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8/9/2019 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
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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
Dec 20149
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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
Previous finalDec 201410
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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
3
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)
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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
C
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
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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