Instructors-solution-manual-to-Introduction-to-Linear-Algebra-with-Applications-James-DeFranza-Daniel-Gagliardi-z-lib.org (1)

1.1 Systems of Linear Equations 1
Solutions to All Exercises
1
Systems of Linear Equations and
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
x2−2x3=−1
x3= 2
,
x1−2x2+x3= 2
−x2+ 2x3=−3,
2x1+x3= 1
x2−x3= 2
0 = 4
have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system,
the variable x3is a free variable, and once assigned any real number the values of x1and x2are 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
2x1+ 2x2+x3=b
2x3=c
which is equivalent to
−−−−−−−−−−−−−−−−→
−x1−x2=a
x3=b+ 2a
0 = c−2b−4a
.
This linear system is consistent only for values a, b and csuch that c−2b−4a= 0.
Solutions to Exercises
1. Applying the given operations we obtain the equivalent triangular system
x1−x2−2x3= 3
−x1+ 2x2+ 3x3= 1
2x1−2x2−2x3=−2
E1+E2→E2
−−−−−−−−−−→
x1−x2−2x3= 3
x2+x3= 4
2x1−2x2−2x3=−2
(−2)E1+E3→E3
−−−−−−−−−−−−−−→

2Chapter 1Systems of Linear Equations and Matrices
x1−x2−2x3= 3
x2+x3= 4
2x3=−8
.Using back substitution, the linear system has the unique solution
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
x1−2x2= 2
E1↔E2
−−−−−−→
x1−3x2+x3=−2
2x1−2x2−x3=−3
x1−2x2= 2
(−2)E1+E2→E2
−−−−−−−−−−−−−−→
x1−3x2+x3=−2
4x2−3x3= 1
x1−2x2= 2
(−1)E1+E3→E3
−−−−−−−−−−−−−−→
x1−3x2+x3=−2
4x2−3x3= 1
x2−x3= 4
E2↔E3
−−−−−−→
x1−3x2+x3=−2
x2−x3= 4
4x2−3x3= 1
(−4)E2+E3→E3
−−−−−−−−−−−−−−→
x1−3x2+x3=−2
x2−x3= 4
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+x2+ 4x4= 3
2x1+x3+ 8x4= 3
x1+x2+x3+ 6x4= 2
(−1)E1+E2→E2
−−−−−−−−−−−−−−→
x1+ 3x4= 2
x2+x4= 1
2x1+x3+ 8x4= 3
x1+x2+x3+ 6x4= 2
(−2)E1+E3→E3
−−−−−−−−−−−−−−→
x1+ 3x4= 2
x2+x4= 1
+x3+ 2x4=−1
x1+x2+x3+ 6x4= 2
(−1)E1+E4→E4
−−−−−−−−−−−−−−→
x1+ 3x4= 2
x2+x4= 1
+x3+ 2x4=−1
x2+x3+ 3x4= 0
(−1)E2+E4→E4
−−−−−−−−−−−−−−→
x1+ 3x4= 2
x2+x4= 1
x3+ 2x4=−1
x3+ 2x4=−1
(−1)E3+E4→E4
−−−−−−−−−−−−−−→
x1+ 3x4= 2
x2+x4= 1
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+x2+ 4x3=−1
2x1+ 2x3+x4=−1
(−1)E1+E2→E2
−−−−−−−−−−−−−−→
x1+x3=−2
x2+ 3x3= 1
2x1+ 2x3+x4=−1
(−2)E1+E3→E3
−−−−−−−−−−−−−−→

1.1 Systems of Linear Equations 3
x1+x3=−2
x2+ 3x3= 1
x4= 3
.Using back substitution, the set of solutions is given by x1=−2−x3, x2= 1 −
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=−2
3.Hence, the linear system has the unique solution x= 0, y =−2 3.
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→E1gives 5x=−1,so x=−1
5.Substitution in equation two, then gives y=−1 5.
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
set S=2t+4
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=5
3y+1 3, y ∈R.
11. The operations E1↔E3, E1+E2→E2,3E1+E3→E3and −8
5E2+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
−2x+ 2y−4z=−1
−y+ 3z= 1
reduces to
−−−−−−−→
x+ 3y+z= 2
8y−2z= 3
−y+ 3z= 1
reduces to
−−−−−−−→
x+ 3y+z= 2
8y−2z= 3
z=1
2
.
So the unique solution is x= 0, y =1
2, z =1 2.
13. The operations E1↔E2,2E1+E2→E2,−3E1+E3→E3and 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+1
2, tt∈R.
14. Reducing the linear system gives
−x+y+ 4z=−1
3x−y+ 2z= 2
2x−2y−8z= 2
reduces to
−−−−−−−→
−x+y+ 4z=−1
2y+ 14z=−1
2x−2y−8z= 2
reduces to
−−−−−−−→
−x+y+ 4z=−1
2y+ 14z=−1
0 = 0
.
There are infinitely many solutions with solution set S=−3t+1
2,−7t−1 2, t|t∈R.
15. Adding the two equations yields 6x1+ 6x3= 4,so that x1=2
3−x3.Substituting this value
in the first equation gives x2=−1
2.The linear system has infinitely many solutions with solution set
S=−t+2
3,−1 2, tt∈R.
16. Reducing the linear system gives

4Chapter 1Systems of Linear Equations and Matrices
(−2x1+x2= 2
3x1−x2+ 2x3= 1 reduces to
−−−−−−−→(−2x1+x2= 2
x2+ 4x3= 8 .
There are infinitely many solutions with solution set S={(−2t+ 3,−4t+ 8, t)|t∈R}.
17. The operation 2E1+E2→E2gives the equation −3x2−3x3−4x4=−9.Hence, the linear system has two
free variables, x3and x4.The two parameter set of solutions is S=3−5
3t, −s−4 3t+ 3, s, ts, t ∈R.
18. The linear system is in reduced form. The solution set is a two parameter family given by
S=1
2s−3t+5 2,3t−2, s, t|s, t ∈R.
19. The operation −2E1+E2→E2gives x=b−2a. Then y=a+ 2x=a+ 2(b−2a) = 2b−3a, so that
the unique solution is x=−2a+b, y =−3a+ 2b.
20. The linear system
(2x+ 3y=a
x+y=breduces to
−−−−−−−→(2x+ 3y=a
y=a−2b,
so the solution is x=−a+ 3b, y =a−2b.
21. The linear system is equivalent to the triangular linear system
−x−z=b
y=a+ 3b
z=c−7b−2a
,
which has the unique solution x= 2a+ 6b−c, y =a+ 3b, z =−2a−7b+c.
22. The linear system
−3x+ 2y+z=a
x−y−z=b
x−y−2z=c−E2+E3→E3, E1+ 3E2→E1reduces to
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
−y−2z=a+ 3b
x−y−z=b
−z=−b+c
,
so the solution is x=−a−3b+c, y =−a−5b+ 2c, z =b−c.
23. Since the operation 2E1+E2→E2gives the equation 0 = 2a+ 2,then the linear system is consistent
for a=−1.
24. Since
(−x+ 3y=a
2x−6y= 3 reduces to
−−−−−−−→(−x+ 3y=a
0 = 3 + 2a,
the linear system is consistent if a=−3
2.
25. Since the operation 2E1+E2→E2gives the equation 0 = a+b, then the linear system is consistent for
b=−a.
26. Since
(6x−3y=a
−2x+y=breduces to
−−−−−−−→(6x−3y=a
0 = 1
3a+b,
the linear system is consistent if b=−1
3a.

1.1 Systems of Linear Equations 5
27. The linear system is equivalent to the triangular linear system
x−2y+ 4z=a
5y−9z=−2a+b
0 = c−a−b
and hence, is consistent for all a, b, and csuch that c−a−b= 0.
28. Since
x−y+ 2z=a
2x+ 4y−3z=b
4x+ 2y+z=c
reduces to
−−−−−−−→
x−y+ 2z=a
6y−7z=b−2a
0 = c−2a−b
,
the linear system is consistent if c−2a−b= 0.
29. The operation −2E1+E2→E2gives the equivalent linear system
(x+y=−2
(a−2)y= 7 .
Hence, if a= 2,the linear system is inconsistent.
30. Since
(2x−y= 4
ax + 3y= 2 reduces to
−−−−−−−→(2x−y= 4
3 + 1
2a= 2 −2a,
the linear system is inconsistent if 3 + 1
2a= 0,that is a=−6.Notice that if a=−6,then 2 −2a6= 0.
31. The operation −3E1+E2→E2gives the equivalent linear system
(x−y= 2
0 = a−6.
Hence, the linear system is inconsistent for all a6= 6.
32. Since
(2x−y=a
6x−3y=areduces to
−−−−−−−→(2x−y=a
0 = −2a
the linear system is inconsistent for a6= 0.
33. To find the parabola y=ax2+bx +cthat passes through the specified points we solve the linear system
c= 0.25
a+b+c=−1.75
a−b+c= 4.25
.
The unique solution is a= 1, b =−3,and c=1
4,so the parabola is y=x2−3x+1 4=x−3 22−2.The
vertex of the parabola is the point 3
2,−2.
34. To find the parabola y=ax2+bx +cthat passes through the specified points we solve the linear system
c= 2
9a−3b+c=−1
0.25a+ 0.5b+c= 0.75
.