Solution Manual Linear Algebra Complete Solution

SOLUTION MANUAL LINEAR ALGEBRA COMPLETE SOLUTION

,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 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 x 2 + x3 =4 (−2)E1 + E3 → E3
−−−−−−−−−→ −−−−−−−−−−−−−→
2x1 — 2x2 — 2x3 = −2 2x 1 — 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


x1 1− −3x
2x 2x2+− xx3 = −3
= −2 E1 ↔ E2 x1 − 3x2 + x3 = −2
2 3
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 E ↔ E
−−−−−−−−−−−→ −−−−2→
−
x1 − 2x2 =2 x2 − x3 =4 3


x1 − 3x2 + x3 = −2 x1 − 3x2 + x3 = −2
x2 − x3 =4 ( — 4)E 2 + E 3 → E 3 x 2 − x 3 = 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 1 + x2 + 4x4 =3 x2 + x4 =1
(−1)E1 + E2 → E2
2x1 + x3 + 8x4 = 3 −−−−−−−−−−−−−→ 2x1 + x3 + 8x4 =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
−−−−−−−−−−−−−→ +x3 + 2x4 = −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 + x 3 4 +3x x3 + 2x4 = −1
=0 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
—4, x2 = 1 x4, x−3 = 1 2x4, −x4 −R.
x1 = 2 3x ∈
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 = 3− 2 .
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 = − 1 . Substitution
5
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 equatio.n has in}finitely many solutions the linear system has infinitely many solutions with solution
,t . t ∈ R .
2t+4
set S = 3
10. Since the first equation is −3 times the second, the equations describe the same line and hence, there are
5 1
infinitely many solutions, given by x = 3 y + 3, y ∈ R.
11. The operations E1 ↔ E3, E1 + E2 → E2, 3E1 + E3 → E3 and − 8 E52 + E3 → E3, reduce the linear system
to the equivalent triangular system

x − 2y + z = −2
−5y + 2z = −5 .
9
5
z =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 r−ed 8y − 2z 8y − 2z =3 .
−u
−c−e−s−→
to = 3 −r−ed
−u
−c−es−−→
to
1
−y + 3z =1 −y + 3z =1 z = 2
1 1
So the unique solution is x = 0, y = , z = .
2 2
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
1
. }
The linear system has infinitely many solutions with solution set S = .
−1 − 5t, 6t + 2 , 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 r−ed
−u
−c−e−s−→
2y + 14z = −1 .
to
2x − 2y − 8z =2 2x − 2y − 8z =2 0 =0
1 }
There are infinitely many solutions with solution set S = −3t + 2, −7t − 21 ,2t | t ∈ R .
15. Adding the two equations 1yields 6x1 + 6x3 = 4, so that x1 = − x3. Substituting this value
in the first equation gives x = − 3

2 1 . }2
S= −t + 3, − ,2t . t ∈ R . 2 . The linear system has infinitely many solutions with solution set
16. Reducing the linear system gives