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  • Linear Algebra - Exercises

    9 Linear Systems. Method of Elimination1. Determine whethera) x = 1, y = 1b) x = 2, y = 3

    are solutions to the linear equation

    2x+ 3y = 1.

    Solution. a)2 1 + 3 (1) = 1,

    1 = 1,so x = 1, y = 1 is a solution.b)

    2 2 + 3 (3) = 1,5 6= 1,

    so x = 2, y = 3 is not a solution.2.Determine ifa) x = 0, y = 2b) x = 1, y = 3

    are solutions to the system

    x + 3y = 8,2x + y = 5.

    Solution. a) Substituting x = 0, y = 2 into Equation 1, we obtain

    1 0 + 3 2 = 8,6 6= 8,

    47

  • LINEAR ALGEBRA - EXERCISES 48

    so x = 0, y = 2 is not a solution.b) Substituting x = 1, y = 3 into Equation 1, we obtain

    1 (1) + 3 (3) = 8,8 = 8,

    and into Equation 2, we obtain

    2 (1) + (3) = 5,5 = 5.

    Therefore, x = 1, y = 3 is a solution.3. Solve the system

    2x + y = 2,4x y = 1.

    by the method of elimination.

    Solution.12 Equation 1

    x + 12y = 1

    4x y = 1Equation 2 4 Equation 1

    x + 12y = 1

    3y = 3>From Equation 2

    y = 1

    Substituting y = 1 into Equation 1

    x+1

    2= 1

    x =1

    2

    The system has the unique solution

    x =1

    2y = 1

  • LINEAR ALGEBRA - EXERCISES 49

    4. Solve the system2x + y = 1,x + 2y = 1,x y = 2.

    by the method of elimination.

    Solution.

    Interchange Equation 1 and Equation 2

    x + 2y = 12x + y = 1x y = 2

    Equation 2 2 Equation 1Equation 3 Equation 1

    x + 2y = 1 3y = 3 3y = 3

    13 Equation 2

    13 Equation 3

    x + 2y = 1y = 1y = 1

    Equation 3 Equation 2x + 2y = 1

    y = 10 = 0

    We substitute y = 1 into Equation 1x+ 2 (1) = 1

    x = 1

    The system has the unique solution

    x = 1

    y = 1

    5. Solve the systemx + y + z = 1,x + 2y + 2z = 1,x y + z = 3.

  • LINEAR ALGEBRA - EXERCISES 50

    by the method of elimination.

    Solution.Equation 2 Equation 1Equation 3 Equation 1

    x + y + z = 1y + z = 0

    2y = 3Equation 3 + 2 Equation 2

    x + y + z = 1y + z = 0

    2z = 2

    >From Equation 3, we havez = 1.

    We substitute z = 1 into Equation 2:

    y + 1 = 0

    y = 1.We substitute y = 1 and z = 1 into Equation 1:

    x+ (1) + 1 = 1x = 1.

    The solution isx = 1

    y = 1z = 1

    3. Solve the system2x + y + z = 2,x + 2y + z = 0,x + y = 1.

    by the method of elimination.

  • LINEAR ALGEBRA - EXERCISES 51

    10 Matrices1. Let

    A =

    2 5 52 2 3

    .

    Writea) the size of A b) Row2(A) c) Col1(A) d) (2, 3) entry of A

    Solution.a) The size of A is 2 3.b)

    Row2(A) =2 2 3

    c)

    Col1(A) =

    22

    d) a23 = 3

    2. Determine which of the following matrices are square matrices. Forsquare matrices, write the order and the diagonal.

    A =

    1 0 02 0 1

    , B =

    2 14 4

    , C =

    1 0 03 1 40 1 0

    .Solution. A is not a square matrix.B is a square matrix of order 2, the diagonal of B is

    2, 4

    C is a square matrix of order 3, the diagonal of C is

    1, 1, 0

    2. Let

    A =

    3 1 02 2 3

    , B =

    1 0 02 5 1

    , C =

    2 11 2

    .

    Find, if possible

    a) A+B b) A+ C c) 2C + 3I2

  • LINEAR ALGEBRA - EXERCISES 52

    Solution.a) A and B are of the same size 2 3, so A+B is defined

    A+B =

    4 1 00 3 4

    b) A+ C is not defined because A and C are not of the same size.c)

    2C + 3I2 = 2 2 11 2

    + 3

    1 00 1

    =

    4 22 4

    +

    3 00 3

    =

    7 22 1

    3. Let

    A =

    1 32 23 01 1

    , B = 1 0 2 02 1 3 1.

    Find, if possible

    a) AT b) A+B c) AT +B

    Solution.a)

    AT =

    1 2 3 13 2 0 1

    b) A+B is not definedc)

    AT +B =

    1 2 3 13 2 0 1

    +

    1 0 2 02 1 3 1

    =

    2 2 5 15 3 3 2

    10.1 Homework Problems

    1. Let

    A =

    3 1 1 02 1 0 11 0 0 11 3 2 00 0 1 3

    .Writea) the size of A b) Row3(A) c) Col4(A) d) (2, 4) entry of A

  • LINEAR ALGEBRA - EXERCISES 53

    2. Let

    A =

    2 11 30 4

    , B = 0 2 15 1 0

    , C =

    1 71 710 5

    .Compute, if possible

    a) AB b) A+ C c) ABT d) AT + C3. Let

    A =

    1 52 4

    , B =

    0 10 2

    , C =

    0 2 41 6 01 6 6

    .Compute

    a) A+ 4B b) 2A 3I c) 12C d) C + 3I

    10.2 Answers to Homework Problems

    1. a) 5 4 b) 1 0 0 1 c)01103

    d) 1

    2. a) Not defined. b)

    3 60 410 1

    c) 2 63 41 4

    d) Not defined.3. a)

    1 92 4

    b)1 104 5

    c)

    0 1 2123 0

    12

    3 3

    d) 3 2 41 9 01 6 9

  • LINEAR ALGEBRA - EXERCISES 54

    11 Matrix Multiplication1. Let

    A =1 0 1 , B =

    210

    , C = 35

    .

    Find, if possible

    a) AB b) AC

    Solution. a)

    AB =1 0 1

    210

    = 2 + 0 + 0 = 2b) AC is not defined because the number of columns of A is not equal to thenumber of rows of C.

    2. Let

    A =

    1 30 11 23 0

    , B =1 12 32 10 1

    , C = 2 0 13 1 0

    Find, if possible

    a) AB b) AC

    Solution. a)A B = AB4 2 4 2 not defined

    The product AB is not defined because the sizes do not match (the numberof columns of A is not equal to the number of rows of B).b)

    A C = AC4 2 2 3 4 3

    Therefore, AC is a 4 3 matrix. Write AC as a blank 4 3 matrix andcompute, one by one, the elements of AC:

    AC =

    1 30 11 23 0

    2 0 13 1 0=

    11 3 13 1 04 2 16 0 3

  • LINEAR ALGEBRA - EXERCISES 55

    3. Let

    A =

    1 10 2

    , B =

    2 01 1

    .

    Determine whether AB = BA.

    Solution.

    AB =

    1 10 2

    2 01 1

    =

    1 12 2

    BA =

    2 01 1

    1 10 2

    =

    2 21 3

    Therefore, AB 6= BA.

    4. LetA =

    1 0 2

    Find, if possible

    a) A2 = AA b) ATA c) AAT

    Solution. a)A A = AA1 3 1 3 not defined

    The product AA is not defined.b)

    AT =

    102

    We have

    A AT = AAT

    1 3 3 1 1 1so AAT is defined,

    AAT =1 0 2

    102

    = [5]c)

    AT A = AAT

    3 1 1 3 3 3

    ATA =

    102

    1 0 2 =1 0 20 0 02 0 4

  • LINEAR ALGEBRA - EXERCISES 56

    11.1 Homework Problems

    1.

    A =0 2 3 , B =

    3102

    , C = 211

    .Find, if possible

    a) AB b) AC

    2. Let

    A =

    1 1 02 0 11 1 1

    , B =1 10 13 1

    , C = 2 0 13 1 0

    Find, if possible

    a) AB b) AC

    3. Let

    A =

    1 10 2

    , B =

    0 0 21 1 3

    .

    Compute

    a) AB b) AT c) BT d) BTAT

    e) Check that (AB)T = BTAT .

    11.2 Answers to Homework Problems

    1. a) Not defined. b) 1.

    2. a)

    1 05 34 3

    b) Not defined.3. a)

    1 1 12 2 6

    b)1 01 2

    c)

    0 10 12 3

    d)1 21 21 6

  • LINEAR ALGEBRA - EXERCISES 57

    12 Solutions of Linear Systems of Equations1. Write the augmented matrix representing the linear system

    x 4y + z = 2,2x + 3y = 1.

    Solution. 1 4 12 3 0

    21

    2. The following matrix represents a linear system in variables x, y and z.2 1 10 1 31 0 11 2 5

    0231

    Write the system.

    Solution.2x + y z = 0

    y + 3z = 2x z = 3x + 2y + 5z = 1

    3. Determine which of the following matrices are in reduced row echelonform:

    A =

    1 0 10 1 20 0 0

    120

    A is not in reduced row echelon form because the pivot in Row1 is not equalto 1.

    B =

    1 0 10 1 2

    B is in reduced row echelon form.

    C =

    1 0 0 40 0 1 20 1 0 3

    501

    C is not in reduced row echelon form because the pivot in Row3 is not to theright of the pivot in Row2.

    D =

    1 2 00 1 10 0 0

  • LINEAR ALGEBRA - EXERCISES 58

    D is not in reduced row echelon form because not all the entries above thepivot in Row2 are zero.

    E =

    1 0 30 1 50 0 0

    530

    E is in reduced row echelon form.

    4. The following augmented matrices represent systems of linear equationsin variables x, y and z. In each case either state the general solution or thatno solution exists.

    a)

    1 0 20 1 30 0 0

    430

    b)1 0 00 1 70 0 0

    053

    c)1 0 00 1 00 0 10 0 0

    1260

    Solution. a) The system corresponding to the matrix is

    x + 2z = 4y + 3z = 3

    0 = 0

    The system has infinitely many solutions. The general solution is

    x = 4 2zy = 3 3z

    where z is any real number.

    b) The equation corresponding to the third row is

    0 = 3.

    Therefore, the system has no solution.

    c) The system has the unique solution

    x = 1y = 2

    z = 6

    5. Solve the system3x + 2y = 1,x + y = 1.

    by Gauss-Jordan reduction.

  • LINEAR ALGEBRA - EXERCISES 59

    Solution. The augmented matrix of the system is3 21 1

    11

    We do Gauss-Jordan reduction to transform the matrix in reduced row ech-elon form.

    Row2Row1

    1 13 2

    11

    Row2 3Row11 10 1

    12

    Row21 10 1

    12

    Row1 Row2

    1 00 1

    12

    The system has the unique solution

    x = 1, y = 2.

    6. Solve the systemx + y + z = 22x y + z = 13x + 2z = 3

    by Gauss-Jordan reduction.Solution. 1 1 12 1 1

    3 0 2

    213

    Row2 2Row1Row3 3Row1

    1 1 10 3 10 3 1

    233

    13Row2

    Row3 Row2

    1 1 10 1 13

    0 0 0

    210

    Row1 Row2

    1 0 230 1 13

    0 0 0

    110

  • LINEAR ALGEBRA - EXERCISES 60

    We readx+

    2

    3z = 1

    y +1

    3z = 1

    The system has infinitely many solutions. The general solution is

    x = 1 23z

    y = 1 13z

    where z is any real number.

    12.1 Homework Problems

    1. Determine which of the following matrices are in reduced row echelonform:

    A =

    1 0 30

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