Math Teacher:
NC
Name:
x 
Describe the transformtation that occurs when you multiply a vector   by each of the
 y
transformation matrices below (putting the transformation matrix on the left). Give a one or two
sentence description as modeled for the first matrix.
(35)
0 1
b. 

1 0 
1 0 
a. 

0 1
This matrix reflects vectors over the
1 0   x   x 
x-axis. Note: 
    
0 1  y    y 
This matrix rotates vectors 90 counter-clockwise
0 1 cos(90 )  sin(90 ) 
about the origin: 


1 0   sin(90 ) cos(90 ) 
 2  2


2 
c.  2
 2
2 


2 
 2
This matrix rotates vectors 45 counter-clockwise
 2  2


2
2   cos(45 )  sin(45 ) 
about the origin: 


 2
2   sin(45 ) cos(45 ) 


 2
2 
Semester Review P 1
 1  3


2 
d.  2
 3
1 


2 
 2
This matrix rotates vectors 60
counter-clockwise about origin.
Math Teacher:
NC
Name:
(36) Prove the following statements using the Princiople of Mathematical Induction:
I.
For all positive integers n, 3  11  19 
Proof (by induction):
Base case (n = 1):
LHS  (8 1  5)  3
RHS  4 12  1  3
 (8n  5)  4n2  n
, so the equality holds when n = 1.
Inductive step: Suppose that for some positive integer k,
3  11  19 
 (8k  5)  4k 2  k .
Then,
3  11  19 
 (8k  5)  (8(k  1)  5)   4k 2  k   (8(k  1)  5) , by the inductive hypothesis
 4k 2  7 k  3
 4(k 2  2k  1)  (k  1)
 4(k  1) 2  (k  1)
Hence, by Mathematical Induction 3  11  19 
integers n.
 (8n  5)  4n2  n holds for all positive
Semester Review P 2
Math Teacher:
Name:
II.
For all positive integers n, 11! 2  2! 3  3! ...........  n  n!  ( n  1)! 1
Proof (by induction):
Base case (n = 1):
LHS  11!  1
, so the equality holds when n = 1.
RHS  (1  1)! 1  1
Inductive step: Suppose that for some positive integer k,
11! 2  2! 3  3! ...........  k  k !  (k  1)! 1 .
Then,
11! 2  2! 3  3! ...........  k  k !  (k  1)(k  1)!  (k  1)! 1  ( k  1)( k  1)!
, by the inductive hypothesis
 (k  1)! (k  1)(k  1)! 1
 (k  1)!1  (k  1)   1
 (k  1)! k  2  1
 (k  2)! 1
Hence, by Mathematical Induction 11! 2  2! 3  3! ...........  n  n!  ( n  1)! 1 holds for all
positive integers n.
Semester Review P 3
Math Teacher:
III.
Name:
For all integers n  0, n3  5n  6 is divisible by 3.
Proof (by induction):
Base case (n = 0): For n = 0, n3  5n  6  6 which is clearly divisible by 3. Therefore,
the base case holds.
Inductive step: Suppose that for some positive integer k, k 3  5k  6 is divisible by 3.
Then there is a positive integer p such that 3 p  k 3  5k  6
Then,
(k  1)3  5(k  1)  6  k 3  3k 2  3k  1  5k  5  6,
remember Pascal's triangle?
= k 3  5k  6  3k 2  3k  6
=3 p  3(k 2  k  2)
= 3( p  k 2  k  2)
This last expression is clearly a multiple of 3.
Hence, by Mathematical Induction n3  5n  6 is a multiple of (divisible by) 3 for all n  0.
Semester Review P 4