Unit 5 Homework Problems MAT150 College Algebra
© Andrew McKintosh 2010
All problems should be done WITHOUT a calculator unless otherwise indicated!!!
(Note: using a calculator to HELP you with basic computations is fine!)
Day 57 Homework: Topics: Gaussian Elimination with Back Substitution for 3x3 systems
Supplemental Reading: There are resources out there on this BUT everyone seems to do it slightly differently
(particularly notation wise) SO, I recommend that you just refer heavily to your class notes!
1. Solve the following systems of equations using Gaussian Elimination with back substitution. Be sure to SHOW
your row operations between steps (on top and bottom of your arrows, just like I did in class). If you find that your
system has an infinite number of solutions be sure to find the infinite generating set.
A)
 x  9 y  2 z  50

 6 x  2 y  9 z  14
5 x  6 y  4 z  43

 x  9 y  2 z  50
 x  9 y  2 z  50
 x  9 y  2 z  50
R3  R2
39 R2



 
56 y  3z  286  
2184 y  117 z  11154   2184 y  117 z  11154
R3  5 R1
56 R3
  39 y  6 z  207
  2184 y  336 z  11592

 219 z  438



So from the last "row" we can see that  219 z  438  z  2 now if we "back substitute" this answer for z into an "earlier"
R2  6 R1
(easier) version of "row" two , 56 y  3z  286  56 y  3  2   286  56 y  6  286  56 y  280  y  5.
Finally, substituting BOTH of these results for z and y into the first "row" we obtain...
x  9 y  2 z  50  x  9  5   2  2   50  x  45  4  50  x  49  50  x  1
So the answer to our system is 1, 5, 2 
B)
1  x  6 y  z  40
4 x  5 y  6 z  26 R  R  x  6 y  z  40 R 5 R  x  6 y  z  40
R2
1
3
3 
2
1



27 y  12 z  186  
9 y  4 z  62
5 x  3 y  7 z  14  5 x  3 y  7 z  14 R 

4R
 x  6 y  z  40
4 x  5 y  6 z  26 3 1  19 y  10 z  134
 19 y  10 z  134




 x  6 y  z  40
 x  6 y  z  40
R3  R2
19 R2


   171y  76 z  1178    171y  76 z  1178
9 R3
 171y  90 z  1206

 14 z  28


So from the last "row" we can see that  14 z  28  z  2. Substituting that into an "earlier" version of "row" two
27 y  12 z  186  27 y  12  2   186  27 y  24  186  27 y  162  y  6. Substituting BOTH of these
into the first "row" we obtain x  6 y  z  40  x  6  6   2  40  x  36  2  40  x  38  40  x  2
So the answer to our system is  2, 6, 2 
 x  2y  z  3

C)  x  4 y
 1
2 x
 4 z  12

D)  x  y  z  7

 3x  2 y  z  3
  x  6 y  3 z  25

x  2 y  z  3

  2y  z  2
R3  2 R1
 4 y  2z  6

R2  R1
x  2 y  z  3

  2y  z  2
NO SOLUTION

02

R3  2 R2
x  y  z  7 R R x  y  z  7
3
2


   5 y  2 z  18    5 y  2 z  18 Infinite generating set is...
R3  R1
  5 y  2 z  18

00


2
18
5 y  2 z  18  5 y  2 z  18  y   z 
5
5
2
18
3
17
x  y  z  7  x   z   z  7  5 x  2 z  18  5 z  35  5 x  3 z  17  x   z 
5
5
5
5
17 2
18 
 3
 z  , z  , z
5
5
5 
 5
R2  3 R1
Day 58 Homework: Topics: Gaussian Elimination with Back Substitution
© Andrew McKintosh 2010
AND Gauss-Jordan Elimination
Supplemental Reading: There are resources out there on this BUT everyone seems to do it slightly differently
(particularly notation wise) SO, I recommend that you just refer heavily to your class notes!
1. Solve problems A and B below using the Gauss – Jordan method and use EITHER Gaussian Elimination OR the
Gauss – Jordan method for problems C and D. Be sure to SHOW your row operations between steps (on top and
bottom of your arrows, just like I did in class). If you find that your system has an infinite number of solutions be
sure to find the infinite generating set.
A)
 3 x  y  z  3

 2 x  3 y  z  8
2 x  y  2 z  2


x

R1  2 R2

 
R3  5 R2




 x  2 y  2z  5

  2 x  3 y  z  8
2 x  y  2 z  2

R1  R2

x
7
R3 
17 





4
1
z
7
7
5
18
y z 
7
7
17
34
z
7
7

x  2 y  2z  5


 
R3  2 R1


R2  2 R1
4
1
z
7
7
5
18
y z 
7
7
z  2
 7 y  5 z  18
5 y  6z  8

x

 
5
R2  R3 
7

1
4
R1  R3
7
y
4
So the solution to our system is 1, 4, 2 
z  2
B) 
x  y  2z  7
3z  6
x  y  2z  7 R R x 
R2  R1
1
2



y  z  1
  x  2 y  z   8 R
 y  z  1 R 

6 R
3  2 R1
 2x  4 y  z  1
 6 y  3z  15 3 2 
 9 z  9



3
x
R1  3 R3

 
y
 2 So the answer to our system is  3, 2,1
R2  R3

z  1

C)  x  2 y  z  4

4 x  3 y  z  8
5 x  y
 12

x  2 y  2z  5

5
18


y z 
7
7

5 y  6z  8

1
 R2
7
3z  6
x 


y  z  1

z 1

1
 R3
9
x  2 y  z  4 R R x  2 y  z  4
3
2


   11y  5 z  8    11 y  5 z  8 must find infinite generating set
R3  5 R1
  11y  5 z  8

00


5
8
11 y  5 z  8  11 y  5 z  8  y  z 
11 11
8
10
16
5
x  2 y  z  4  x  2  z    z  4  x  z   z  4  11x  10 z  16  11z  44
11
11
11
11


R2  4 R1
 11x  z  28  x 
D)  x  3 y  4 z  1

 3x  4 y  5 z  3
 x  8 y  11z  2

1
28
z
so our answer is
11
11
x  3y  4z  1

   5y  7z  0
R3  R1

5y  7 y  1

R2  3 R1
28 5
8 
1
 z  , z  ,z
11 11 11 
 11
x  3y  4z  1

   5y  7z  0
NO SOLUTION!

0 1

R3  R2
Day 59 Homework: Topics: Gaussian Elimination with Back Substitution
© Andrew McKintosh 2010
AND Gauss-Jordan Elimination
Supplemental Reading: There are resources out there on this BUT everyone seems to do it slightly differently
(particularly notation wise) SO, I recommend that you just refer heavily to your class notes!
1. Solve problem B using the Gauss – Jordan method and do the others using EITHER Gaussian Elimination with
Back Substitution OR the Gauss – Jordan method. Be sure to SHOW your row operations between steps (on top and
bottom of your arrows, just like I did in class). If you find that your system has an infinite number of solutions be
sure to find the infinite generating set.
A)  5 x  y  3z  2
 11
 11
 11
 x  4y
 x  4y
 x  4y
R1  3 R3
R2  5 R3
R2  9 R1




 3
 9 x  2 y  5 z   7  9 x  2 y  5 z   7 R 
 34 y  5 z  92    y
3  2 R1
 2x  y  z  3
 2x  y  z  3
  7 y  z  19
  7 y  z  19




This problem is unusual as we can now solve the system. Take the second "row" and solve for y  y  3
Now substitute that into the first and third rows to find x and z respectively.
x  4y  11  x  4  3 =11  x  1
 7 y  z  19  7  3   z  19  21  z  19  z  2
and
So the solution to our system is  1,3, 2 
B)  3x  y  z   5
 x  2y  4 z   6 R  2 R  x  2y  4 z   6  R  x  2y  4 z   6
R1  R3
2
1
2




2
x

3
y

z

11

 y  9 z  23  
y  9 z  23

 2 x  3 y  z  11 R 

3  2 R1
2 x  y  3 z   1
2 x  y  3z   1


5 y  11z  13
5 y  11z  13




 14 z  40 1 R3  x
 14 z  40
 2
x
x
R1 14 R3
R1  2 R2
34 


 
y  9 z  23  
y  9 z  23  
y
4
R3  5 R2
R2  9 R3



34 z  102
z 3
z 3



So the solution to our system is  2, 4,3
 x  2y  z  8

C) 2 x  y  z  4
8 x  y  z  2

D)
 2 x  y  3z  1

 x  4y  z  6
 4 x  7 y  z 13

x  2 y  z   8

   5 y  3z  20
R3 8 R1
  15 y  9 z  66

R2  2 R1
 x  4y  z  6

 2 x  y  3z  1
4 x  7 y  z 13

R1  R2
x  2 y  z   8

   5 y  3z  20 No Solution!

06

R3  3 R2
 x  4 y  z  6
1
x  4 y  z  6
R2  
9 
5
11

  9 y  5 z  11  
y z 
R3  4 R1
9
9
 9 y  5 z  11


 9 y  5 z  11
R2  2 R1
11
10

 z
x
9
9

R1  4 R2
5
11

 
y z 
So we can see that we have an infinite number of solutions since we got 0 = 0
R3  9 R2
9
9

00



in the bottom "row". Note that I chose to do Gauss-Jordan here to demonstrate that when you end up with
an infinite number of solutions Gauss-Jordan allows you to almost immediately come up with the infinite
11
10
11 10
z
 x z
9
9
9
9
11 
 11 10 5
So our solution is  z  , z  , z 
9 9
9 
9
generating set. x

5
11
5
11
y z   y  z
9
9
9
9
Day 60 Homework: Topics: Graphing linear inequalities
and Systems of linear inequalities.
Supplemental Reading:
1. http://www.purplemath.com/modules/ineqgrph.htm
2. http://www.purplemath.com/modules/syslneq.htm
© Andrew McKintosh 2010
1. Sketch the graph of each individual inequality.
A. y  2x  4
B) 2x  3 y  6
C)
 x  1
2
2. Sketch the graph for the following systems of inequalities.
3x  2 y  6

A.  x  4 y  2
 2x  y  3

x  2 y  4
x  y  1

B. 
x  0
 y  0
3. List the “vertices” in each of your graphs in 2A and 2B above.
 2,0  0,3  2, 1
A. ______________________________________
 0,0  0, 2  2,1 1,0
B. ___________________________________
  y  2  4
2
Day 61 Homework: Topics : Linear Programming
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.purplemath.com/modules/linprog.htm
1. Sketch the region determined by the constraints. Then find the minimum and maximum values of the objective
function and where they occur, subject to the indicated constraints.
A) Objective function: z = 6x + 10y
Constraints:
x  0

y  0
2 x  5 y  10

z  6 x  10 y
 0, 0   z  6  0   10  0   0  Minimum value!
 0, 2   z  6  0   10  2   20
 5, 0   z  6  5  10  0   30  Maximum value!
B) Objective function: z = 4x + 5y
Constraints:
x  0
y  0


x  y  8
3 x  5 y  30
z  4x  5 y
8, 0   z  4 8  5  0   32  Minimum value!
 5,3  z  4  5  5  3  35
10, 0   z  4 10   5  0   40  Maximum value!
(Day 61 Homework continued)
© Andrew McKintosh 2010
2. A manufacturer wants to maximize the profit for two types of nuts that he sells. A box of Peanuts yields a profit
of $1.50 per box and a box of Cashews yields a profit of $2.00 per box. Market tests and available resources have
indicated the following constraints……
*** The combined production level should not exceed 1200 boxes per month
***The demand for a box of Cashews is no more than one half the demand for a box of Peanuts.
***The production level of a box of Peanuts should be less than or equal to 600 boxes plus three times the
production level of a box of Cashews.
Find the number of boxes of Peanuts and the number of boxes of Cashews that will maximize the profit. Let x = the
number of boxes of Peanuts and y = the number of boxes of Cashews. Also, let each marking on the x and y axes of
the graph below represent 100 boxes.
Objective Function (Profit) P  1.5 x  2 y
Constraints

x  0

y  0

 x  y  1200

1
y  x
2

 x  600  3 y
1200
600
800, 400
1050,150
600
1200
P  1.5 x  2 y
 0, 0   P  1.5  0   2  0   $0
800, 400   P  1.5 800   2  400   $2000 
1050,150   P  1.5 1050   2 150   $1875
 600, 0   P  1.5  600   2  0   $900
Max Profit!!!!
The profit will be maximized with 600 boxes of Peanuts and 600 boxes of Cashews
Day 62 Homework: Topics: Matrix Size (or order), Augmented Matrices,
© Andrew McKintosh 2010
Gauss-Jordan with Matrices
Supplemental Reading:
1. http://www.purplemath.com/modules/matrices.htm (first two pages only)
2. http://www.purplemath.com/modules/mtrxrows.htm (you can read this BUT realize their notation is slightly
different than what we have been doing)
1. Write the size (or order) of the following matrices at the lower right corner of each matrix.
1 3 5 
 2 7 1

 2 x3
1 3 
5 2 


7 1  3 x 2
3 
 2
 
1  3 x1
0
 2
 1  2

2 x2
1
2 31x3
2
3
2. Write the following system as an augmented matrix…
 x  2y  z  7
 1 2 1


 3x  y  2 z  11   3 1  2
2 x  y  7 z  10
 2 1  7

7
 11
10 
3. Write the following augmented matrix as a system of equations…
3
 1 2
 0 3 2

 1 4
0
 5   x  2 y  3x  5

1   
3y  2z  1

3  x  4 y
 3
4. Use matrices and the Gauss-Jordan method to solve the following systems of equations.
A)  2 x  3 y  3
3  R1  R2  1  1
 2 3

 

2
 1
2
 x  2 y  1  1
 1
R1  R2
0
3
 1
 
So our solution is  3,1
1
1 
 0
2  R2  R1  1
 
 1 
 0
1
1
2
1 
B)
3
 x  2 y  3z  11
 1 2

 2 3

2
x

3
y

5
z

1

5


 3x  3z  z  4
 3 3  1

0
 19
 35  1 R3 1
1
R1  2 R2
89

 0
1
 11
 23  0
R3  9 R2
0
0
0
89
178 
11
1
4 
0
1
 0
R3  3 R1
0
R2  2 R1
 19
1
 11
0
1
2
1
9
 35 
 23 
2 
11 
1
 R2
23   0
 10
 29 
0
0
0
1
R1 19 R3

 0
1
0
R2 11R3 
0
0
1
3
11
1  11
9  10
11 
 23 
 29 
3
 1 So our solution is  3, 1, 2 
2 
Day 63 Homework: Topics: Solving systems using Matrices.
© Andrew McKintosh 2010
Supplemental Reading: (None) We will be following “similar” steps to what we did in our notes (and homework)
on days 57 – 59 BUT we will create an augmented matrix first, do the work without writing all of the variables
during every step and re-insert the variables at the end.
1. Use matrices and the Gauss-Jordan method to solve the following systems of equations.
A)
 4 x  2 y  3z  5

8 x  y  z   5
 2x  y  2z  5

 2
R1  R2
  0
R3  4 R2
 0
 2 1 2
  8  1 1
 4  2  3
0 1
1
3
0
5
5
 5 
5 
R1  R3
0
5 
15 
 2
  0
 0
1
R3
5
 2 1 2
  0 3 9
R3  2 R1 
 0  4  7
5  1 R2  2 1 2
3
15    0 1 3
 0  4  7
 5 
R2  4 R1
0 1
1
3
0
1
0
5 
3 
 2
R1  R3
  0
R2  3 R3
 0
0
0
1
0
0
1
5
5 
 5

1
3  1 R1 
2 
 4    0
 0
3 


0
0
1
0
0
1
3

So our solution is  , 4,3 
2

B)
3
 x  2 y  3 z  11
 1 2


2 3
 4 x  2 y  3z  4   4
3 x  3 y  z  4

1

 3  3

 1

R1  2 R2
  0
R3  3 R2 

 0

7

1
3 2 
 R3 
5
20  5 
1 
 0

2
3


5
 0
0
9


2
 73 7 18 
So our solution is  ,  , 
 15 3 5 
C)
0
2

2 1
 x  2 y  z  8
 1

 2 1
2
x

y

z

4

1


8 x  y  z  2
 8  1  1


 1

R1  2 R2
  0
R3 15 R2 

 0

1
5
3
1 
5
0
0
0
11
3
 1 2
R2  4 R1


4    0  6  15
R3  3 R1
 0 3
 4 
10
0
2
1

0
 8
4 
 2 
5
2
1
7
3

20 
3

18 
5 


 1

R1  2 R3
  0
5 
R2  R3
2 
 0

2 1
 1
  0  5
3
R3  8 R1
 0 15  9
R2  2 R1
 1 2

 0 1

 0 3

73 
0
15 

7
0 
3

18 
1
5 
11
 40 
29 
1
 R2
6
0
1
0
8 
20 
 66 
 1

 0

 0

1
 R2
5
11 
20 
3
29 
3
5

2
10
2 1
3
5
15  9
1 
8 

4 

 66 

 2

 4  So our system has NO SOLUTION !


6

D) 3x  4 y  z  6
6
3  4  1


1
 1 
2 x  y  z   1   2  1
4 x  7 y  3z  18
 4  7  3
18

1 1  3
2
7
0
1
1
R2
R1  3 R2
5
 0 1
1
 2    0 1
1
R3  5 R2
0 5
0 0
5  10 
0
So our solution is   z  1,  z  2, z 
1  3  2
  2  1
1
 4  7  3
R1  R2
1
 2 
0
7
 1 
18
1  3  2
 0 5
5
R3  4 R1 
 0 5
5
R2  2 R1
y  z  2  y   z  2
7
 10 
 10
x  z  1  x  z 1
3
2

 4
3


Day 64 Homework: Topics: Equal Matrices,
© Andrew McKintosh 2010
Addition and Subtraction of Matrices, Multiplication of Matrices.
Supplemental Reading: 1. http://www.purplemath.com/modules/matrices3.htm
2. http://www.purplemath.com/modules/mtrxadd.htm
3. http://www.purplemath.com/modules/mtrxmult.htm
4. http://www.coolmath.com/algebra/24-matrices/02-adding-subtracting--01.htm
5. http://www.coolmath.com/algebra/24-matrices/03-scalar-multiplication-01.htm
6. http://www.coolmath.com/algebra/24-matrices/04-multiplying-matrices-01.htm
1. Find x, y and z
3
0

a b 
2. M  

c d 
2y 5
7  3

1
3x  2   4  2 z
e f
N 
h i
7
1
7 
 1 
g
j 
A) Find M  N
B) Find N  M
 a b  e f g 
c d    h i j 

 

 ae  bh af  bi ag  bj 
ce  dh cf  di cg  dj 


e f g   a b 
 h i j   c d 

 

Can't do it! (# of columns in N  # of rows in M)
 1 2 
3. A  

 3 5
 2  3
B
4 
0
0 2
1
C
2

3
1 

 1 3 
D   2 5
 1 2 
I. Find 2A – 3B
4  6  9   8 13
 1 2 
 2  3  2
2
 3





4   6 10  0 12   6  2 
 3 5
0
II. Find A  B
 1 2   2  3  2 11

 3 5   0
4   6 11

 
III. Find B  A
 2  3  1 2   11  11


0
4   3 5  12
20 

IV. Find B  C
0 2   4
9 1
 2  3 1
0
   2  3 1   8  12 4 
4

 
 

V. Find C  B
0 2  2  3
1
 2  3 1   0
4 

 
Can't do it! (# of columns in C  # of rows in B)
VI. Find C  D
 1 3 
0 2 
7
1
  1
 2  3 1    2 5   7  7 

 


 1 2 
VII.
 1
 2

 1
Find D  C
3
 5  9 1
0 2 
1


5  
  12  15 9
2

3
1

 
2 
 5  6 4 
Day 65 Homework: Topics: Inverse Matrices
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.purplemath.com/modules/mtrxinvr.htm (first page only) (remember, their notation is slightly
different than what we have been using)
2. http://www.coolmath.com/algebra/24-matrices/06-inverse-matrices-01.htm
3. http://www.coolmath.com/algebra/24-matrices/07-solving-systems-using-matrices-01.htm
1 3 
1
1. A  
 Find A
2
4


1 3
2 4

1

0

3

2

2
A1  

 1 1

2 
1 0  R2  2 R1 1
3
 
0 1
0

2


1
 
0

3
2

1
1  
2 
2
0
R1  3 R2
1
1
 R2
2
1 0
 2 1
0
1
1  
2
3
1
1
2. Use your result from problem 1 to solve the following system.
 x  3y  9

 2 x  4 y  14
1 3  x 
 9
 2 4    y   14 

  
 
AX  B
A1 AX  A1 B
IX  A1 B
3

2
x  
2 9
  
 y  
   1  1  14 

2 
 x  3 
 y    2  So our answer is  3, 2 
   
2
5
3. M  
3

1 


1
 
0

5
R3
11
4.
1

0
5

3
5

11
11 
2
5
1
 6  9
N 
6 
 4
 6 9
 4
6

M 1
1
2
R 
1 0  5 1 1

5

0 1 
3  1
2
5
3  1


Find
Find
1

0
5

0 1 

1
 
0

2
R1  R2
5
0
1
2

1
5
 
11
0 

5
1
2
11
11 

3
5

11
11 
R2  3 R1
1

0
5

3

1

5
M 1
2
1
11
11 


3  5 
11
11 
N 1
1
3
R 
1 0 6 1  1 

2

0 1
6
 4
N 1 DOES NOT EXIST !
X  A 1 B
1

0
6

0 1 
3

 1 2
 
 0
0

R2  4 R1
1

0
6

2 
1
3 
Day 66 Homework: Topics: Finding Inverses for 3x3 Matrices
Supplemental Reading: (None)
1  1 0 
1. A  1
0  1  Find
6  2  3
1  1 0
1
0 1

6  2  3
0 1
 1


0
1 1
R3  4 R2 
 0
0
1
 2  3 1
1
A   3  3 1
 2  4 1 
R1  R2
A1
 1 1 0
  0
1 1
R3  6 R1 
 0
4 3
0
1 0
 1
R1  R3

1
1 0    0
R2  R3
 0
 2  4 1 
1 0 0
0 1 0 
0 0 1
A1
2
0
 1
 3 1
2

 2
3 2
2
0
 1
  0  7
2
R3  2 R1
 0
7 2
 1
1
 R2 
7
 0

 0

A1
2
0
1
2
7
7 2
3
7
2
1 
DOES NOT EXIST !
1 0 0
 1 1 0 
 6 0 1
R2  R1
2
0
 1

2. A   3  1
2  Find
 2 3  2 
1 0 0
0 1 0 
0 0 1 
© Andrew McKintosh 2010
R2  3 R1
0 0

1

0
7

0 1

 1

R1  2 R2
  0
R3  7 R2 

 0

0
0
1
0
0
1
 2  3 1
 3  3 1
 2  4 1
1 0 0
 3 1 0 
2 0 1 
4
1
2
0
7
7
7
2
3
1
1 

7
7
7
0
0
1 1

0

0


1

Day 67 Homework: Topics: Finding Determinants for 2x2 and 3x3 matrices.
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.purplemath.com/modules/determs.htm (Note: on the second page they show a “short cut”
specifically for 3x3 determinants BUT that method does not work for 4x4 and bigger SO, please do the 3x3
problems using the “expanding by cofactors” method that I taught…see the next reading!)
2. http://www.purplemath.com/modules/minors.htm
1. Find the determinant of each matrix below…
 2 3
A) 

 1 5 
6
 2
B) 

 1  3 
0 0
C) 

0 0
1 0 
D) 

 0 1
 5 2    3 1
 2  3   6  1
 0  0    0  0 
11   0  0 
10  3
6  6
00
1 0
7
0
0
1
2. Find the determinant of the matrix as indicated in each part.
2 1
0
3  1 2 


 4 0 1
A) Find the determinant by expanding about row 2
3  C21  1 C22  2  C23
3  1
2 1
3  1
3
M 21  1 1
2 1
0 1
 1 1
2 2
4
M 22  2  1
0 1
4 1
 2  1
5
23
M 23
0 2
4 0
3  1 2   11 4   2  1 8 
6  4  16
14
B) Find the determinant by expanding about column 2
2  C12  1  C22  0  C32
2  1
1 2
2  1
3
M 12  1 1
3 2
4 1
 1 1
2 2
4
2  1 5   11 4 
10  4
14
M 22  0
0 1
4 1
Day 68 Homework: Topics: Cramer’s Rule (using Determinants to solve systems)
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.purplemath.com/modules/cramers.htm
2 x  3 y  1
1. Solve 
by using Cramer’s Rule if possible (if not then use another method)
 x  3y  5
D
2 3
  2  3   31  6  3  9
1
3
1 3
Dx 
Dy 
x
 1 3   3 5   3  15  18
5
3
2 1
  2  5   11  10  1  9
1 5
Dx 18

2
D
9
y
Dy
D

9
1
9
So our solution is  2,1
3 x  2 y  7
2. Solve 
by using Cramer’s Rule if possible (if not then use another method)
 2 x  3 y  4
3 2
  3 3   2  2   9  4  13
2
3
D
Dx 
Dy 
x
7 2
4
3
3
7
2 4
  7  3   2  4   21  8  13
  3 4    7  2   12  14  26
Dx 13
 1
D 13
y
Dy
D

26
 2
13
So our solution is 1, 2 
2 x  2 y  7
3. Solve 
by using Cramer’s Rule if possible (if not then use another method)
3x  3 y  4
D
2 2
  2  3   2  3  6  6  0
3 3
So Cramer's Rule won't work!!!!
* 3

2 x  2 y  7  
6 x  6 y  21


*2
6 x  6 y  8
3x  3 y  4  

0  29 No Solution !
 6x  2 y  8
4. Solve 
by using Cramer’s Rule if possible (if not then use another method)
9 x  3 y  12
6 2
D
  6  3   2  9   18  18  0 So Cramer's Rule won't work!!!!
9
3

 6x  2 y  8

9 x  3 y  12
*3
  18 x  6 y  24

*2
18 x  6 y  24
 
0  0 we must find the infinite generating set !
6x  2 y  8  6x  2 y  8  x 
1
4
y
3
3
4 
1
 y  , y
3 
3