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    Section 9.1: Exercises 6, 12, 20, 28, 32, 50

    6.) Explain why each graph is that of a function.

    Each graph is that of a function because at each point in a graph we get a value for ycorresponding to a value for x. This is just the pictorial reflection of what we get in case of a

    function.

    Solve each system by substitution

    12.) 1154 yx

    52 yx

    Solution

    2x + y = 5 y = 5- 2x

    Substituting in first equation

    4x5y = -11

    4x5(5-2x) = -11

    4x25 + 10x = -11

    14 x = -11 + 25

    14x = 14

    x = 1

    Therefore, we get :

    So, y = 5-2x = 5 -2*1 = 3

    x = 1, y =3

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    Solve each system by elimination.

    20.) 234 yx

    172 yx

    Solution

    4x +y = -23 eq 1

    x - 2y = -17.eq 2

    Performing 2*eq 1 + eq 2 we get :

    8x + 2y = -46

    x2y = -17

    This Gives us :

    8x + x = -46 -17

    9x = -63

    x = -7

    So,

    x2y = -17

    -7 -2y = -17

    2y = 17-7 =10

    y = 5

    So, X = -7, Y =5

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    28.) 222

    3

    yx

    022

    yx

    Solution

    3x/2 + y/2 = -2

    3x + y = -4 .eq 1

    x/2 + y/2 = 0

    x + y = 0 ..eq2

    Performing eq 1eq 2

    3x + y = -4

    (-)x + (-)y = (-)0

    3xx = -4

    2x = -4 X = -2

    So,

    X + y = 0

    -2 + y = 0 Y = 2

    So, x = - 2, y = 2

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    Solve each system. State whether it is inconsistent or has infinitely many solutions. If the system has

    infinitely many solutions, write the solution set and y arbitrary.

    32) 523 yx

    846 yx

    Solution

    3x + 2y = 5 .eq 1

    6x + 4y = 8 .eq2

    3x + 2y = 4 ..eq 2

    This system is inconsistent as the coefficients of one equation does not match the other equations

    coefficients

    Solve each system.

    50.) 234 zyx

    1553 xyx

    1442 zyx

    Solution

    X -1

    Y = 4

    Z 2

    Section 9.2: Exercises 4, 8, 10, 20, 32

    Use the given row transformation to change each matrix as indicated.

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    4.)

    3

    4

    2

    7

    1

    5

    1

    2

    6

    ; -6 times row 3 added to row 1

    Solution

    3

    4

    6*32

    7

    1

    7*65

    1

    2

    6*16

    3

    4

    16

    7

    1

    37

    1

    2

    0

    Write each augmented matrix for each system and give its size. Do not solve the system.

    8.) 04324 zyx

    0745

    0753

    zyx

    zyx

    Solution

    5

    3

    4

    1

    5

    2

    4

    1

    3

    7

    7

    4

    Write the system of equations associated with each augmented matrix. Do not solve.

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    10.)

    5

    4

    2

    0

    3

    1

    4

    0

    3

    11

    10

    12

    Solution

    01232 zyx

    01145

    01034

    zx

    yx

    Use the Gauss-Jordan method to solve each of equations. For systems in two variables with infinitely

    many solutions, give the solution with y arbitrary; for systems in three variables with infinitely many

    solutions, give the solution with z arbitrary.

    20.)0153

    01052

    yx

    yx

    Solution

    Writing in augmented form

    3

    2

    1

    5

    15

    10

    R1= R1

    3

    1

    1

    2/5

    15

    5

    R23R1 = R2

    0

    1

    2/17

    2/5

    0

    5

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    2/17* R2= R2

    0

    1

    1

    2/5

    0

    5

    R1 + 5/2 R2 = R1

    0

    1

    1

    0

    0

    5

    So, X = 5, Y = 0

    32.) 1yx

    22

    02

    zy

    zx

    Solution

    Writing in augmented form

    0

    2

    1

    1

    0

    1

    2

    1

    0

    2

    0

    1

    R22R1 = R2

    0

    0

    1

    1

    2

    1

    2

    1

    0

    2

    2

    1

    -1/2 *R2 = R2

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    0

    0

    1

    1

    1

    1

    2

    2/1

    0

    2

    1

    1

    R1R2 = R1

    R3R2 = R3

    0

    0

    1

    0

    1

    0

    2/3

    2/1

    2/1

    3

    1

    0

    2/3*R3 = R3

    0

    0

    1

    0

    1

    0

    1

    2/1

    2/1

    2

    1

    0

    R1 + R3 = R1

    R2 R3 = R2

    0

    0

    1

    0

    1

    0

    1

    0

    0

    2

    2

    1

    So,

    X = -1

    Y = 2

    Z = -2

    Section 9.3: Exercises 6, 14, 18, 40, 52, 62, 74, 82

    Find the value of each detriment.

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    6.)1

    0

    5

    2

    Determinant = 5*02*1 = -2

    Find the cofactor of each element in the 2nd

    row for each detriment.

    14.) 2 -1 4

    3 0 1

    -2 1 4

    Co factors for each of the elements in the 2 ndrow :-

    For 3 = - (-1*41*4) = +8

    For 0 = 2*4(-2)*(4) = +16

    For 1 = - (2*1(-2)(-1) ) = 0

    Find the value of each detriment.

    18.) 2 1 -1

    4 7 -2

    2 4 0

    Determinant = 2(7 * 0 - 4 *(-2)) - 4(1*0 (4) *(-1))+2(1 *(-2) - 7*(-1)) = 10

    Solve each equation for x.

    40.) 4 3 0

    2 0 1 = 5

    -3 x -1

    Determinant = 4*(0*-1x*1)3(2*-1(-3)*1) +0 = -4x -3

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    -4x3 = 5

    -4x = 8

    X = -2

    Use the determinant theorems to find the value of each detriment.

    52.) 4 8 0

    1 -2 1

    2 4 3

    Determinant = 4(-2*3-4*1)-1(8*3-4*0)+2(8*12*0) = -48

    Use Cramers rule to solve each equation. If D = 0, use another method to determine the solution set.

    62.) 423 yx

    52 yx

    SOLN :

    D = -7

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    D1 = 14

    D2 = -7

    X = D1/D 14/-7 = -2

    Y = D2/D = -7/-7 = 1

    74.) x + y + z4 = 0

    2xy + 3z4 = 0

    4x + 2yz + 15 = 0

    SOLUTION:

    D=17

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    D1 = -68

    D2 = 51

    D3=85

    X= D1/D = -68/17 = -4

    Y=D2/D = 51/17 = 3

    Z = D3/D = 85/17 = 5

    82.) 3x + 5y = -7

    2x + 7z = 2

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    4y + 3z = -8

    Here the determinant is 0, so we use Gauss Jordan Rule :

    Section 9.5: Exercises 12, 22, 48

    Give all solutions of each nonlinear system of equations, including those with nonreal complex

    components.

    12.) 962 xxy

    22 yx

    Using the value of y in eqn 2 we get :

    X+ 2x2+12x+18 = -2

    Or, 2x2+13x+20 = 0

    Or, 2x2+8x+5x+20 =0

    Or, 2x(x+4) + 5(x+4) = 0

    Or, (2x+5)(x+4) = 0

    Or, x = -5/2, -4

    Therefore y = , 1

    22.) 92 22 yx

    2743 22 yx

    Multiply eqn 1 by 2 to get :1842

    22 yx

    Adding eqn (ii) with eqn(i) we get: 5x2= 45

    Solving this we get, X = (+

    - 3)

    And putting the value of x in eqn (i) we get : Y = 0

    Therefore, X=+-3 and Y=0

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    Solve each problem using a system of equations in two variables.

    48.) Unknown numbers Find two numbers whose sum is 10 and who squares differ by 20.

    Let the numbers be x and y

    X + y = 10 .eq 1

    X2y2= 20. Eq2

    (x+y)(x-y) = 20Since x+y =10, we replace it in the eqn : (x+y)(x-y) =20 , to get

    10*(x-y) = 20 X-y = 2 . Eq 3

    Now,

    We add eqn1 and Eqn 3 to get :

    Eq1 + eq 3

    X+ y + Xy = 10 - 2

    2x = 12

    X = 6

    X + y = 10

    Y = 10x = 10-6 = 4

    Section 9.6: Exercises 4, 32, 38

    Graph each inequality.

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    4.) 4y3x < 5

    Graph the solution set of each system of inequalities.

    32.) yx 34 < 12

    y + 4x > -4

    SOLUTION :

    X< 3-3y/4 and x> -1/4y -1

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    38.) x + y 9

    x 2

    y

    SOLUTION :

    X 9-y

    X -y2

    Section 9.7: Exercises 6, 14, 22, 36, 52, 58, 66, 88

    Find the values of the variables for which each statement is true, if possible.

    6.)

    6

    2

    5

    0

    3

    4

    x

    5

    8

    9

    =

    6

    4

    3

    z

    y

    0

    3

    2

    w

    8

    9

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    X-4 = 2 . Therefore x = 6

    Y+3 = 5 . Therefore Y = 2

    Z+4 = 2 . Therefore Z = -2

    W = 5 . Given

    Find the size of each matrix. Identify any square, column, or row matrices.

    14.)

    4

    9

    1

    6

    8

    2

    Size = 2*3. It is neither a row or a column or a square matrix.

    Perform each operation when possible.

    22.)

    8

    9

    2

    4

    4

    3

    7

    2 = 6 6

    -12 9

    36.)

    nm

    ak

    xz

    yk

    nm

    ak

    xz

    yk

    24

    64

    52

    65

    24

    52

    36

    84

    =

    Find each matrix product when possible.

    -k -14y

    4z -8x

    -2k -a

    -8m 4n

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    52.)

    7

    1

    2

    6

    0

    5 =

    58.)

    2

    7

    28

    0

    2

    32

    6

    7 =

    Given A =

    3

    4

    1

    2, B =

    3

    0

    5

    7

    2

    1

    , and C =

    0

    5

    3

    4

    6

    1, find each product when possible.

    66.) AC

    Yes it is possible and the solution is :

    4*-5 -2*0 4*4-2*3 1*1-2*6 = -20 10 -8

    3*-5+1*0 3*4+1*3 3*1+1*6 -15 15 9

    A =

    21

    11

    a

    a

    22

    12

    a

    a, B =

    21

    11

    b

    b

    22

    12

    b

    b, and C =

    21

    11

    c

    c

    22

    12

    c

    c, where all the elements are real numbers.

    Use these materials to show each statement is true for 2 x 2matrices.

    88.) A + (B + C) = (A + B) + C (associative property)

    B+C = b11 + c11 b12+c12

    b21+c21 b22+c22

    -1*6 + 5*2 4

    7*6 0*2 = 42

    7*23+ 0*2 7*-7 + 0*6 221 -7

    2*23 + 28*2 2*-7 +28*6 = 4(7+3) 107

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    A+(B+C)= a11+b11+c11 a12+b12+c12

    a21+b21+c21 a22+b22+c22

    A+B = a11+b11 a12+b12

    A21+b21 a22+b22

    a11+b11+c11 a12+b12+c12

    (A+B)+C = a21+b21+c21 a22+b22+c22

    Therefore , (A+B)+C = A+(B+C)