3 = −3 , x2 − x3 = 2

 
 

x3 = 2 0 =4

have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system,
the variable x3 is a free variable, and once assigned any real number the values of x1 and x2 are determined.
In this way the linear system has infinitely many solutions. If a linear system has the same form as the second
system, but also has the additional equation 0 = 0, then the linear system will still have free variables. The
third system is inconsistent since the last equation 0 = 4 is impossible. In some cases, the conditions on the
right hand side of a linear system are not specified. Consider for example, the linear system
 
−x1 − x2
 =a −x1 − x2
 =a
2x1 + 2x2 + x3 = b which is equivalent to x3 = b + 2a .

 −−−−−−−−−−−−−−−−→  
2x3 = c 0 = c − 2b − 4a

This linear system is consistent only for values a, b and c such that c − 2b − 4a = 0.

Solutions to Exercises
1. Applying the given operations we obtain the equivalent triangular system
 
 x1 − x2 − 2x3
 =3  x1 − x2 − 2x3
 =3
−x1 + 2x2 + 3x3 = 1 E1 + E2 → E2 x2 + x3 = 4 (−2)E1 + E3 → E3

 −−−−−−−−−−→ 
 −−−−−−−−−−−−−−→
2x1 − 2x2 − 2x3 = −2 2x1 − 2x2 − 2x3 = −2

,2 Chapter 1 Systems of Linear Equations and Matrices

 x1 − x2 − 2x3
 =3
x2 + x3 = 4 . Using back substitution, the linear system has the unique solution


2x3 = −8

x1 = 3, x2 = 8, x3 = −4.

2. Applying the given operations we obtain the equivalent triangular system
 
2x1 − 2x2 − x3 = −3
 x1 − 3x2 + x3
 = −2
x1 − 3x2 + x3 = −2 E1 ↔ E2 2x1 − 2x2 − x3 = −3 (−2)E1 + E2 → E2

 −−−−−−→ 
 −−−−−−−−−−−−−−→
x1 − 2x2 =2 x1 − 2x2 =2
 
x1 − 3x2 + x3
 = −2 x1 − 3x2 + x3
 = −2
4x2 − 3x3 = 1 (−1)E1 + E3 → E3 4x2 − 3x3 = 1 E2 ↔ E3

 −−−−−−−−−−−−−−→   −−−−−−→
x1 − 2x2 =2 x2 − x3 =4
 
x1 − 3x2 + x3
 = −2 x1 − 3x2 + x3 = −2

x2 − x3 = 4 (−4)E2 + E3 → E3 x2 − x3 = 4 .

 −−−−−−−−−−−−−−→ 
4x2 − 3x3 = 1 x3 = −15
Using back substitution, the linear system has the unique solution x1 = −20, x2 = −11, x3 = −15.
3. Applying the given operations we obtain the equivalent triangular system
 

 x1 + 3x4 =2 
 x1 + 3x4 =2

x + x 

1 2 + 4x4 =3 x2 + x4 =1
(−1)E1 + E2 → E2

 2x 1 + x 3 + 8x4 = 3 −−−−−−−−−−−−−−→   2x 1 + x3 + 8x 4 = 3

 

x1 + x2 + x3 + 6x4 =2 x1 + x2 + x3 + 6x4 = 2


 x1 + 3x4 =2

 x2 + x4 =1
(−2)E1 + E3 → E3 (−1)E1 + E4 → E4
−−−−−−−−−−−−−−→  +x 3 + 2x 4 = −1 −−−−−−−−−−−−−−→


x1 + x2 + x3 + 6x4 = 2
 

 x1 + 3x4 = 2 
x1 + 3x4 = 2

 

x2 + x4 = 1 x2 + x4 = 1
(−1)E2 + E4 → E4

 +x3 + 2x4 = −1 −−−− −− −−−−− −− −→ 
 x3 + 2x4 = −1

 

x2 + x3 + 3x4 = 0 x3 + 2x4 = −1


 x1 + 3x4 = 2

 x2 + x4 = 1
(−1)E3 + E4 → E4 .
−−−−−−−−−−−−−−→   x3 + 2x4 = −1


0 =0
The final triangular linear system has more variables than equations, that is, there is a free variable. As a
result there are infinitely many solutions. Specifically, using back substitution, the solutions are given by
x1 = 2 − 3x4 , x2 = 1 − x4 , x3 = −1 − 2x4 , x4 ∈ R.
4. Applying the given operations we obtain the equivalent triangular system

 
x1
 + x3 = −2 x1
 + x3 = −2
x1 + x2 + 4x3 = −1 (−1)E1 + E2 → E2 x2 + 3x3 = 1 (−2)E1 + E3 → E3

 −−−−−−−−−−−−−−→  −−−−−−−−−−−−−−→
2x1 + 2x3 + x4 = −1 2x1 + 2x3 + x4 = −1

, 1.1 Systems of Linear Equations 3

x1
 + x3 = −2
x2 + 3x3 = 1 . Using back substitution, the set of solutions is given by x1 = −2 − x3 , x2 = 1 −


x4 = 3
3x3 , x4 = 3, x3 ∈ R.
5. The second equation gives immediately that x = 0. Substituting the value x = 0 into the first equation,
we have that y = − 23 . Hence, the linear system has the unique solution x = 0, y = − 23 .
6. From the second equation y = 1 and substituting this value in the first equation gives x = −4.
7. The first equation gives x = 1 and substituting this value in the second equation, we have y = 0. Hence,
the linear system has the unique solution x = 1, y = 0.
8. The operation 3E2 + E1 → E1 gives 5x = −1, so x = − 15 . Substitution in equation two, then gives y = − 15 .
9. Notice that the first equation is three times the second and hence, the equations have the same solutions.
Since each
 equation
has infinitely many solutions the linear system has infinitely many solutions with solution
2t+4
set S = 3 ,t t∈R .
10. Since the first equation is −3 times the second, the equations describe the same line and hence, there are
infinitely many solutions, given by x = 53 y + 13 , y ∈ R.
11. The operations E1 ↔ E3 , E1 + E2 → E2 , 3E1 + E3 → E3 and − 85 E2 + E3 → E3 , reduce the linear system
to the equivalent triangular system

 x − 2y + z
 = −2
−5y + 2z = −5 .

 9
5z =0
The unique solution is x = 0, y = 1, z = 0.
12. Reducing the linear system gives

  

 x + 3y + z =2 
 x + 3y + z =2 
 x + 3y + z =2
−2x + 2y − 4z = −1 reduces to
−−−−−−−→  8y − 2z = 3 reduces to
−−−−−−−→  8y − 2z =3 .

  
−y + 3z =1 −y + 3z =1 z = 12

So the unique solution is x = 0, y = 21 , z = 12 .
13. The operations E1 ↔ E2 , 2E1 + E2 → E2 , −3E1 + E3 → E3 and E2 + E3 → E3 , reduce the linear system
to the equivalent triangular system

 x
 + 5z = −1
−2y + 12z = −1 .


0 =0


The linear system has infinitely many solutions with solution set S = −1 − 5t, 6t + 12 , t t ∈ R .
14. Reducing the linear system gives

  
−x + y + 4z
 = −1 −x + y + 4z
 = −1 −x + y + 4z
 = −1
3x − y + 2z = 2 reduces to
−−−−−−−→ 2y + 14z = −1 reduces
−−−−−−−→to 2y + 14z = −1 .

 
 

2x − 2y − 8z =2 2x − 2y − 8z =2 0 =0


There are infinitely many solutions with solution set S = −3t + 12 , −7t − 12 , t | t ∈ R .
15. Adding the two equations yields 6x1 + 6x3 = 4, so that x1 = 32 − x3 . Substituting this value
1
in the
 first equation
gives x2 = − 2 . The linear system has infinitely many solutions with solution set
2 1
S = −t + 3 , − 2 , t t ∈ R .
16. Reducing the linear system gives