Linear Algebra and Its Applications 4th Edition 1.3 Solutions

Instructor Solutions Manual

for

Introduction to

Linear Algebra with Applications

Jim DeFranza St. Lawrence University

Contents

1 Systems of Linear Equations and Matrices
- Exercise Set 1.1 Systems of Linear Equations
- Exercise Set 1.2 Matrices and Elementary Row Operations
- Exercise Set 1.3 Matrix Algebra
- Exercise Set 1.4 The Inverse of a Matrix
- Exercise Set 1.5 Matrix Equations
- Exercise Set 1.6 Determinants
- Exercise Set 1.7 Elementary Matrices andLUFactorization
- Exercise Set 1.8 Applications of Systems of Linear Equations
- Review Exercises
- Chapter Test
2 Linear Combinations and Linear Independence
- Exercise Set 2.1 Vectors inRn
- Exercise Set 2.2 Linear Combinations
- Exercise Set 2.3 Linear Independence
- Review Exercises
- Chapter Test
3 Vector Spaces
- Exercise Set 3.1 Definition of a Vector Space
- Exercise Set 3.2 Subspaces
- Exercise Set 3.3 Basis and Dimension
- Exercise Set 3.4 Coordinates and Change of Basis
- Exercise Set 3.5 Application: Differential Equations
- Review Exercises
- Chapter Test
4 Linear Transformations
- Exercise Set 4.1 Linear Transformations
- Exercise Set 4.2 The Null Space and Range
- Exercise Set 4.3 Isomorphisms
- Exercise Set 4.4 Matrix Transformation of a Linear Transformation
- Exercise Set 4.5 Similarity
- Exercise Set 4.6 Application: Computer Graphics
- Review Exercises
- Chapter Test
5 Eigenvalues and Eigenvectors
- Exercise Set 5.1 Eigenvalues and Eigenvectors
- Exercise Set 5.2 Diagonalization
- Exercise Set 5.3 Application: Systems of Linear Differential Equations
- Exercise Set 5.4 Application: Markov Chains

1.1 Systems of Linear Equations 1

Solutions to All Exercises

1 Systems of Linear Equations and Matrices

Exercise Set 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 resultin 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 isat 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  



x 1 −x 2 +x 3 = 2 x 2 − 2 x 3 =− 1 x 3 = 2

,







x 1 − 2 x 2 +x 3 = 2 −x 2 + 2x 3 =− 3 ,







2 x 1 +x 3 = 1 x 2 −x 3 = 2 0 = 4

have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system, the variablex 3 is a free variable, and once assigned any real number the values ofx 1 andx 2 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. Considerfor example, the linear system

 



−x 1 −x 2 =a 2 x 1 + 2x 2 +x 3 =b 2 x 3 =c

which is equivalent to −−−−−−−−−−−−−−−−→







−x 1 −x 2 =a x 3 =b+ 2a 0 =c− 2 b− 4 a

.

This linear system is consistent only for valuesa, bandcsuch thatc− 2 b− 4 a= 0.

Solutions to Exercises

1.Applying the given operations we obtain the equivalent triangular system

 



x 1 −x 2 − 2 x 3 = 3 −x 1 + 2x 2 + 3x 3 = 1 2 x 1 − 2 x 2 − 2 x 3 =− 2

E 1 +E 2 →E 2

−−−−−−−−−−→







x 1 −x 2 − 2 x 3 = 3 x 2 +x 3 = 4 2 x 1 − 2 x 2 − 2 x 3 =− 2

(−2)E 1 +E 3 →E 3

−−−−−−−−−−−−−−→

2 Chapter 1 Systems of Linear Equations and Matrices

 



x 1 −x 2 − 2 x 3 = 3 x 2 +x 3 = 4 2 x 3 =− 8

.Using back substitution, the linear system has the unique solution

x 1 = 3, x 2 = 8, x 3 =− 4.

2.Applying the given operations we obtain the equivalent triangular system

 



2 x 1 − 2 x 2 −x 3 =− 3 x 1 − 3 x 2 +x 3 =− 2 x 1 − 2 x 2 = 2

E 1 ↔E 2

−−−−−−→







x 1 − 3 x 2 +x 3 =− 2 2 x 1 − 2 x 2 −x 3 =− 3 x 1 − 2 x 2 = 2

(−2)E 1 +E 2 →E 2

−−−−−−−−−−−−−−→







x 1 − 3 x 2 +x 3 =− 2 4 x 2 − 3 x 3 = 1 x 1 − 2 x 2 = 2

(−1)E 1 +E 3 →E 3

−−−−−−−−−−−−−−→







x 1 − 3 x 2 +x 3 =− 2 4 x 2 − 3 x 3 = 1 x 2 −x 3 = 4

E 2 ↔E 3

−−−−−−→







x 1 − 3 x 2 +x 3 =− 2 x 2 −x 3 = 4 4 x 2 − 3 x 3 = 1

(−4)E 2 +E 3 →E 3

−−−−−−−−−−−−−−→







x 1 − 3 x 2 +x 3 =− 2 x 2 −x 3 = 4 x 3 =− 15

.

Using back substitution, the linear system has the unique solutionx 1 =− 20 , x 2 =− 11 , x 3 =− 15. 3.Applying the given operations we obtain the equivalent triangular system

  

 

x 1 + 3x 4 = 2 x 1 +x 2 + 4x 4 = 3 2 x 1 +x 3 + 8x 4 = 3 x 1 +x 2 +x 3 + 6x 4 = 2

(−1)E 1 +E 2 →E 2

−−−−−−−−−−−−−−→











x 1 + 3x 4 = 2 x 2 +x 4 = 1 2 x 1 +x 3 + 8x 4 = 3 x 1 +x 2 +x 3 + 6x 4 = 2

(−2)E 1 +E 3 →E 3

−−−−−−−−−−−−−−→











x 1 + 3x 4 = 2 x 2 +x 4 = 1 +x 3 + 2x 4 =− 1 x 1 +x 2 +x 3 + 6x 4 = 2

(−1)E 1 +E 4 →E 4

−−−−−−−−−−−−−−→











x 1 + 3x 4 = 2 x 2 +x 4 = 1 +x 3 + 2x 4 =− 1 x 2 +x 3 + 3x 4 = 0

(−1)E 2 +E 4 →E 4

−−−−−−−−−−−−−−→











x 1 + 3x 4 = 2 x 2 +x 4 = 1 x 3 + 2x 4 =− 1 x 3 + 2x 4 =− 1

(−1)E 3 +E 4 →E 4

−−−−−−−−−−−−−−→











x 1 + 3x 4 = 2 x 2 +x 4 = 1 x 3 + 2x 4 =− 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 x 1 = 2− 3 x 4 , x 2 = 1−x 4 , x 3 =− 1 − 2 x 4 , x 4 ∈R. 4.Applying the given operations we obtain the equivalent triangular system







x 1 +x 3 =− 2 x 1 +x 2 + 4x 3 =− 1 2 x 1 + 2x 3 +x 4 =− 1

(−1)E 1 +E 2 →E 2

−−−−−−−−−−−−−−→







x 1 +x 3 =− 2 x 2 + 3x 3 = 1 2 x 1 + 2x 3 +x 4 =− 1

(−2)E 1 +E 3 →E 3

−−−−−−−−−−−−−−→

4 Chapter 1 Systems of Linear Equations and Matrices

{

− 2 x 1 +x 2 = 2 3 x 1 −x 2 + 2x 3 = 1

reduces to−−−−−−−→

{

− 2 x 1 +x 2 = 2 x 2 + 4x 3 = 8

.

There are infinitely many solutions with solution setS={(− 2 t+ 3,− 4 t+ 8, t)| t∈R}. 17.The operation 2E 1 +E 2 →E 2 gives the equation− 3 x 2 − 3 x 3 − 4 x 4 =− 9 .Hence, the linear system has two free variables,x 3 andx 4 .The two parameter set of solutions isS=

{(

3 − 53 t,−s− 43 t+ 3, s, t

)∣

∣ s, t∈R

}

.

18.The linear system is in reduced form. The solution set is a twoparameter family given by S=

{( 1

2 s− 3 t+

5 2 , 3 t− 2 , s, t

)

| s, t∈R

}

.

19.The operation− 2 E 1 +E 2 →E 2 givesx=b− 2 a.Theny=a+ 2x=a+ 2(b− 2 a) = 2b− 3 a,so that the unique solution isx=− 2 a+b, y=− 3 a+ 2b.

20.The linear system

{ 2 x+ 3y =a x +y =b

reduces to−−−−−−−→

{

2 x+ 3y =a y =a− 2 b

,

so the solution isx=−a+ 3b, y=a− 2 b. 21.The linear system is equivalent to the triangular linear system

 



−x −z =b y =a+ 3b z =c− 7 b− 2 a

,

which has the unique solutionx= 2a+ 6b−c, y=a+ 3b, z=− 2 a− 7 b+c. 22.The linear system







− 3 x+ 2y +z =a x −y −z =b x −y− 2 z =c

−E 2 +E 3 →E 3 , E 1 + 3E 2 →E 1 reduces to −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→







−y − 2 z =a+ 3b x −y −z =b −z =−b+c

,

so the solution isx=−a− 3 b+c, y=−a− 5 b+ 2c, z=b−c. 23.Since the operation 2E 1 +E 2 →E 2 gives the equation 0 = 2a+ 2,then the linear system is consistent fora=− 1. 24.Since

{ −x+ 3y =a 2 x− 6 y = 3

reduces to−−−−−−−→

{

−x+ 3y =a 0 = 3 + 2a

,

the linear system is consistent ifa=− 32. 25.Since the operation 2E 1 +E 2 →E 2 gives the equation 0 =a+b,then the linear system is consistent for b=−a. 26.Since

{ 6 x− 3 y =a − 2 x +y =b

reduces to−−−−−−−→

{

6 x− 3 y =a 0 = 13 a+b

,

the linear system is consistent ifb=− 13 a.

1.1 Systems of Linear Equations 5

27.The linear system is equivalent to the triangular linear system  



x− 2 y+ 4z =a 5 y− 9 z =− 2 a+b 0 =c−a−b

and hence, is consistent for alla, b,andcsuch thatc−a−b= 0.

28.Since

 



x −y + 2z =a 2 x + 4y − 3 z =b 4 x + 2y +z =c

reduces to−−−−−−−→







x −y + 2z =a 6 y − 7 z =b− 2 a 0 =c− 2 a−b

,

the linear system is consistent ifc− 2 a−b= 0. 29.The operation− 2 E 1 +E 2 →E 2 gives the equivalent linear system { x+y =− 2 (a−2)y = 7

.

Hence, ifa= 2,the linear system is inconsistent.

30.Since

{ 2 x−y = 4 ax+ 3y = 2

reduces to−−−−−−−→

{

2 x−y = 4 3 + 12 a = 2− 2 a

,

the linear system is inconsistent if 3 + 12 a= 0,that isa=− 6 .Notice that ifa=− 6 ,then 2− 2 a 6 = 0. 31.The operation− 3 E 1 +E 2 →E 2 gives the equivalent linear system { x−y = 2 0 =a− 6

.

Hence, the linear system is inconsistent for alla 6 = 6.

32.Since

{ 2 x−y =a 6 x− 3 y =a

reduces to−−−−−−−→

{

2 x−y =a 0 =− 2 a

the linear system is inconsistent fora 6 = 0. 33.To find the parabolay=ax 2 +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 isa= 1, b=− 3 ,andc= 14 ,so the parabola isy=x 2 − 3 x+ 14 =

(

x− 32

) 2

− 2 .The vertex of the parabola is the point

( 3

2 ,− 2

)

.

34.To find the parabolay=ax 2 +bx+cthat passes through the specified points we solve the linear system

 



c = 2 9 a− 3 b+c =− 1 0. 25 a+ 0. 5 b+c = 0. 75

.

1.2 Matrices and Elementary Row Operations 7

a.Letx 4 =s,andx 5 =t,so thatx 3 = 2 + 2s− 3 t, x 2 =−1 +s+t,andx 1 =−2 + 3t. b.Letx 3 =s andx 5 =t,so thatx 4 =−1 + 12 s+ 32 t, x 2 =−2 + 12 s+ 52 t,andx 1 =−2 + 3t.

43.ApplyingkE 1 →E 1 , 9 E 2 →E 2 ,and−E 1 +E 2 →E 2 gives the equivalent linear system { 9 kx+ k 2 y = 9k (9−k 2 )y =− 27 − 9 k

.

Whether the linear system is consistent or inconsistent cannow be determined by examining the second equation. a.Ifk= 3,the second equation becomes 0 =− 54 ,so the linear system is inconsistent.b.Ifk=− 3 ,then the second equation becomes 0 = 0,so the linear system has infinitely many solutions.c.Ifk 6 =± 3 ,then the linear system has a unique solution.

44.The linear system

 



kx+y+z = 0 x+ky+z = 0 x+y+kz = 0

reduces to−−−−−−−→







x+ ky+ z = 0 (1−k)y+ (k−1)z = 0 (2 +k)(1−k)z = 0

.

a. The linear system has a unique solution ifk 6 = 1 andk 6 =− 2 .b. Ifk=− 2 ,the solution set is a one parameter family.c.Ifk= 1,the solution set is a two parameter family.

Exercise Set 1.

Matrices are used to provide an alternative way to representa linear system. Reducing a linear system to triangular form is then equivalent to row reducing the augmented matrix corresponding to the linear system to a triangular matrix. For example, the augmented matrix for the linear system  



−x 1 −x 2 −x 3 − 2 x 4 = 1 2 x 1 + 2x 2 +x 3 − 2 x 4 = 2 x 1 − 2 x 2 +x 3 + 2x 4 =− 2





− 1 − 1 − 1 − 2 1

2 2 1 − 2 2

1 −2 1 2 − 2



.

The coefficient matrix is the 3×4 matrix consisting of the coefficients of each variable, thatis, the augmented

matrix with the augmented column





1

2 − 2



deleted. The first four columns of the augmented matrix cor-

respond to the variablesx 1 , x 2 , x 4 ,andx 4 ,respectively and the augmented column to the constants on the right of each equation. Reducing the linear system using thethree valid operations is equivalent to reducing the augmented matrix to a triangular matrix using the row operations:

Interchange two rows.
Multiply any row by a nonzero constant.
Add a multiple of one row to another.

In the above example, the augmented matrix can be reduced to either   

-1 − 1 − 1 − 2 1

0 -3 0 0 − 1

0 0 -1 − 6 4





 or







1 0 0 − 4 8 / 3

0 1 0 0 1 / 3

0 0 1 6 − 4







The left matrix is in row echelon form and the right is in reduced row echelon form. The framed terms are the pivots of the matrix. The pivot entries correspond todependent variables and the non-pivot entries correspond to free variables. In this example, the free variable isx 4 andx 1 , x 2 ,andx 3 depend onx 4 .So the linear system has infinitely many solutions given byx 1 = 83 + 4x 4 , x 2 = 13 , x 3 =− 4 − 6 x 4 ,andx 4 is an

8 Chapter 1 Systems of Linear Equations and Matrices

arbitrary real number. For a linear system with the same number of equations as variables, there will be a unique solution if and only if the coefficient matrix can be rowreduced to the matrix with each diagonal entry 1 and all others 0.

Solutions to Exercises

1. [

2 − 3 5

−1 1 − 3

]

2. [

2 − 2 1

3 0 1

]

3. 



2 0 − 1 4

1 4 1 2

4 1 − 1 1



 4.





−3 1 1 2

0 0 − 4 0

−4 2 − 3 1





5. [

2 0 − 1 4

1 4 1 2

]

6. [

4 1 − 1 1

4 −4 2 − 2

]

7. 



2 4 2 2 − 2

4 − 2 − 3 − 2 2

1 3 3 − 3 − 4



 8.





3 0 −3 4 − 3

−4 2 − 2 − 4 4

0 4 −3 2 − 3





9.The linear system has the unique solutionx= − 1 , y= 12 , z= 0.

The linear system has the unique solution x= 2, y= 0, z=− 32.

11.The linear system is consistent with free vari- ablez.There are infinitely many solutions given byx=− 3 − 2 z, y= 2 +z, z∈R.

12.The linear system is consistent with free vari- ablez.There are infinitely many solutions given byx= 4 + 13 z, y= 43 − 3 z, z∈R.

The variablez= 2 andyis a free variable, so the linear system has infinitely many solutions given byx=−3 + 2y, z= 2, y∈R.
The variablesyandzare free variables, so the linear system has infinitely many solutions given byx=− 1 − 5 y− 5 z, y∈R, z∈R.
The last row of the matrix represents the impossible equation 0 = 1,so the linear system is inconsistent.
The last row of the matrix represents the impossible equation 0 = 1,so the linear system is inconsistent.
The linear system is consistent with free variableszand w.The solutions are given by x= 3 + 2z− 5 w, y= 2 +z− 2 w, z∈R, w∈R.
The linear system is consistent with free variablesy and z.The solutions are given by x= 1− 3 y+ 3z, w= 4, y∈R, z∈R.
The linear system has infinitely many so- lutions given byx= 1 + 3w, y = 7 +w, z = − 1 − 2 w, w∈R.
The linear system has infinitely many solu- tions given byx=− 1 − 25 z, y= 1+3z, w= 45 , z∈ R.

21.The matrix is in reduced row echelon form. 22.The matrix is in reduced row echelon form.

Since the matrix contains nonzero entries above the pivots in rows two and three, the ma- trix is not in reduced row echelon form.

24.Since the pivot in row two is not a one, the matrix is not in reduced row echelon form.

25.The matrix is in reduced row echelon form. 26.The matrix is in reduced row echelon form.

Since the first nonzero term in row three is to the left of the first nonzero term in row two, the matrix is not in reduced row echelon form.
Since the matrix contains nonzero entries above the pivots in rows two and three, the ma- trix is not in reduced row echelon form.
[ To find the reduced row echelon form of the matrix we first reduce the matrix to triangular form using 2 3 −2 1

]

R 1 +R 2 →R 2

−−−−−−−−−−→

[

2 3

0 4

]

.The next step is to make the pivots 1, and eliminate the term above the

pivot in row two. This gives [ 2 3 0 4

]

1

4 R 2 →R 2

−−−−−−−→

[

2 3

0 1

]

(−3)R 2 +R 1 →R 1

−−−−−−−−−−−−−−→

[

2 0

0 1

]

1

2 R 1 →R 1

−−−−−−−→

[

1 0

0 1

]

.

30.The matrix

[

−3 2

3 3

]

reduces to−−−−−−−→

[

1 0

0 1

]

.

10 Chapter 1 Systems of Linear Equations and Matrices

The linear system is inconsistent. 42.The augmented matrix

 

3 0 − 2 − 3

−2 0 1 − 2

0 0 − 1 2



reduces to −−−−−−−→





1 0 0 0

0 1 0 0

0 0 0 1



.

The linear system is inconsistent. 43.The augmented matrix for the linear system and the reduced row echelon form are

[ 3 2 3 − 3 1 2 − 1 − 2

]

−→

[

1 0 2 − 12

0 1 − 32 − 34

]

.

As a result, the variablex 3 is free and there are infinitely many solutions to the linear system given by x 1 =− 12 − 2 x 3 , x 2 =− 34 + 32 x 3 , x 3 ∈R. 44.The augmented matrix

[ 0 − 3 − 1 2 1 0 1 − 2

]

reduces to−−−−−−−→

[

1 0 1 − 2

0 1 13 − 23

]

.

As a result, the variablex 3 is free and there are infinitely many solutions to the linear system given by x 1 =− 2 −x 3 , x 2 =− 23 − 13 x 3 , x 3 ∈R. 45.The augmented matrix for the linear system and the reduced row echelon form are

 

−1 0 3 1 2

2 3 −3 1 2

2 − 2 − 2 − 1 − 2



−→





1 0 0 121

0 1 0 121

0 0 1 121



.

As a result, the variablex 4 is free and there are infinitely many solutions to the linear system given by x 1 = 1− 12 x 4 , x 2 = 1− 12 x 4 , x 3 = 1− 12 x 4 , x 4 ∈R. 46.The augmented matrix

 

− 3 −1 3 3 − 3

1 −1 1 1 3

−3 3 −1 2 1



reduces to −−−−−−−→





1 0 0 344

0 1 0 946

0 0 1 525



.

As a result, the variablex 4 is free and there are infinitely many solutions to the linear system given by x 1 = 4− 34 x 4 , x 2 = 6− 94 x 4 , x 3 = 5− 52 x 4 , x 4 ∈R. 47.The augmented matrix for the linear system and the reduced row echelon form are

 

3 −3 1 3 − 3

1 1 − 1 − 2 3

4 −2 0 1 0



−→





1 0 − 13 − 121

0 1 − 23 − 322

0 0 0 0 0



.

As a result, the variablesx 3 andx 4 are free and there are infinitely many solutions to the linearsystem given byx 1 =− 8 −x 3 − 8 x 4 , x 2 =− 11 −x 3 − 11 x 4 , x 3 ∈R, x 4 ∈R. 48.The augmented matrix

 

−3 2 − 1 − 2 2

1 −1 0 − 3 3

4 −3 1 − 1 1



reduces to −−−−−−−→





1 0 1 8 8

0 1 1 11 − 11

0 0 0 0 0



.

As a result, the variablesx 3 andx 4 are free and there are infinitely many solutions to the linearsystem given byx 1 =− 8 −x 3 − 8 x 4 , x 2 =− 11 −x 3 − 11 x 4 , x 3 ∈R, x 4 ∈R. 49.The augmented matrix for the linear system and the row echelon form are

1.3 Matrix Algebra 11





1 2 − 1 a 2 3 − 2 b − 1 −1 1 c



−→





1 2 − 1 a 0 −1 0 − 2 a+b 0 0 0 −a+b+c



.

a.The linear system is consistent precisely when the last equation, from the row echelon form, is consistent. That is, whenc−a+b= 0.b.Similarly, the linear system is inconsistent whenc−a+b 6 = 0.c.For those values ofa, b,andcfor which the linear system is consistent, there is a free variable, so that there are infinitely many solutions.d.The linear system is consistent ifa= 1, b= 0, c= 1.If the variables are denoted byx, y andz,then one solution is obtained by settingz= 1,that is,x=− 2 , y= 2, z= 1. 50.Reducing the augmented matrix gives

[ a 1 1 2 a− 1 1

]

→

[

2 a− 1 1 a 1 1

]

→

[

1 a− 21 12 a 1 1

]

→

[

1 a− 21 12 0 −a(a 2 −1)+ 1 1 − 12 a

]

.

a. Ifa 6 =− 1 ,the linear system is consistent. b. If the linear system is consistent anda 6 = 2,then the solution is unique. If the linear system is consistent anda= 2 there are infinitely many solutions. c. Let a= 1.The unique solution isx= 12 , y= 12. 51.The augmented matrix for the linear system and the reduced row echelon form are

 

−2 3 1 a 1 1 − 1 b 0 5 − 1 c



−→





1 0 − 45 − 12 a+ 103 b 0 1 − 1515 c 0 0 0 a+ 2b−c



.

a. The linear system is consistent precisely when the last equation, from the reduced row echelon form, is consistent. That is, whena+2b−c= 0.b.Similarly, the linear system is inconsistent whena+2b−c= 0 6 = 0. c. For those values ofa, b,andcfor which the linear system is consistent, there is a free variable, so that there are infinitely many solutions.d.The linear system is consistent ifa= 0, b= 0, c= 0.If the variables are denoted byx, yandz,then one solution is obtained by settingz= 1,that is,x= 45 , y= 15 , z= 1.

52.

[

1 0

0 1

]

,

[

1 0

0 0

]

,

[

1 1

0 0

]

,

[

0 1

0 0

]

.

Exercise Set 1.

Addition and scalar multiplication are defined componentwise allowing algebra to be performed on ex- pressions involving matrices. Many of the properties enjoyed by the real numbers also hold for matrices. For example, addition is commutative and associative, the matrix of all zeros plays the same role as 0 in the real numbers since the zero matrix added to any matrixAisA.If each component of a matrixAis negated, denoted by−A,thenA+ (−A) is the zero matrix. Matrix multiplication is also defined. The matrixAB that is the product ofAwithB,is obtained by taking the dot product of each row vector ofAwith each column vector ofB.The order of multiplication is important since it is not always the case thatABandBA are the same matrix. When simplifying expressions with matrices, care is then needed and the multiplication of matrices can be reversed only when it is assumed or known that the matrices commute. The distributive property does hold for matrices, so thatA(B+C) =AB+AC.In this case however, it is also necessary to note that (B+C)A=BA+CAagain since matrix multiplication is not commutative. The transpose of a matrixA,denoted byAt,is obtained by interchanging the rows and columns of a matrix. There are important properties of the transpose operation you shouldalso be familiar with before solving the exercises. Of particular importance is (AB)t=BtAt.Other properties are (A+B)t=At+Bt,(cA)t=cAt,and (At)t=A.A class of matrices that is introduced in Section 1.3 and considered throughout the text are the symmetric matrices. A matrixAis symmetric it is equal to its transpose, that is,At=A.For example, in the case of 2×2 matrices,

[

a b c d

]

=At=

[

a c b d

]

⇔b=c.

1.3 Matrix Algebra 13

14.(A+B)C=

[

0 − 3

1 0

][

2 0

− 1 − 1

]

=

[

3 3

2 0

]

15. 2 A(B− 3 C) =

[

10 − 18

− 24 0

]

16.(A+ 2B)(3C) =

[

21 9

−6 0

]

To find the transpose of a matrix the rows and columns are reversed. SoAtandBtare 3× 2 matrices and the operation is defined. The result

is 2At−Bt=





7 5

−1 3

− 3 − 2



.

18.SinceBtis 3×2 and 2Ais 2× 3 ,the expression Bt− 2 Ais not defined.

SinceAis 2×3 andBtis 3× 2 ,then the

productABtis defined withABt=

[

− 7 − 4

−5 1

] 20.BA

[

− 7 − 5

−4 1

]

21.(At+Bt)C=





−1 7

6 8

4 12



 22. SinceCis 2×2 andAt+Btis 3× 2 ,the expression is not defined.

23.(AtC)B=





0 20 15

0 0 0

− 18 − 22 − 15



 24.SinceAtis 3×2 andBtis 3× 2 ,the expression is not defined.

25.AB=AC=

[

− 5 − 1

5 1

]

IfA=

[

0 2

0 5

]

andB=

[

1 1

0 0

]

,then

AB=

[

0 0

]

.

27.The product

A 2 =AA=

[

a b 0 c

][

a b 0 c

]

=

[

a 2 ab+bc 0 c 2

]

=

[

1 0

0 1

]

if and only ifa 2 = 1, c 2 = 1,andab+bc=b(a+c) = 0.That is,a=± 1 , c=± 1 ,andb(a+c) = 0,so thatA

has one of the forms

[

1 0

0 1

]

,

[

1 b 0 − 1

]

,

[

− 1 b 0 1

]

,or

[

−1 0

0 − 1

]

.

28.SinceAM=

[

2 a+c 2 b+d a+c b+d

]

andM A=

[

2 a+b a+b 2 c+d c+d

]

,thenAM=M Aif and only if 2a+c=

2 [a+b, 2 b+d=a+b, a+c= 2c+d, bed=c+d.That is,b=candd=a−b,so the matrices have the form a b b a−b

]

.

LetA=

[

1 1

0 0

]

andB=

[

− 1 − 1

1 1

]

.ThenAB=

[

0 0

]

,and neither of the matrices are the

zero matrix. Notice that, this can not happen with real numbers. That is, if the product of two real numbers is zero, then at least one of the numbers must be zero.

30.LetA=

[

a b c d

]

andB=

[

e f g h

]

,so thatAB−BA=

[

1 0

0 1

]

if and only if [ bg−cf (af+bh−(be+f d) (ce+dg)−(ag+ch) cf−bg

]

=

[

1 0

0 1

]

.Sobg−cfandcf−bgmust both be 1, which

is not possible.

31.The product

[ 1 2 a 0

][

3 b −4 1

]

=

[

− 5 b+ 2 3 a ab

]

14 Chapter 1 Systems of Linear Equations and Matrices

will equal

[

−5 6

12 16

]

if and onlyb+ 2 = 6, 3 a= 12,andab= 16.That is,a=b= 4.

32.LetA=

[

a b c d

]

andB=

[

e f g h

]

.Since

AB−BA=

[

bg−cf (af+bh−(be+f d) (ce+dg)−(ag+ch) cf−bg

]

,

then the sum of the terms on the diagonal is (bg−cf) + (cf−bg) = 0.

33.Several powers of the matrixAare given by

A 2 =





1 0 0

0 1 0

0 0 1



, A 3 =





1 0 0

0 −1 0

0 0 1



, A 4 =





1 0 0

0 1 0

0 0 1



,andA 5 =





1 0 0

0 −1 0

0 0 1



.

We can see that ifnis even, thenAnis the identity matrix, so in particularA 20 =





1 0 0

0 1 0

0 0 1



.Notice also

that, ifnis odd, thenAn=





1 0 0

0 −1 0

0 0 1



.

34.Since (A+B)(A−B) =A 2 −AB+BA−B 2 ,then (A+B)(A−B) =A 2 −B 2 whenAB=BA.

35.We can first rewrite the expressionA 2 BasA 2 B=AAB.SinceAB=BA,thenA 2 B=AAB=ABA= BAA=BA 2.

a. SinceAB=BAandAC=CA,then (BC)A=B(CA) =B(AC) =A(BC) and henceBCandA

commute. b.LetA=

[

1 0

0 1

]

,so thatAcommutes with every 2×2 matrix. Then select any two matrices

that do not commute. For example, letB=

[

1 0

]

andC=

[

0 1

]

.

Multiplication ofAtimes the vectorx=







1

0 ..

.

0 





gives the first column vector of the matrixA.Then

Ax= 0 forces the first column vector ofAto be the zero vector. Then letx=







0

1 ..

.

0 





and so on, to show

that each column vector ofAis the zero vector. Hence,Ais the zero matrix.

38.LetAn=

[

1 −n −n n 1 +n

]

andAm=

[

1 −m −m m 1 +m

]

.Then

AnAm=

[

(1−n)(1−m)−nm (1−n)(−m)−(1 +m)n n(1−m) +m(1 +n) −mn+ (1 +n)(1 +m)

]

=

[

1 −(m+n) −(m+n) m+n 1 + (m+n)

]

=Am+n.

39.LetA=

[

a b c d

]

,so thatAt=

[

a c b d

]

.Then

AAt=

[

a b c d

][

a c b d

]

=

[

a 2 +b 2 ac+bd ac+bd c 2 +d 2

]

=

[

0 0

]

if and only ifa 2 +b 2 = 0, c 2 +d 2 = 0,andac+bd= 0.The only solution to these equations isa=b=c=d= 0, so the only matrix that satisfiesAAt= 0 is the 2×2 zero matrix.

16 Chapter 1 Systems of Linear Equations and Matrices





− 2 −2 1 1 0 0

1 − 1 − 2 0 1 0

2 1 − 2 0 0 1



→





1 0 0 −4 3 − 5

0 1 0 2 −2 3

0 0 1 −3 2 − 4





︸︷︷︸

A− 1 but 



0 − 2 − 2 1 0 0

− 1 −1 0 0 1 0

2 1 − 1 0 0 1



→





1 0 − 1 0 1 1

0 1 1 0 − 2 − 1

0 0 0 1 − 4 − 2



.

The inverse of the product of two invertible matricesAandBcan be found from the inverses of the individual matricesA− 1 andB− 1 .But as in the case of the transpose operation, the order of multiplication is reversed, that is, (AB)− 1 =B− 1 A− 1.

Solutions to Exercises

Since (1)(−1)−(−2)(3) = 5 and is nonzero, the inverse exists and

A− 1 = 15

[

−1 2

−3 1

]

.

Since (−3)(2)−(1)(1) =−7 and is nonzero, the inverse exists and A− 1 =− 17

[

2 − 1

− 1 − 3

]

.

3.Since (−2)(−4)−(2)(4) = 0,then the matrix is not invertible.

Since (1)(2)−(1)(2) = 0,then the matrix is not invertible.
To determine whether of not the matrix is invertible we row reduce the augmented matrix

 

0 1 − 1 1 0 0

3 1 1 0 1 0

1 2 − 1 0 0 1



R 1 ↔R 3

−−−−−−→





1 2 − 1 0 0 1

3 1 1 0 1 0

0 1 − 1 1 0 0



(−3)R 1 +R 2 →R 2

−−−−−−−−−−−−−−→





1 2 − 1 0 0 1

0 −5 4 0 1 − 3

0 1 − 1 1 0 0



R 2 ↔R 3

−−−−−−→





1 2 − 1 0 0 1

0 1 − 1 1 0 0

0 −5 4 0 1 − 3



(−2)R 2 +R 1 →R 1

−−−−−−−−−−−−−−→





1 0 1 −2 0 1

0 1 − 1 1 0 0

0 −5 4 0 1 − 3



(5)R 2 +R 3 →R 3

−−−−−−−−−−−−−→





1 0 1 −2 0 1

0 1 − 1 1 0 0

0 0 − 1 5 1 − 3



(−1)R 3 →R 3

−−−−−−−−−−→





1 0 1 −2 0 1

0 1 − 1 1 0 0

0 0 1 − 5 −1 3



(−1)R 3 +R 1 →R 1

−−−−−−−−−−−−−−→





1 0 0 3 1 − 2

0 1 − 1 1 0 0

0 0 1 − 5 −1 3



(1)R 3 +R 2 →R 2

−−−−−−−−−−−−−→





1 0 0 3 1 − 2

0 1 0 − 4 −1 3

0 0 1 − 5 −1 3



.Since the original matrix has been reduce to the identity matrix, the inverse

exists andA− 1 =





3 1 − 2

− 4 −1 3

− 5 −1 3



.

6.Since 



0 2 1 1 0 0

−1 0 0 0 1 0

2 1 1 0 0 1



reduces to −−−−−−−→





1 0 0 0 −1 0

0 1 0 1 − 2 − 1

0 0 1 −1 4 2





the matrix is invertible andA− 1 =





0 −1 0

1 − 2 − 1

−1 4 2



.

1.4 The Inverse of a Matrix 17

Since the matrixAis row equivalent to the

matrix





1 −1 0

0 0 1

0 0 0



,the matrixAcan not be

reduced to the identity and hence is not invertible.

Since the matrixAis not row equivalent to the identity matrix, thenAis not invertible.

9.A− 1 =







1 / 3 − 1 −2 1/ 2

0 1 2 − 1

0 0 −1 1/ 2

0 0 0 − 1 / 2





 10.A

− 1 =







1 −3 3 0

0 1 −1 1/ 2

0 0 1/2 1/ 2

0 0 0 1/ 2







11.A− 1 = 13







3 0 0 0

−6 3 0 0

1 − 2 −1 0

1 1 1 1





 12.

A− 1 =







1 0 0 0

2 1 0 0

− 1 / 2 − 1 / 2 − 1 /2 0

1 1 0 1/ 2







13.The matrixAis not invertible. 14.The matrixAis not invertible.

15.A− 1 =







0 0 −1 0

1 − 1 −2 1

1 − 2 −1 1

0 − 1 −1 1





 16.The matrixAis not invertible.

17.Performing the operations, we have thatAB+A=

[

3 8

10 − 10

]

=A(B+I) andAB+B=

[

2 9

6 − 3

]

=

(A+I)B.

18.Since the distributive property holds for matrix multiplication and addition, we have that (A+I)(A+I) = A 2 +A+A+I=A 2 + 2A+I.

19.LetA=

[

1 2

−2 1

]

.a.SinceA 2 =

[

−3 4

− 4 − 3

]

and− 2 A=

[

− 2 − 4

4 − 2

]

,thenA 2 − 2 A+ 5I= 0. b.

Since (1)(1)−(2)(−2) = 5,the inverse exists andA− 1 = 15

[

1 − 2

2 1

]

= 15 (2I−A).

c.IfA 2 − 2 A+5I= 0,thenA 2 − 2 A=− 5 I,so thatA

( 1

5 (2I−A)

)

= 25 A− 15 A 2 =− 15 (A 2 − 2 A) =− 15 (− 5 I) =I.

HenceA− 1 = 15 (2I−A).

20. Applying the operations (−3)R 1 +R 2 →R 2 and (−1)R 1 +R 3 →R 3 gives



1 λ 0 3 2 0 1 2 1



reduces to −−−−−−−→





1 λ 0 0 2− 3 λ 0 1 2 1



.So ifλ= 2 3 ,then the matrix can not be reduced to the identity

and hence, will not be invertible.

The matrix is row equivalent to





1 λ 0 0 3−λ 1 0 1− 2 λ 1



.Ifλ=− 2 ,then the second and third rows are

identical, so the matrix can not be row reduced to the identity and hence, is not invertible.

Since the matrix is row equivalent to





1 2 1

0 λ− 4 − 1 0 4− 2 λ 0



,ifλ= 2,then the matrix can not be row

reduced to the identity matrix and hence, is not invertible.

a.Ifλ 6 = 1,then the matrixAis invertible.

b.Whenλ 6 = 1 the inverse matrix isA− 1 =





−λ− 11 λ−λ 1 −λλ− 1 1 λ− 1 −

1 λ− 1

1 λ− 1 0 0 1



.

Linear Algebra and Its Applications 4th Edition 1.3 Solutions

Source: https://www.studocu.com/row/document/american-university-of-beirut/linear-algebra/solution-manual/7075603