COLLEGE ALGEBRA II (MATH 010)
FALL 2014
PRACTICE FOR EXAM II
SOLUTIONS
1. Solve using substitution.



x − y = 1
4x + 3y = 18
Solution:
x=1+y
4(1 + y) + 3y = 18
4 + 4y + 3y = 18
7y = 14
y=2
x=1+y =3
(3, 2)
2. Solve using elimination.



3x + 4y = 10
x − 4y = −2
Solution:
2 −
3x + 4y = 10
x − 4y = −2
4x
= 8
x
= 2
1
4y
−4y
y
= −2
= −4
=
1
(2, 1)
3. Solve each system, or show that it has no solution.
If it has infinitely many solutions express them in
terms of a single parameter (t).
(a)



x + 4y = 8
3x + 12y = 2
(b)



2x − 6y = 10
−3x + 9y = −15
Solution:
(a)
−3x + −12y = −24
3x + 12y =
2
0 +
0 = −22
Inconsistent
(b)
6x − 18y = 30
−6x + 18y = −30
0 +
0 =
0
Let x = t, so that 2t − 6y = 10, and y = − 35 + 3t .
(t, − 53 + 3t )
2
4. Solve the system, or show that it is inconsistent.









x −
y − z = 4
2y + z = −1
−x + y − 2z = 5
Solution:
EQ1+EQ3=EQ3





x −
y − z = 4
2y + z = −1
− 3z = 9
x −
y − z = 4
2y + z = −1
z = −3




− 31 EQ3=EQ3









2y + (−3) = −1, y = 1
x − 1 − (−3) = 4, x = 2
(2, 1, −3)
5. Solve the system, or show that it is inconsistent.
(a)









y − 2z = 0
2x + 3y
= 2
−x − 2y + z = −1
(b)









x + 2y − z = 1
2x + 3y − 4z = −3
3x + 6y − 3z = 4
3
Solution:
(a)
EQ1↔EQ3









−x − 2y + z = −1
2x + 3y
= 2
y − 2z = 0
2EQ1+EQ2=EQ2









−x − 2y + z = −1
− y + 2z = 0
y − 2z = 0
EQ2+EQ3=EQ3









−x − 2y + z = −1
− y + 2z = 0
0 + 0 = 0
Let z = t
−y + 2t = 0, y = 2t
−x − 2(2t) + t = −1, x = 1 − 3t
(1 − 3t, 2t, t)
(b)
-3EQ1+EQ3=EQ3









x + 2y − z = 1
2x + 3y − 4z = −3
0 + 0 + 0 = 1
Inconsistent
4
6. Solve the system using row echelon operations (a matrix).



x − 2y + z = 1


y + 2z = 5



 x +
y + 3z = 8
Solution:





1 −2 1 1
0 1 2 5
1 1 3 8


R3−3R2=R3 

→





 R3−R1=R3 


→


1 −2 1 1
0 1 2 5
0 0 −4 −8
1 −2 1 1
0 1 2 5
0 3 2 7


 − 1 R3
 4
 →










1 −2 1 1
0 1 2 5
0 0 1 2





z=2
y = 5 − 2z = 5 − 4 = 1
x = 1 + 2y − z = 1 + 2(1) − 2 = 1
(1, 1, 2)
7. Solve the system using row echelon operations (a matrix).
(a)









x + y + z = 2
y − 3z = 1
2x + y + 5z = 0
(b)









2x − 3y − 9z = −5
x
+ 3z = 2
−3x + y − 4z = −3
5
Solution:
(a)





1 1 1 2
0 1 −3 1
2 1 5 0




 R3−2R1=R3 


→




R3+R2=R3 

→

1 1 1 2
0 1 −3 1
0 −1 3 −4
1 1 1 2
0 1 −3 1
0 0 0 −3










Inconsistent
(b)





2 −3 −9 −5
1 0 3 2
−3 1 −4 −3


R3+3R1=R3 

→


− 31 R2 

→ 





 R1↔R2 
 → 


1 0 3 2
2 −3 −9 −5
0 1 5 3
1 0 3 2
0 1 5 3
0 1 5 3

1 0 3 2
2 −3 −9 −5
−3 1 −4 −3




 R2−2R1=R2 


→





 R3−R2=R3 


→


Let z = t
y = 3 − 5z = 3 − 5t
x = 2 − 3z = 2 − 3t
(2 − 3t, 3 − 5t, t)
6





1 0
3 2
0 −3 −15 −9
0 1
5 3
1 0 3 2
0 1 5 3
0 0 0 0










8. Perform the indicated operation, if possible.

A=
D=

H=
(a) AD
Solution:
(b) DA

2 −5
0 7
7 3

3 1
2 −1


(c) AH
(a) Impossible
(b)

7 3

2 −5
0 7


=
14 −14
(c)


2 −5
0 7


3 1
2 −1



=
−4 7
14 −7


9. (a) Write as a matrix equation.
(b) Calculate the determinant of the coefficient matrix.
(c) Is the coefficient matrix invertible? Explain briefly.









x − 2y + z = 1
y + 2z = 5
x + y + 3z = 8
Solution:
7
(a)





1 −2 1
0 1 2
1 1 3





x
y
z







=


1
5
8





(b)
|A| =
(1) 0 1
0 2 1 2 + (1) − (−2) 1 1
1 3 1 3
= (3 − 2) + 2(−2) + 1(−1) = −4
(c) It is invertible because |A| =
6 0.
10. Solve the system by converting to a matrix equation
and using the inverse of the coefficient matrix.



x − 4y = −2
−2x + y = −3
Solution:


1 −4
−2 1


x
y



=
−2
−3


|A| = 1(1) − (−4)(−2) = −7

A


−1
x
y
1 1 4
=− 
7 2 1



=
− 71 − 47
− 72 − 17
8



=


− 17 − 47
− 27 − 17
−2
−3



=


2
1


11. Solve using Cramer’s Rule.


x + 2y = 7
5x − y = 2

Solution:
1 2 = −1 − 10 = −11
|D| =
5 −1 7 2 = −7 − 4 = −11
|Dx | =
2 −1 1 7 = 2 − 35 = −33
|Dy | =
5 2 x=
y=
|Dx |
|D|
|Dy |
|D|
=
=
−11
−11
−33
−11
=1
=3
9
12. Solve using Cramer’s Rule.









x − y + 2z = 7
3x
+ z = 11
−x + 2y
= 0
Solution:
7 −1 2
1 −1 2 3 0 1 , |Dx | = 11 0 1
|D| =
0
2 0
−1 2 0 1 −1 7
1 7 2 3 11 1 , |Dz | = 3 0 11
|Dy | =
−1
2 0
−1 0 0 |D| =
(1) 0 1
2 0
− (−1) 3 0
= 11
−1 2
|Dx | =
(7) 11 1 11 0 0 1 = 30
− (−1) + (2) 0 0 0 2 2 0 |Dy | =
(1) 11 1 3 1 3 11 −(7) = 15
+(2) −1 0 −1
0 0 0 |Dz | =
(1) 0 11 3 11 3 0 −(−1) +(7) = 31
−1
−1 2 2 0 0 x=
|Dx |
|D|
=
3 1
−1 0
+ (2) 30
11 ,
y=
|Dy |
|D|
=
10
15
11 ,
z=
|Dz |
|D|
=
31
11
13. Use row operations to find
ing matrix.

2


−1
A=

1
the inverse of the follow4 1
1 −1
4 0





Solution:










2 4 1 1 0 0
−1 1 −1 0 1 0
1 4 0 0 0 1




 R1↔R3 
 → 


1 4 0 0 0 1
−1 1 −1 0 1 0
0 −4 1 1 0 −2


R2−R3−>R2 

→



R1−4R2−>R1 

→




 R2−R1−>R2 


→


1 4 0 0 0 1
0 1 0 1 1 −1
0 −4 1 1 0 −2

1 4 0 0 0 1
−1 1 −1 0 1 0
2 4 1 1 0 0




 R3+4R2−>R3 


→



A−1 = 


−4 −4 5
1 1 −1
5 4 −6
11






 R3−2R1−>R3

→

1 4 0 0 0 1
0 5 −1 0 1 1
0 −4 1 1 0 −2
1 0 0 −4 −4 5
0 1 0 1 1 −1
0 0 1 5 4 −6












1 4 0 0 0 1
0 1 0 1 1 −1
0 0 1 5 4 6




