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Self-Assessment Exam C
SYSTEMS, MATRICES AND DETERMINANTS
3x

1. Use the elimination method to solve the system  x
5 x

3 3x  4y  2z  25
 9x  12y  6z


4  x  3y  4z  11   4x  12y  16z
 20x  12y  24z
4 5x  3y  6z
3

 4 y  2 z  25
 3 y  4z
 11
 3 y  6z
3
 75 eliminate y
 44
 12

13x  10z  31

 29x  30z  63
3 13x  10z  31
39x  30z  93
156
39
 

 68x  156  x  

68
17
 29x  30z  63
 29x  30z  63
2
 39 
13x  10z  31  13    10z  31  507  170z  527  170z  20  z 
17
 17 
39
47
 2
 3y  4   11  
 3z  11  47  51z  187
17
17
 17 
234 78
 51y  234  y 

51 17
39
78
2
x  , y  , z 
17
17
17
x  3y  4z  11  
 3z  68
5 x

2. Use the elimination method to solve the system 
 2 y  5 z  71
7 x  3 y
 25

 3z  68
 3z  68
 3z  68
 5x
 5x
 5x



 2y  5z  71   3 
 2y  5z  71  
 6y  15z  213

7 x  3y
14x  6y
 25
2 7 x  3y
 25
 50


 5x  3z  68  5  5x  3z  68   25x  15z  340




263
14x  15z 263
 14x  15z
14x  15z 263
 11x  77  x  7
5x  3z  68  35  3z  68  3z  33  z  11
 6y  15z  213   6y  165  213  6y  48  y  8
x  7, y  8, z  11
-1
3. The sum of three numbers is 86. If the third number is subtracted from twice the sum of the
first two numbers, the result is 148. The first number is 14 less than three times the second
number. Find the numbers. Show the work.
Solution: let the numbers be: first = x, second = y and third = z
x + y + z = 86
2(x + y) –z = 148
x = 3y – 14
 x  y  z  86
 x  y  z  86
3x  3y  234


 2x  2y  z  148   2x  2y  z  148  
 x  3y  14
 x  3y
 x  3y
 14
 14


 4x  220  x  55
x  3y  14  55  3y  14  3y  69
 y  23
x  y  z  86  55  23  z  86  78  z  86  z  8
The numbers are 55, 23 and 8
4. Write the augmented matrix for the system:
 3 y  z  4w  10
 x
 3x  2 y  3z  10w  10


5y  7z
8w
 26

 3x  y  5 z
4
answer:
3 1
 1
4 10 


 3  2 3 10  10 
 0
5  7 8 26 


5
0
4 
  3 1
5. The row-reduced echelon form for the matrix above is
1 0 0 0  2
0 1 0 0 3 


0 0 1 0  1  , give the solution of the system

1 
0 0 0 1 2 
 x  2 ___
 y  3 ___

 z  1 ___

1
 w  ____
2

6. Find the missing numbers for the given product of matrices.
 4 1  2
3 
5 3  1 
 31 12
 4  4 3   4 3 4    3  2  27




1 2  1
7 5
 3  5
7. IF A= 
and B= 

 , show that A and B are inverse matrices.
4 3
 4 7 
7 5  3  5 1 0
4 3  4 7  = 0 1 They are inverses because their product is the identity matrix.



 
-2
8. Write the augmented matrix for the following system of equation:
 x  y  2 z  w  11
3x  y  z  2w  16


2 y  3z  w  9

2 x  y
3w  14
1
1  1 2
3 1  1 2
Answer: 
0 2  3  1

0
3
2 1
11 
16 
 9

14 
9. Use the elementary row transformations to convert the matrix in exercise 8 to row reduced
echelon form. Give the solution of the system.
1 11 
1
11 
1  1 2
1  1 2
 3 1  1 2 16   first row  3 and add to second row 0 4  7  1  17 




 0 2  3  1  9
 first row  2 and add to fourth row 0 2  3  1  9 




0
3 14 
2 1
0 3  4 1  8 
1 11 
1  1 2
 0 1  3  2  9

  fourth row by  1 and add to second row 
 0 2  3  1  9


 0 3  4 1  8
1 0  1  1 2 
add second row to first row
0 1  3  2  9

  second row by  2 and add to third row 
0 0 3
3
9 
 second row by  3 and add to fourth row 

7 19 
0 0 5
1 0  1  1 2 
 0 1  3  2  9

 divide third row by 3  
0 0 1
1
3


7 19 
0 0 5
1 0 0 0 5 
1 0 0
Add third row to first row
0 1 0 1 0
0 1 0
divide fourth


  third row by 3 and add to second row


 0 0 1 1 3
0 0 1
row by 2
 third row by  5 and add to fourth row 


 0 0 0 2 4
0 0 0
1
 fourth row by  1 and add to second row 0

 fourth row by  1 and add to third row 0

0
-3
5 
 x5

 y  2
1 0 0  2



0 1 0 1
 z 1

 w  2
0 0 1 2 
0 0 0
0 5
1 0
1 3

1 2
10. Use a calculator to find the product of the following matrices:
 1 3
2  3 4  
  20 14 
7 2  1    2 4   0 24

 


 3 5
2 8
11. Use a calculator to find the inverse of the matrix 

1 6 
Without a calculator:
2 8
Step 1. Augment the 
 by the 2  2 identity matrix
1 6 
 2 8 1 0


1 6 0 1
Step 2. Convert the left 2  2 matrix to the identity matrix by using the elementary row
transformations.
 2 8 1 0
Interchange the two rows of the augmented matrix 
 to obtain
1 6 0 1
1 6 0 1
1 6 0 1 

   first row by  2 and add to second row 

 2 8 1 0
 0  4 1  2
1 6 0 1 
1 1
 Divide the second row by  4 , 
0 1  4 2 



3
1 0 2
  the second row by  6 and add to first row 
1
0 1 
4

 3

The inverse matrix is  2
1

 4
 3

 2 8   2  2
Check : 


 1 6   1 1 
 4 2 

 2
1 

2 
1 0
0 1


-4

 2
1 

2 
12. Use a calculator to find the value of the determinant of the following matrix:
5
1
2  1 3
1 0
2  3 5 

7 2
4
3
1


1 6  3 2  4 
2  1 5
0
2 
answer: ___2162____________
1 2  1
2 
13. Find the inverse of the matrix  3 1
 2  1 0 
1 2  1
3 1
2 

 2  1 0 
1
1
 2
 15 15
 4
2

 15 15
 1 1
  3 3
1 
3 
1
 
3
1
 
3
14. Find the value of the determinant
3 5
3
7
3 5
3
7
= (-3)(7) – (3)(-5) = -21 + 15 = -6
. Do not use a calculator. Show the work.
Answer: -6
15. Use Cramer’s rule to find the value of z in the system:
8
2 x  3 y  z

 x  4 y  2 z  3
2 x
 4z  1

Use a calculator to find the values of the determinants involved.
2 3
1
z
4
8
3
2 0
1
Dz


2 3 1
D
1 4 2
2
0
 35
48
4
-5
16. Find the product of the following matrices. You can use a calculator.
 8 21 
 2 3 1 0  1 3  11 12

  1 4 3 4  2 5   11 16
 7 17 


 
17. Find the inverse of the matrix A. Give answer in fractional form. Use a calculator.
 2 1 0
A=  1 3 2
 1 0 1 
 2 1 0
  1 3 2


 1 0 1 
1
1
 1
 3 9
 1
2

9
 3
 1 1
  3 9
2 
9 
 4

9 
7 
9 
18. Use elementary row transformations to change the given matrix to an equivalent matrix
with a11  1 and zeros in every entry below a11  1
 1
 3 6  12  3

 4 2

1
0   Add the third row to first row  4

 2
  2 1
2
 4

 1

  first row by 2 and add to third row  4
 0

7
 10
2
1
15
 18
 1 7

  first row by  4 and add to second row  0  26
 0 15

 10
41
 18
7
 10
2
1
1
2


0 
 18 

7


28 
 18 

7
8
2 x  3 y  z

19. Write the augmented matrix for the system:  x  4 y  2 z  3
2 x
 4z  1

8
2  3 1
ANSWER: 1 4  2  3
 2 0
4
1 
-6
7 

0 
4 

20. Use a calculator to find the solution of the system below by using the row reduced
echelon form of the matrix.
x
 2 y  z  2w  3

 2x
 y  3z  4w  9


2
 3 x  2 y  z
 2 x  3 y  2 z  4w  10
a) Augmented matrix:
2 1 2
3
1
 2 1 3
4  9

R
 3  2 1
0
2


3
2  4 10 
2
Row reduced echelon form of the matrix in part a)
 x  3
1 0 0 0  3 

0 1 0 0 4 
4
 y




0 0 1 0 1 
z
1


1
1
 w 
0 0 0 1  
2

2

20. Use determinants to find the solution of the following system:
5
 3x  2 y  5 z

 4y
1
 x
 2 x 6 y
3z  12

2 5
3
D 1
0 112 , D x  1
3
 12
4
2
6
3
5
5
Dy  1
1
0  56,
 2  12 3
x
2 5
5
4
0 336
6
3
3
Dz  1
2
2
5
4
1
6
 12
  112
D y  56
D x 336
D
1
 112

 3, y 

 , z x
 1
D 112
D
112
2
D
112
21. Add the following 2 by 4 matrices
6 12
3  5 3 4 3  5 3 8  6  10
6 2  5 0  6 2  5 0  12
4
 10 0 

 
 
-7
For the following two word problems, write the system of equations, do not solve
the system. Leave answer in standard form
22. Pablo has 25 coins: nickels, dimes and quarters. The value of all the coins is $2.80
and the number of nickels is twice the number of quarters. Find how many nickels (x),
dimes (y) and quarters (z) he has.
Coins: n: nickels, d: dimes, q: quarters
Equation 1: 25 coins
Equation 2: value = $2.80
Equation 3: The number of
Each nickel = 5 cents
nickels is twice the number of
Each dime = 10 cents
quarters.
Each quarter = 25cents
Each dollar = 100 cents
n + d + q =25
5n + 10d +25q = 280
n = 2q
d
q
 25
n

System : 5n  10d  25q  280
n
 2q
0

 10  n
d
q
 25
  10n  10d  10q  250
  5n  15q  30


 10d  25q  280  
5n  10d  25q  280   5n
 2q  0
 n
n
 n
 2q
0
 2q
0


  n  3q  6
Divide the first equation by 5  
q6
 n  2q  0
n  2q  n  12 and n  d  q  25  12  d  6  25  d  7
Answer: Pablo has 12 nickels, 7 dimes and 6 quarter.
23. A certain job can be done in 5 days by three men, John, David and Carlos, if they
work together. John and Carlos can do the job by themselves in 8 days. If Carlos and
David work by themselves on the job for 6 days, then John can finish the job in 2 more
days working alone. Find how long it will take John working alone (x), how long it will
take David working along (y) and how long it will take Carlos working alone (z)?
John can do the job in x hours
David can do the job in y hours
Carlos can do the job in z hours
1
of the job per hour
x
1
 David is able to do of the job per hour
y
1
 Carlos is able to do of the job per hour
z

John is able to do
-8
1)All three together do the job in 5 days  Together they do
Therefore : first equation is
1
of job per hour
5
1 1 1 1
5 5 5
   or    1 where 1 represents the whole job
x y z 5
x y z
2)John and Carlos can do the job by themselves in 8 days  John and Carlos do
1
of
8
the job per hour
1 1 1
8 8
  or   1
x z 8
x z
3) Carlos and David work alone for 6 days, then John finishes in 2 days by himself
Therefore : second equation is
Therefore : third equation is
6 6 2
  1
y z x
5 5 5
 1
x  y  z 1
 x a
 5a  5b  5c  1

1
 8 8


System : 
 1
 8z  1
Let   b  8a
x
z
y


 2a  6b  6c  1

 6  6  2 1
1 c
 y z x
 z
6  5a  5b  5c  1
 30a  30b  30c  6
1
 20a



 8z  1   8a
 8z
1  
8a
 8a  8c  1
  10a  30b  30c  5
 5  2a  6b  6c  1

1
8
2
3
3
 20a  1  a 
, 8a  8c  1 
 8c  1   8c  1  8c   c 
20
20
5
5
40
5
15
25
15
3
5a  5b  5c  1 
 5b 
 1  5b 
 1  5b 
b
20
40
40
8
40
1


 a  20
 x  20 days


3
40
1
 y 
 13 days
b 
40
3
3


3
40
c 
z 
days


40
3
John can do the job by himself in 20 days .
1
David can do the job by himself in 13 days .
3
1
Carlos can do the job by himself in 13 days .
3
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