Final Review Part II

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    GE 111

    Study ProblemsPart II

    Dec 20141

    Note: this is a collection of questions and solutions for study purposes. Not all solutions are

    present. Solutions have not been checked for accuracy.

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    A. Matrix Basics

    Dec 20142

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    1. Linear equations

    Which of the following are not linear equations?

    Dec 20143

    a) x + 2y2 = 7

    b) y – sinx = 0

    c) 2sinx – ½ siny + 11 sinz = 20d) x + 2.43y – 3z = 4

    e) log(100)x – e3y +103z = 14

    f) 3x + 2y + xz = 4g) x1

    2 + 2x2 + x3 = 1

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    2. Consistent or inconsistent? Solve.

    Dec 20144

     x + 2y = 3 x + y = 1

     x + 2y = 3

     x + y = 8

    2x + 3y = 3

    4 x + 6y = 2

    2x + 3y = 3

    1 x + 6y = 31½ x + 31/3y = 3

    22/3 x - 21/3y = -3

    a)

    b)

    c)

    d)

    e)

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    3. Dependent or Independent? Solve

    Dec 20145

     x + 2y = 3y = 1

     x + 2y = 3

     x + y = 8

     x + 2y = 3

    - x - 2y = -3

    4 x + 6y = 3

    2 x + 3y = 1.5

    10 x + 12y = 23

    -2 x - 3y = 10

    a)

    b)

    c)

    d)

    e)

    A. Matrix Basics

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    4. Material Balance

    A material balance question done in class on a balancedsystem yielded the following equations:

    0.75Y = 23.36 + 0.04X

    0.25Y = 1.41 + 0.34X

    Z = 12.42 + 0.63X

    Z + Y = 37.20 + X

    Write the augmented matrix for this set of equations. Whatis the rank of this matrix? Explain your answer.

    Previous midterm

    Dec 20146

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    Organize Equations:

    Write Augmented Matrix:

    The rank is 3 as the system is balanced and so 3 unknowns should yield 3

    independent equations and 1 dependant equation. Also, the sum of thefirst 3 equations yields the fourth. Gauss elimination can also be used to

    reduced the last equation to all 0s.

    SOLUTION

    4. Material Balance

    Dec 20147

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    1. Gauss

    Use Gauss elimination to solve the following set ofindependent linear equations. Write the complete

    augmented matrix with each step of your work.

    25 + 3X – 6Y = 710X = -12Y + 9X + 64

    Previous midtermDec 20148

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    Organize Equations:

    Write Augmented Matrix:

    Use back substitution to find solution:

    Y = 5

    X – 2Y = -6

    X – 2(5) = -6

    X = 4

    SOLUTION

    1. Gauss

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    2. Solve by substitution

    Show the set of linear equations represented by the

    following matrix. Solve the equations by substitution

    (the correct work and method must be shown for full

    marks). Show your work and box your answer

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    6x+y-3z=46

    7x-2y=42

    2x-z=13

    This question is actually quite easy – x is common in both the last 2

    equations, so solve for the other variables in terms of x and thensubstitute back into the first equation...

    x=8

    y=7

    z=3

    SOLUTION

    2. Solve by substitution

    Dec 201411

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    3. Solve

    You are given the following set of equations.2x1 + 3x2 -4x3 =-3

    x1 +2x2 +3x3 = 3

    3x1 - x2 - x3 = 4

    a) Using the above system of equations, rewrite them in the

    matrix form Ax=b.

    b) Show the augmented matrix for this system of equations.

    c) Solve the system of equations using matrix operations to

    determine the values of x1, x2, and x3. You MUST show ALL of

    your work to obtain ANY marks for this section of the question.

    Dec 201412

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    Question 3 Solution

    3. Solve

    2 3 -41 2 3

    3 -1 -1

    x1x2 =

    x3

    -33

    4

    2 3 -41 2 3

    3 -1 -1

    -33

    4

    1 2 32 3 -4

    3 -1 -1

    3-3

    4

    interchange

    R1 and R2

    1 2 3

    2 3 -4

    3 -1 -1

    3

    -3

    4

    1 2 3

    0 -1 -10

    0 -7 -10

    3

    -9

    -5

    -2R1 + R2 = R2’

    -3R1 + R3 = R3’

    1 2 3

    0 -1 -10

    0 0 60

    3

    -9

    58

    -7R2’ + R3’ = R3’’

    Solve x3:

    60x3 = 58 → x3 = 58/60

    Substituting x3 into R2’:

    -x2 – 10x3 = -9 → x2 = 9 - 10x3 → 9 - 580/60 → (540-580)/60 → -40/60Substitute x2 and x3 into R1:

    x1 + 2x2 + 3x3 = 3 → x1 = 3 - 2 (-40/60) – 3(58/60) → (180+ 80- 174) /60 → 86/60

    A x = B

    x1 = 86/60, x2 = -40/60, x3 = 58/60

    x1 = 1.43, x2 = -0.67, x3 = 0.97

    Dec 201413

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    4. Gauss

    a) Solve the following set of linearly independent equations usingGauss elimination. First draw the matrix in the form Ax=b, then

    write the augmented matrix and solve (the correct work and

    method must be shown for full marks). Show your work and box

    your answer.

    x+2y+3z = 13

    y = 1

    3x+3y+3z = 18

    b) Find A-1 using Gauss-Jordan elimination. Check that youranswer is correct by showing that A•A-1 = I. Show your work

    and box your answer.

    Previous finalDec 201414

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    Solution

    1 2 30 1 03 3 3

    13118

     

    3 = 3

    1

    − 3  

    → 

    1 2 3

    0 1 0

    0 3 6

    13

    1

    21 

    3 = 32 − 3  →  1 2 30 1 00 0 −6

    13

    1−18 3 = 3−6  →  

    1 2 3

    0 1 0

    0 0 1

    1313

     

    + 2 + 3 = 13  = 1  = 3  + 21+ 33 = 13  = 2 

    4. Gauss

     

    From ‘a’: 1 2 30 1 00 0 1

    1313

     

    1 =

    1

    −3

    3

    → 

    1 2 0

    0 1 0

    0 0 14

    1

    1 = 1 − 22 →  1 0 00 1 00 0 1

    213

     

    a) Gauss

    a) Gauss-Jordan

    Dec 201415

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    5. Canadian coins.

    Dec 201416

    Coin X Y Z $total

    CanA 5 5 5 16.25

    CanB 1 10 2 14.25

    CanC 20 2 3 13.00

    [A] Number of coins of different denominations

    A charity organization was collecting at three different locations within

    the city; A, B, and C. During one day of collection the same three types

    of coins were given (x, y, and z) at each location.

    The number of coins, grouped as to their worth, and the total amount

    from each location are given in the above table.

    What are the value of the three types of coins collected?

    Use all three methods: adjoint, Gauss-Jordan, and Cramer’s

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    5 Coins: Gauss Elimination

    Dec 201417

    5 5 5 16.25

    1 10 2 14.25

    20 2 3 13.00

    exchange R1and R2

    1 10 2 14.25

    5 5 5 16.25

    20 2 3 13.00

    1 10 2 14.25

    0 -45 -5 -55

    0 -198 -37 -272

    -5R1+R2-2R1+R3

    1 10 2 14.250 -45 -5 -55

    0 0 -15 -557

    -198/45R2+R3

    1 10 2 14.250 1 0.11 1.22

    0 0 1 2

    -1/45 R2-1/15 R3

    Backward substitution

    z = 2$y + 0.111(2) = 1.22, y = 1$

    x + 10(1) + 2(2) = 14.25, x = 0.25$

    Note: if you are calculating each step

    carrying over only 3 decimal places

    then the answer may be

    z = 1.995

    y = 0.993

    x = 0.233

    solution

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    5. Coins: Gauss-Jordan Elimination

    Dec 201418

    5 5 5 16.25

    1 10 2 14.25

    20 2 3 13.00

    exchange R1and R2

    1 10 2 14.25

    5 5 5 16.25

    20 2 3 13.00

    1 10 2 14.25

    0 -45 -5 -55

    0 -198 -37 -272

    -5R1+R2-20R1+R3

    1 10 2 14.250 -45 -5 -55

    0 0 -15 -557

    -198/45R2+R3

    1 10 2 14.250 1 0.11 1.22

    0 0 1 2

    -1/45 R2-1/15 R3

    1 10 0 10.250 1 0 1

    0 0 1 2

    -2R3+R1

    -0.11R3+R2

    1 0 0 0.250 1 0 1

    0 0 1 2

    -10R2+R1 xy

    z

    solution

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    5. Coins: Adjoint-Inverse Method

    Dec 201419

    5 5 5

    1 10 2

    20 2 3

    M11 = 10(3)-2(2) = 26

    26 -37 -198

    5 -85 -90

    -40 5 45

    |A| = a(ei-fh) – b(di-fg) + c(dh-eg)

    5(10(3)-2(2))-5(1(3)-2(20))+5(1(2)-10(20)) = -675

    A-1 = 1/(|A|) adj(A)

    A x = b, x = A-1 b

    Find determinant (using minor and cofactor matrices)

    Mij =

    Cij = (-1)i+j Mij

    26 37 -198

    -5 -85 90

    -40 -5 45Cij = | A| = a1,j C 1,j = -675

    n

     j=1

    or find determinant (using equation for 3by3) a b cd e f 

    g h i

    adj (A) = CofactorT

    26 -5 -40

    37 -85 -5-198 90 45

    A-1 = -1/675

    -0.039 0.007 0.059

    -0.055 0.126 0.0070.293 -0.133 -0.067

    A-1 =

    -0.039 0.007 0.059

    -0.055 0.126 0.007

    0.293 -0.133 -0.067

    =16.25

    14.25

    13.00

    0.25

    1.00

    2.00

    solution

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    5. Coins: Gauss-Jordan Inverse Method

    Dec 201420

    5 5 5 1 0 0

    1 10 2 0 1 0

    20 2 3 0 0 1

    exchange R1and R2

    1 10 2 0 1 0

    5 5 5 1 0 0

    20 2 3 0 0 1

    1 10 2 0 1 0

    0 -45 -5 1 -5 0

    0 -198 -37 0 -20 1

    -5R1+R2-20R1+R3

    1 10 2 0 1 00 -45 -5 1 -5 0

    0 0 -15 -4.4 2 1

    -198/45R2+R3

    1 10 0 -0.59 1.27 0.130 -45 0.11 2.47 -5.67 -0.33

    0 0 -15 -4.4 2 1

    -5/15 R3+R22/15 R3+R1

    1 0 0 -0.385 0.007 0.059

    0 1 0 -0.055 0.126 0.007

    0 0 1 0.293 -0.133 -0.067

    10/45 R2+R1

    -1/45 R2

    1/15 R3

    -0.385 0.007 0.059

    -0.055 0.126 0.007

    0.293 -0.133 -0.067

    16.25

    14.25

    13.00

    0.25

    1.00

    2.00

    =

    A x = b, x = A-1 b

    solution

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    6. Practice

    21

    31835

    7753

    36542

     z  y x

     z  y x

     z  y x

    Find a solution for the linear systems a) and b) described as follows:

    Rewrite into a matrix form: Ax=b and define A, x , and b.

    Solve using Gauss Elimination, Gauss-Jordan Elimination: [ A b].

    Solve using Cramer’s Rule.

    Find Inverse matrix: x=A-1b .

    Minors, Cofactors, Adj, Determinant.

    Gauss-Jordan Elimination: [ A I ]→[I A-1]

    Verify your answer.

     2 x1  3 x

    2   3 x

    3   2

    5 x2  5 x

    3   2

    6 x1  9 x

    2  8 x

    3   5

    a) b)

    Dec 2014

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    7. Grocery shopping (previous final)

    Mrs. Simpson has $50.00 to buy oranges, apples and mangos for

    her family. If she buys 10 oranges, 13 apples and 17 mangos the

    total cost is $45.70. If she buys 15 oranges, 12 apples and 16

    mangos the total cost would be $49.40. If she buys 9 oranges,

    16 apples and 10 mangos the total cost is $41.90. Find the price

    of each fruit. Set up a matrix equation in form of Ax =b and usethe adjoint matrix to find the matrix inverse. Show the minors

    and the cofactors, too.

    22 Dec 2014

    solution

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    7. Grocery ShoppingMrs. Simpson has $50.00 to buy oranges, apples and mangos for her family. If she buys

    10 oranges, 13 apples and 17 mangos the total cost is $45.70. If she buys 15 oranges, 12apples and 16 mangos the total cost would be $49.40. If she buys 9 oranges, 16 applesand 10 mangos the total cost is $41.90.

    x, y, and z are the price of an orange, an apple, and a mango, respectively.

    23

    90.41

    40.49

    70.45

    10169

    161215

    171310

     z 

     y

     x

    b Ax   1361016

    161211    M    6

    109

    161512    M    132

    169

    121513    M 

    75954

    4353142

    1326136

     M 

    75954

    4353142

    1326136

    754313295536

    4142136

    )(

      T 

    C  A Adj

    80613217)6(13)136(10)det(    A

    754313295536

    4142136

    806

    1

    )()det(

    11

     A Adj A A

     

    95.0

    35.1

    20.1

    90.41

    40.49

    70.45

    7543132

    95536

    4142136

    806

    11b A x   b Ax  

    90.41

    40.49

    70.45

    95.0

    35.1

    20.1

    10169

    161215

    171310Verify your answerSolve for x, y, z

    solution

    Dec 2014

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    8. Bus Trip

    On the day of the GE111 final exam, Dec 17th, some people from rural

    Saskatchewan are on a bus tour to Saskatoon. In the evening, one half of thetour participants (a mixture of men, women and children) go to the CreditUnion Center to attend a Saskatoon Blades vs. Swift Current Broncos hockeygame. The rest of the people go to TCU Place to attend the “Wizard of Oz”family musical.

    The number of women on the bus is twice the number of children, and the

    number of men on the bus is equal to twice the sum of the number of womenand children. Given that 18 people go to the Blades game, you mustdetermine the total men, women, and children that went on the bus tour.

    a) Set up necessary equations to solve this problem, and write them in matrixform. Show your work and box your answer.

    b) Use Gauss-Jordan Elimination to solve the system of equations establishedin part (a). Show your work and box your answer.

    Previous final

    24 men

    Dec 201424

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    Solution

    (a)

    childrenof #=

     womenof #=

    menof #=

    where

    0

    0

    36

    221

    210

    111

     z 

     y

     x

     z 

     y

     x

    bAx 

    (b) There are 24 men, 8 women and 4 children.

    8. Bus Trip

    Dec 201425

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    The sodium intake of 3 students was measured from the daily

    amount of pop, pizza, and donuts they ate:

    What is the sodium content of pop, pizza, and a donut?

    Calculate using

    1. Gauss Elimination

    2. Gauss-Jordan Elimination

    3. Adjoint method of matrix inversion

    4. Gauss-Jordan matrix inversion

    Problem 9: Solving for Sodium

    Dec 201426

    Student Pop Pizza Donut Sodium (mg)

    Engineer 1 2 2 1250

    Arts 2 3 2 1800

    Comm 2 2 2 1300

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    Sodium 9. Gauss-Elimination:1

    Dec 201427

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    Get rid of x in R2 and R3

    -2R1 add to R3-2R1 add to R2

    1 2 2 1250

    0 -1 -2 -7000 -2 -2 -1200

    Get rid of y in R3

    -2R2 add to R31 2 2 1250

    0 -1 -2 -700

    0 0 2 200

    Convert leading values to 1’s

    -1xR2

    ½ x R3

    1 2 2 1250

    0 1 2 700

    0 0 1 100

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    Sodium 9. Gauss-Elimination: 2

    Dec 201428

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    Convert leading values to 1’s

    -1xR2½ x R3

    1 2 2 1250

    0 1 2 7000 0 1 100

    Solve using backward substitution

    z(donuts) = 100 mg each

    y (pizza) + 2(100) = 700, pizza = 500 mg each

    x (pop) + 2(500) + 2(100) = 1250, pop = 50 mg each

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    Sodium 9. Gauss-Jordan Elimination

    Dec 201429

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    End of Gauss-Elimination

    -1xR2

    ½ x R3

    1 2 2 1250

    0 1 2 700

    0 0 1 100

    Get rid of z in R2 and R1 using R2 and R3

    -2R3 onto R2

    -2R3 onto R1

    1 2 0 1050

    0 1 0 500

    0 0 1 100

    Get rid of y R1 using R2

    -2R2 onto R1

    1 0 0 50

    0 1 0 500

    0 0 1 100

    mg/pop

    mg/pizza

    mg/donut

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    Sodium 9. Adjoint inverse method

    Given A x = b then x = A-1

    b

    Dec 201430

    2 by 2: |A| = ad - bc

    3 by 3: |A| = +a(ei-fh) – b(di-fg) + c(dh-eg)

    adj (A) = [cofactor (A)]TCij = (-1)i+j Mij

    A-1 = 1/(|A|) adj(A) | A| = a1,j C 1,j

    n

     j=1

    e f 

    h iM11 = = (ei – fh)

    a b c

    d e f 

    g h i

    A =

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    Sodium 9. Adjoint-Inverse:1

    Dec 201431

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    1 2 22 3 2

    2 2 2

    3 2

    2 2M11 = = 3(2) – 2(2) = 2

    1. Find Minors and cofactors

    Minor matrix

    2 0 -20 -2 -2

    -2 -2 -1

    2 0 -20 -2 2

    -2 2 -1

    Cofactor matrix CofactorT

    2 0 -20 -2 2

    -2 2 -1

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    Sodium 9. Adjoint-Inverse:2

    Dec 201432

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    adj(A)2 0 -2

    0 -2 2

    -2 2 -1

    A-1 = 1/(|A|) adj(A), where |A| = -2, then-1 0 10 1 -1

    1 -1 0.5

    A-1 =

    A x = b, x = A-1 b

    -1 0 1

    0 1 -1

    1 -1 0.5

    1250

    1800

    1300

    50 mg/pop

    500 mg/pizza

    100 mg/donut=

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    Sodium 9. Gauss-Jordan Inverse:1

    Dec 201433

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    Get rid of x in R2 and R3

    -2R1 add to R3-2R1 add to R2

    Get rid of y in R3

    -2R2 add to R3

    Convert leading values to 1’s

    -1xR2

    ½ x R3

    1 2 2 1 0 02 3 2 0 1 0

    2 2 2 0 0 1

    1 2 2 1 0 0

    0 -1 -2 -2 1 00 -2 -2 -2 0 1

    1 2 2 1 0 0

    0 -1 -2 -2 1 0

    0 0 2 2 -2 1

    1 2 2 1 0 0

    0 1 2 2 -1 0

    0 0 1 1 -1 0.5

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    Sodium 9. Gauss-Jordan Inverse:2

    Dec 201434

    The sodium intake of 3 studentswas measured from the daily

    amount of pop, pizza, and donuts

    they ate:

    1 2 2 12502 3 2 1800

    2 2 2 1300

    Get rid of z in R2 and R1

    -2R3 onto R2-2R3 onto R1

    Get rid of y R1 using R2

    -2R2 onto R1

    1 2 0 -1 2 -1

    0 1 0 0 1 -10 0 1 1 -1 0.5

    1 0 0 -1 0 1

    0 1 0 0 1 -1

    0 0 1 1 -1 0.5

    -1 0 1

    0 1 -1

    1 -1 0.5A-1 =

    Extract Inverse matrix

    Solve x = A-1 b

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    C. Determinants

    Dec 201435

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    1. Find the determinant

    Dec 201436

    A =

    det(A) = 3(-1)(1)(-2)(3) = 18

    Solution

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    1. Find the determinant

    Dec 201437

    Expand R6C1 as

    R6 has most ‘0’sC61 is -1, (-1)

    7 = -1

    det(A) = 3(-1) [A]

    Expand R1C5 as

    C5 has most ‘0’s

    C15 is 1, (-1)6 = 1

    det(A) = 3 (-1) [A]

    Expand R2C2 as

    C2 has most ‘0’s

    C22 is 1, (-1)4 = 1

    det(A) = 3 (-1) (1) [A]

    A =

    Expand R2C2 asC2 has most ‘0’s

    C22 is 1, (-1)4 = 1

    det(A) = 3 (-1) (1) (-2) [A]

    det of a 2x2 matrix is

    (1x1) – (-1x2) = 3

    det(A) = 3(-1)(1)(-2)(3) = 18

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    2. Determinant Conversions

    38

    A = det(A) = 2

    Using row reduction methods convert the following matrix to it’s identitymatrix. For each step calculate the determinant to show that it remains

    constant.

    Dec 2014

    Solution

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    2. Determinant Conversions

    39

    A =det(A) = 2 The value of a determinant

    remains the same if the

    elements of one row (or

    column) are altered by

    adding them to any constant

    multiple of the

    corresponding elements in

    any other row or column)

    Add -2R1 to R2

    Add -3R1 to R3

    Add 2R2 to R3

    Add -3R3 to R1

    Add -2R2 to R1

    det(A) = 2

    det(A) = 2

    det(A) = 2

    det(A) = 2

    Dec 2014

    Solution

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    More on determinants

    40

    det(A) = 2 If any 2 rows or 2 columns areinterchanged then the resulting

    determinant will be times -1

    det(A) = (-1)2 = -2

    0.5R1 If any row or column is multiplied

    or divided by a constant, the

    resulting determinant will also be

    scaled the same

    det(A) = 1

    R1 switched with R2

    Dec 2014

    Solution

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    3. Cramer’s Rule

    41

    Use Cramer’s Rule to solve for z. Show your work and boxyour answer

    4 x + y + z +w = 6

    3 x +7 y  –  z + w = 1

    7 x +3 y  – 5 z +8w = -3

     x + y + z +2w = 3

    z = 2

    Dec 2014

    Solution

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    3. Cramer’s Rule

    42

    x y z w b

    4 1 1 1 6

    3 7 -1 1 1

    7 3 -5 8 -3

    1 1 1 2 3

    Reduce C2 to ‘0’s

    except R1 (can be any

    C but C3 as that is ‘z’)

    Add -7R1 to R2

    Add -3R1 to R3

    Add – R1 to R4

    x y z w b

    4 1 1 1 6

    -25 0 -8 -6 -41

    -5 0 -8 5 -21

    -3 0 0 1 -3

    Find determinant of R1

    C2 (note that the

    cofactor is ‘-’)

    -25 -8 -6

    -5 -8 5

    -3 0 1

    -1

    C3 can be reducedAdd – R1 to R2

    -25 -8 -6

    20 0 11

    -3 0 1

    CalculatingDeterminant

    20 11

    -3 1

    -1 x 8 x (20+33) = -424

    Find det (Az)

    x y b w

    4 1 6 1

    -25 0 -41 -6

    -5 0 -21 5

    -3 0 -3 1

    Substitute ‘b’into ‘z’ column Reduce to 3x3 with ‘-1’

    -25 -41 -6

    -5 -21 5

    -3 -3 1

    -1

    Reduce to 2x2:

    C3 to ‘0’ in R2 and R3

    Add 5/6R1 to R2Add 1/6R1 to R3

    -25.8 -55.2

    -7.17 -9.83-1 (-6)

    CalculatingDeterminant

    -1x-6x-141.3 = -848

    z = det(Az)/Det(A) = -848/-424 = +2

    Find det (A)

    Dec 2014

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    D. Eigenvalues

    Dec 201443

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    1. Eigenvalues & Eigenvectors

    Dec 201444

    1 2 2

    0 1 2

    0 0 1

    K =

    2 2 0

    0 2 20 0 2

    L =

    3 4 5

    0 3 8

    0 0 3

    N =

    2 0 2

    0 2 0

    2 0 2

    M =

    4 -2

    -3 3A =

    -2 4

    3 -1B =

    -1 3

    4 -2C =

    3 -1

    -2 4D =

    0.3 0.2

    0.1 0.4E = , E = A-1

    λ1 = 6, λ2 = 1

    v1 = 1, -1, v2 = 1, 1.5

    3 2

    1 4F =

    2 00 2

    G = 0 22 0

    H = 2 -1-4 2

    J = , is J invertible?

    2,2,2

    Find eigenvalues and eigenvectors for the following Find eigenvalues

    Solution

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    1. Eigenvalues & Eigenvectors

    Dec 201445

    1 2 20 1 2

    0 0 1

    K =

    Find eigenvalues and eigenvectors for the following

    2 2 0

    0 2 20 0 2

    L =

    3 4 5

    0 3 8

    0 0 3

    N =

    2 0 2

    0 2 0

    2 0 2

    M =

    4 -2

    -3 3A =

    -2 4

    3 -1B =

    -1 3

    4 -2C =

    3 -1

    -2 4D =

    0.3 0.2

    0.1 0.4E = , E = A-1

    2, -5

    1, 1; 1, -0.8

    2, -5

    1, 1; 1, -1.33

    5, 2

    1, -2; 1, 1

    0.5, 0.2

    1, 1; 1, -0.5

    3 2

    1 4F =

    5, 2

    1, 1; 1, -0.5

    2 00 2

    G = 0 22 0

    H =

    2, 2

    1, 0; 1, 02, -2

    1, 1; 1, -1

    2 -1-4 2

    J = , is J invertible?

    4, 0

    1, -2; 1, 2

    1,1,1

    1,0,0; -1,0,0; 1,0,0

    2,2,2

    4, 0, 2

    λ1 = 6, λ2 = 1

    v1 = 1, -1, v2 = 1, 1.5

    Find eigenvalues

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    E. Complex Numbers

    Dec 201446

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    -1 =

    -2 =

    -18 =

    -5 =-75 =

    1. Complex Numbers: Basics

    Dec 201447

    i2 =

    i3 =

    i4 =

    i5 =

    i6 =

    i20 =

    i783 =

    Solution

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    -1 = i

    -2 = i 2

    -18 = 3i 2

    -5 = i 5-75 = 5i 3

    1. Complex Numbers: Basics

    Dec 201448

    i2 = -1

    i3 = -i

    i4 = 1

    i5 = i

    i6 = -1

    i20 = 1, given i4n = 1 and 20= 5(4)

    i783 = -i, given i780 i3; 783 = 195(4)+ 3

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    2. CN: Addition and Subtraction

    Dec 201449

    a) (3 + 4i) + (2 – 3i)b) (-4 – 3i) + (-2 – i)

    c) 3 – 3 √-1 + 2 + i

    d) (3a + 4i) – (-2a – i)e) (a + bi) – (c + di)

    Solution

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    2. CN: Addition and Subtraction

    Dec 201450

    a) (3 + 4i) + (2 – 3i) = 5 + ib) (-4 – 3i) + (-2 – i) = -6 -4i

    c) 3 – 3 √-1 + 2 + i = 3 - 2√-1 or 3 – 2i

    d) (3a + 4i) – (-2a – i) = 5a + 5i

    e) (a + bi) – (c + di) = a – c + (b – d)i

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    3. CN: Multiplication

    Dec 201451

    a) (√2) (√3) = √6 + 0i

    b) (√2) (√-3) = √-9 or 3√-1 or 3i

    c) (√-2) (√-3) = √2i √3i = 6i2 = - 6

    d) (4 + i) (2 + i) = 8 + 6i + i2 = 7 + 6i

    e) (5x + 2i) (3x – 2i) = 15x2 -10xi + 6xi – 4i2 = 15x2 +4 – 4xi

    f) i(2 + 5i) (3 – i) = i(6 – 2i + 15i -6i) = i(12 + 13i) = -13 + 12i

    Solution

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    3. CN: Multiplication

    Dec 201452

    a) (√2) (√3) = √6 + 0i

    b) (√2) (√-3) = √-9 or 3√-1 or 3i

    c) (√-2) (√-3) = √2i √3i = 6i2 = - 6

    d) (4 + i) (2 + i) = 8 + 6i + i2 = 7 + 6i

    e) (5x + 2i) (3x – 2i) = 15x2 -10xi + 6xi – 4i2 = 15x2 +4 – 4xi

    f) i(2 + 5i) (3 – i) = i(6 – 2i + 15i -6i2) = i(12 + 13i) = -13 + 12i

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    4. CN: Represent on a Complex Plane

    Dec 201453

    Im

    Re

    3+ 2ii(3 + 2i)

    2(3 + 2i)

    2i(3 + 2i)

    -1 -8i

    -5 + 2i1 – 8i

    6

    5i

    5i2

    5i3

    5i4

    4i10

    i(2-2i)(4+3i)

    5 10-5-10

    -5

    -10

    5

    10

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    5. CN: Conjugate and Modulus

    Dec 201454

    2 + 3i = 2 – 3i

    5 – 2i = 5 + 2i

    -5i = 5i

    12 = 12

    Find the conjugate and the modulus

    Find |z| when z =(1 + i)4

    (1 + 6i)(2 − 7i)

    |1 + i|

    4

    (1

    2

    + 1

    2

    )

    4

    |1 + 6i| |2 − 7i| 12 + 62 22 + (−7)2==

    4

    37 53=

    2 + 3i = 42 + 32 = 25 = 5

    5 – 2i = 29

    -5i = 5

    12 = 12

    Solution

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    5. CN: Conjugate and Modulus

    Dec 201455

    2 + 3i = 2 – 3i

    5 – 2i = 5 + 2i

    -5i = 5i

    12 = 12

    Find the conjugate and the modulus

    Find |z| when z =(1 + i)4

    (1 + 6i)(2 − 7i)

    |1 + i|

    4

    (1

    2

    + 1

    2

    )

    4

    |1 + 6i| |2 − 7i| 12 + 62 22 + (−7)2==

    4

    37 53=

    2 + 3i = 42 + 32 = 25 = 5

    5 – 2i = 29

    -5i = 5

    12 = 12

    Solution

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    6. CN: Division 1

    Dec 201456

    Given:

    z1 = 4 + 7i

    z2 = 2 + 5i

    Solve:

    a) z1/z2 =

    b) z2/z1 =

    c) z2/z2 = 1

    (4 + 7i) (2-5i) 8-20i+14i-35i2 43 – 6i

    (2 + 5i) (2-5i) 4-10i+10i-25i 29= =

    Find 1/z

    d) z = i

    e) z = 1 – 5i

    f) z = -i/7

    1/z = -i

    1/z = (1 +5i)/26

    1/z = 7i

    43 + 6i

    65

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    6. CN: Division 1

    Dec 201457

    Given:

    z1 = 4 + 7i

    z2 = 2 + 5i

    Solve:

    a) z1/z2 =

    b) z2/z1 =

    c) z2/z2 = 1

    (4 + 7i) (2-5i) 8-20i+14i-35i2 43 – 6i

    (2 + 5i) (2-5i) 4-10i+10i-25i 29= =

    Find 1/z

    d) z = i

    e) z = 1 – 5i

    f) z = -i/7

    1/z = -i

    1/z = (1 +5i)/26

    1/z = 7i

    43 + 6i

    65

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    7. CN: More Division

    Dec 201458

    Solve, expressing answer as a + bi

    2(1-i) (3+i)

    i

    1+i

    1

    (3+4i)2

    (1 + i)

    (1 – i)= i

    = 0.5 + 0.5i

    = = = 0.40 + 0.20i2 (8+4i)(4-2i) 20

    1 (-7 – 24i)

    (-7 + 24i) 625= = = -7/625 – 24i/625

    a)

    b)

    c)

    d)

    e)

    Solution

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    7. CN: More Division

    Dec 201459

    Solve, expressing answer as a + bi

    2(1-i) (3+i)

    i

    1+i

    1

    (3+4i)2

    (1 + i)

    (1 – i)= i

    = 0.5 + 0.5i

    = = = 0.40 + 0.20i2 (8+4i)(4-2i) 20

    1 (-7 – 24i)

    (-7 + 24i) 625= = = -7/625 – 24i/625

    a)

    b)

    c)

    d)

    e)

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    8. Polar and Exponential Coordinates

    Dec 201460

    r = |z| = (12 + 32)0.5 = 100.5; = atan (3/1) = 1.25 radians or 71.6°

    z1 = 100.5(cos (1.25) + isin(1.25)); z1 = 10

    0.5ei1.25 ; I

    r5 = 20.5; = -0.785 radians or -45°;

    r = 3.16; = 1.89 radians or 108.4°

    z2 = 3.16 (cos(1.89) + i sin(1.89)); z2 = 3.16ei1.89; II

    r6 = 20.5

    ,

    = 0.785 radians or 45°;

    r3 = 3.16; = 4.39 radians or 252°; III

    r4 = 3.16; = -1.25 radians or -71.6°’ IV

    Find r, arg, polar and exponential coordinates, and indicate which quadrant

    Find by polar and represent as exponential and Cartesian forms: z1 z6; z1/z6

    g) z1 z6 = 4.47ei2.03 h) z1/z6 = 2.24e

    i0.46

    a) z1 = 1 + 3i

    b) z2 = -1 + 3i

    c) z3 = -1 - 3i

    d) z4 = 1 - 3i

    e) z5 = 1 – i

    f) z6 = 1 + i

    Solution

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    8. Polar and Exponential Coordinates

    r = |z| = (12 + 32)0.5 = 100.5; = atan (3/1) = 1.25 radians or 71.6°

    z1 = 100.5(cos (1.25) + isin(1.25)); z1 = 10

    0.5ei1.25 ; I

    r5 = 20.5; = -0.785 radians or -45°;

    r = 3.16; = 1.89 radians or 108.4°

    z2 = 3.16 (cos(1.89) + i sin(1.89)); z2 = 3.16ei1.89; II

    r6 = 20.5

    ,

    = 0.785 radians or 45°;

    r3 = 3.16; = 4.39 radians or 252°; III

    r4 = 3.16; = -1.25 radians or -71.6°’ IV

    Find r, arg, polar and exponential coordinates, and indicate which quadrant

    Find by polar and represent as exponential form: z1 z6; z1/z6

    g) z1 z6 = 4.47ei2.03 h) z1/z6 = 2.24e

    i0.46

    a) z1 = 1 + 3i

    b) z2 = -1 + 3i

    c) z3 = -1 - 3i

    d) z4 = 1 - 3i

    e) z5 = 1 – i

    f) z6 = 1 + i