Week 6 Homework Key

Math 111
Week 6 Homework
Due 2/16/2016
9.5 Problems #75 to 80
75. Write the matrix here:
76. Answer the question. _4______ Describe how
the matrix shows that this page can be reached
by one click from any other web page:
The four 1s in column 4 show that page 4 can
reach each of the other pages in a single click.
77. Write 𝐴2 here:
78. Answer the question. ____________________
Pages 1 and 2 cannot be reached using two
click path from any other page in the network.
79. __There are two different 2-click paths from page 2 to page 3._
80. __You can get from pge 4 back to page 4 in 2 clicks._
9.6 #72 on page 779
(a) Nothing to write here, just read the traffic map and see why the equations work.
(b) Show the matrix equation:
Show what you enter in your calculator to solve this system:
You can show either of these:
π‘₯1
1
0 0 0
1
0 0 0
5
5
π‘₯2
βˆ’1
1 0 0
βˆ’1
1 0 0
βˆ’3
βˆ’3
[
]βˆ™[π‘₯ ] = [ ]
[
] β†’ [𝐴] [ ] β†’ [𝐡] , [𝐴]βˆ’1 βˆ™ [𝐡] =
0 βˆ’1 1 0
0 βˆ’1 1 0
3
3
3
π‘₯
0
0 1 1
4
0
0 1 1
11
11
or
1
0
0
1
1
0
[
1
1
1
βˆ’1 βˆ’1 βˆ’1
π‘₯1
0
5
π‘₯
0
βˆ’3
2
]βˆ™[ ]=[π‘₯ ]
0
3
3
π‘₯4
1
11
Show the matrix result:
5
2
[ ]
5
6
(c) What does your result mean about traffic?
Between intersection A and B there are 5 cars per
minute, 2 cars per minute between B and C, 5 cars
per minute between C and D and 6 cars per
minute between A and D.
9.7 #36 on page 787
Create three triangles, and yours may be different than mine.
The left-hand triangle gives first matrix, the middle triangle
gives the middle matrix, and the last triangle goes with the
third matrix. After saving each matrix, I completed the
calculation below on my calculator:
Left Triangle:
Middle Triangle:
Right Triangle:
The area of the shaded pentagon is 26. No units were given.
9.7 Extended and Discovery #10
Show the matrix before you take the determinant. And I am going to show my work:
π’™πŸ + π’šπŸ 𝒙 π’š 𝟏
𝟏 𝟎 𝟏] = 𝟎
𝐝𝐞𝐭 [ 𝟏
πŸ“
βˆ’πŸ 𝟐 𝟏
πŸπŸ‘
πŸ‘ 𝟐 𝟏
π’™πŸ + π’šπŸ
𝐝𝐞𝐭 [ 𝟏
πŸ“
πŸπŸ‘
𝒙
𝟏
βˆ’πŸ
πŸ‘
π’š
𝟎
𝟐
𝟐
𝟏
𝟏]
𝟏
𝟏
𝟏 𝟎 𝟏
𝟏 𝟎 𝟏
𝟏
𝟏 𝟏
𝟏
𝟏 𝟎
= (π’™πŸ + π’šπŸ ) [βˆ’πŸ 𝟐 𝟏] βˆ’ 𝒙 [ πŸ“ 𝟐 𝟏] + π’š [ πŸ“ βˆ’πŸ 𝟏] βˆ’ 𝟏 [ πŸ“ βˆ’πŸ 𝟐]
πŸ‘ 𝟐 𝟏
πŸπŸ‘ 𝟐 𝟏
πŸπŸ‘ πŸ‘ 𝟏
πŸπŸ‘ πŸ‘ 𝟐
=𝟎
= (π’™πŸ + π’šπŸ )(βˆ’πŸ–) βˆ’ 𝒙(βˆ’πŸπŸ”) + π’š(πŸ‘πŸ) βˆ’ 𝟏(πŸ–) = 𝟎 Evaluating the determinants
βˆ’πŸ–π’™πŸ + πŸπŸ”π’™ βˆ’ πŸ–π’šπŸ + πŸ‘πŸπ’š βˆ’ πŸ– = 𝟎
Reorganizing in order to complete the squares
π’™πŸ βˆ’ πŸπ’™ + π’šπŸ βˆ’ πŸ’π’š + 𝟏 = 𝟎
Divide both sides by – πŸ–.
(π’™πŸ βˆ’ πŸπ’™ + 𝟏) + (π’šπŸ βˆ’ πŸ’π’š + πŸ’) = πŸ’
Complete both squares. Set equal to constant
What is the equation of the circle?
(𝒙 βˆ’ 𝟏)𝟐 +(π’š βˆ’ 𝟐)𝟐 = πŸ’
The circle has center (-1, -2) and the radius is 2.