Assignment 14

course mth164

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Sorry for all the delays.

????D?J??????assignment #014014.

Precalculus II

01-15-2009

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21:05:00

Query problem 10.1.22 (5th ed 10.1.24) solve the equation 3x - y = 7, 9x - 3y = 21.

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RESPONSE -->

the equations are the same so it is not possible to solve this equation

confidence assessment: 1

** The two equations coincide. This however doesn't mean that there are no solutions; there are in fact infinitely many solutions.

If we subtract 3 times the first from the second we get 0 = 0.

This tells us that one equation is a multiple of the other, and that they are therefore equivalent. Their graphs coincide.

A solution to one equation is a solution to the other. So the solution set consists of all (x, y) satisfying the first equation 3 x - y = 7. These solutions lie on the line y = 3x - 7. **

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21:05:18

What is your solution to the equation?

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RESPONSE -->

the solution i think is not possible

confidence assessment: 1

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21:05:56

Explain in detail how you obtained your solution.

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RESPONSE -->

i didnt because the equations are the same and it is not possible to remove one variable

confidence assessment: 1

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21:44:00

Query problem 10.2.12 (5th edition 10.1.42) solve the equation 3x-2y+2z=6, 7x-3y+2x=-1, 2x-3y+4z=0.

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RESPONSE -->

3 -2 2 6

7 -3 2 -1

2 -3 4 0

x=-43

y=-45

z=22.5

confidence assessment: 1

** If we row reduce the coefficient matrix [[3,-2,2,6],[7,-3,2,-1],[2,-3,4,0]] we get [[1, 0, -0.4] [0; 0, 1, -1.6] [0; 0, 0, 0, 1]], indicating inconsistent system.

Using the equations:

From the equations

3 x - 2 y + 2 z = 6 and

2 x - 3 y + 4 z = 0

we can eliminate z (add -2 times the first to the second) to get

-4 x + y = -12 or

y = 4x - 12.

From the equations

3 x - 2 y + 2 z = 6 and

7x-3y+2z=-1

we eliminate z (just subtract the equations) to get

y = 4 x + 7 .

y cannot be equal to 4x - 12 at the same time it's equal to 4x + 7. The two expressions could only be equal if -12 = 7.

We conclude that the system is inconsistent. No solution exists.

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21:44:19

What is your solution to the equation?

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RESPONSE -->

-43

-45

22.5

confidence assessment: 2

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21:45:45

Explain in detail how you obtained your solution.

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RESPONSE -->

by creating a system of matrices and then using multiplication and adding 2 systems at a time for the end result untill you had all zeros and a line of 1 for placeholders in each colum

confidence assessment: 1

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21:56:08

Query problem 10.1.44 (5th ed 10.1.60) ticket $9 adult $7 senior, 325 people paid $2495.

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RESPONSE -->

110 senior

215 adults

confidence assessment: 2

You need to show your steps, but your answer is correct.

Good student solution:

We can set up a formula for the price of the tickets by saying

x= adult ticket price and

y=senior citizen ticket price

so we have 9x+7y=2495

We can then write an equation for the amount of tickets purchased by saying

x+y=325

We can then multiply the second equation by -9 and we get

-9x-9y=-2925

so when we add the sum of these two equations we get

-2y=-430 so

y=215

so there were 215 senior citizens and 110 adults.

INSTRUCTOR COMMENT:

Good. x adults and y seniors at $9 per adult and $7 per senior yields 9 x + 7 y dollars. This is equal to the $2495, giving us the equation

9x + 7y = $2495.

The total number of tickets is

x + y = 325,

giving us the second equation.

This equation is solved as indicated by the student.

We could also row reduce the matrix

[ [9, 7, 2495], [1, 1, 325] ]

and obtain the same result.

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21:56:23

How many adults and how many seniors bought tickets?

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RESPONSE -->

110 a

215 s

confidence assessment: 2

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21:56:37

What system of equations did you solve to obtain your answer?

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RESPONSE -->

matrices

confidence assessment: 2

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22:29:48

Query problem 10.3.20 (5th ed 10.2.20) R2=-2 r1 + r3, R3 = 3r1+r3, R3 = 6r2 + r1 [ [1, -3, -1], [2, -5, 2], [-3, -6, 4] ].

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RESPONSE -->

dont understand the problem

confidence assessment: 0

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22:30:12

Give the three rows of the matrix you obtained from the first row operation.

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RESPONSE -->

1

-3

-1

confidence assessment: 0

** R2=-2 r1 + r2 tells us to replace Row 2 by -2 * row 1 + row 3.

Row 1 is [ 1 -3 -1 | 2]

Row 2 is [ 2 -5 2 | 6 ]

-2 r1 = -2 * [ 1 -3 -1 | 2] = [ -2 6 2 | -4 ]

so -2 r1 + r2 = [ -2 6 2 | -4 ] + [ 2 -5 2 | 6 ] = [ 0 1 4 | 2 ].

Replacing row 2 by this sum we have

[[ 1 -3 -1 | 2] , [ 0 1 4 | 2 ] , [ -3 -6 4 | 6 ].

R3 = 3r1+r3 tells us to now replace Row 3 by 3 * row 1 + row 3. We get

3 * [ 1 -3 -1 | 2] + [ -3 -6 4 | 6 ] = [ 0 -15 1 | 12 ]. Our matrix is now

[[ 1 -3 -1 | 2] , [ 0 1 4 | 2 ] , [ 0 -15 1 | 12 ]. **

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22:30:54

Give the three rows of the matrix you obtained from the second row operation.

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RESPONSE -->

0-3 -1

2 -5 2

-3 -6 4

confidence assessment: 1

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22:31:03

Give the three rows of the matrix you obtained from the third row operation.

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RESPONSE -->

already did

confidence assessment: 1

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22:36:35

Query problem 10.3.30 (5th ed 10.2.30) reduced echelon form of [ [1, 0, 0, 0, 1], [0, 1, 0, 2, 2], [0, 0, 1, 3, 0] ].

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RESPONSE -->

it is this form

confidence assessment: 1

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22:36:47

What is the reduced echelon form of the matrix?

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RESPONSE -->

it is in it

confidence assessment: 1

The matrix translates into the equations

x1 + 0 x2 + 0 x3 + 0 x4 = 1, i.e., just x1 = 1

0 x1 + 1 x2 + 0 x3 + 2 x4 = 2, i.e., just x2 + 2 x4 = 2

0 x1 + 0 x2 + 1 x3 + 2 x4 = 3, i.e., just x3 + 2 x 4 = 3.

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22:37:31

What series of row operations did you perform obtained this form?

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RESPONSE -->

it was in this form to my knowledge with the all zeros on the left side of the numbers

confidence assessment: 1

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22:57:13

Query problem 10.3.40 (5th ed 10.2.54) system 2x + y - 3z = 0, -2x + 2y + z = -7, 3x - 4y - 3z = 7.

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RESPONSE -->

the equation has no absolute answer due to the uneven size

confidence assessment: 2

** The augmented matrix is

[2, 1, -3, 0; -2, 2, 1, -7; 3, -4, -3, 7]

where the rows are separated by semicolons and the | in the augmented matrix is not included.

We want 0's in the first column of the second and third rows, which we accomplish by R2 = r1 + r2, R3 = -3 r1 + 2 r2. We obtain the matrix

[2, 1, -3, 0; 0, 3, -2, -7; 0, -11, 3, 14]

We now want 0's in the second column of the first and third rows, which we obtain by the operations R1 = 3*r1 - r2 and R3 = 11 * r2 + 3 * r3 to obtain

[6, 0, -7, 7; 0, 3, -2, -7; 0, 0, -13, -35].

We now need 0's in the third column of the first and second rows. Using R1 = -13 r1 + 7 r3 and R2 = -13 r2 + 2 r3 we obtain

[-78, 0, 0, -336; 0, -39, 0, 21; 0, 0, -13, -35].

Now that we have 0's in all the necessary positions we simply multiply row 1 by -1/78, row 2 by -1/39 and row 3 by -1/13 to obtain

[1, 0, 0, 56/13; 0, 1, 0, - 7/13; 0, 0, 1, 35/13],

which is approximated by

[1, 0, 0, 4.307692307; 0, 1, 0, -0.5384615384; 0, 0, 1, 2.692307692]. **

** HERE'S ANOTHER SOLUTION TO ILLUSTRATE AN ALTERNATIVE PROCESS:

Starting with the system

[2, 1, -3, 0

-2, 2, 1, -7

3, -4, -3, 7]

divide the first row by 2 to get

[ 1, .5, -1.5, 0

-2, 2, 1, -7

3, -4, -3, 7]

then add double the first row to the second and subtract three times the first row from the third to get

[1, .5, -1.5, 0

0, 3, -2, -7

0, -5.5, 1.5, 7]

then divide the second row by 3 (and convert everything to fractions) to get

[1, 1/2, -3/2, 0

0, 1, -2/3, -7/3

0, -11/2, 3/2, 7]

then add half the second row to the first and 11/2 the second to the third to get

[ 1, 0, -11/6, -7/6

0, 1, -2/3, -7/3

0, 0, -13/6, -35/6].

Take -6/13 of the third row to get

[1, 0, -11/6, -7/6

0, 1, -2/3, -7/3

0, 0, 1, 35/13]

then add 11/6 of the third row to the first and 2/3 of the third row to the second to get

[ 1, 0, 0, 56/13

0, 1, 0, -7/13

0, 0, 0, 35/13 ].

The system has a solution: x = 56/13, y = -7/13 and z = 35/13. **

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22:57:31

What is your solution to the system?

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RESPONSE -->

it is used by echlongs formula

confidence assessment: 2

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22:57:46

Give the first row of the matrix you reduced to obtain your solution.

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RESPONSE -->

1 0 -5 7

confidence assessment: 2

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22:58:13

Give the second row of the matrix you reduced to obtain your solution.

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RESPONSE -->

0 -5 -3 -14

confidence assessment: 2

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22:58:25

Give the third row of the matrix you reduced to obtain your solution.

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RESPONSE -->

3 -4 -3 7

confidence assessment: 2

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23:01:07

Query problem 10.3.79 (5th ed 10.2.90) Kirchoff system -4 + 8 - 2 I2 = 0, 8 = 5 I4 + I1, 4 = 3 I3 + I1, I3 + I4 = I1.

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RESPONSE -->

ok

confidence assessment: 1

** The equations can be rearranged to give you the following:

0 I1 - 2 *2 + 0 I3 + 0 I4 = -4

I1 + 0 I2 + 0 I3 + 0 I4 = 8

I1 + 0 I2 + 3 I3 + 0 I4 = 4

- I1 + 0 I2 + I3 + I4 = 0

which gives you the matrix

[ [0, -2, 0, 0, -4], [1, 0, 0, 0, 8], [1, 0, 3, 0, 4], [-1, 0, 1, 1, 0]].

The second row is already in the form we want for the first row so we switch rows 1 and 2 to get

[[1, 0, 0, 0, 8]; [0, -2, 0, 0, -4]; [1, 0, 3, 0, 4]; [-1, 0, 1, 1, 0]].

Row 2 has 0 in the first position; we do R3 = -r1 + r3 and R4 = r1 + r4 to eliminate the first-column zeros in rows 3 and 4, obtaining the matrix

[[1, 0, 0, 0, 8]; [0, -2, 0, 0, -4]; [0, 0, 3, 0, -4]; [0, 0, 1, 1, 8]]

Row 2 is in the desired form so except for the -2 where we need 1; so we multiply row 2 by -1/2. Row 3 is in the desired form except for the 3 where we need 1 so we also multiply row 3 by 1/3. We get

[[1, 0, 0, 0, 8]; [0, 1, 0, 0, 2]; [0, 0, 1, 0, -4/3]; [0, 0, 1, 1, 8]].

Adding -r3 to r4 and replacing r4 we get

[[1, 0, 0, 0, 8]; [0, 1, 0, 0, 2]; [0, 0, 1, 0, -4/3]; [0, 0, 0, 1, 28/3]]. **

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23:01:28

What are your four currents?

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RESPONSE -->

didnt record problem and cant see now

confidence assessment: 2

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23:01:31

What are your four currents?

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RESPONSE -->

confidence assessment:

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23:01:46

Give the matrix you solved to obtain your solution.

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RESPONSE -->

didnt record problem

confidence assessment: 2

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23:02:21

Comm on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

i liked the use of the matrices and it simplifies harder equations with lots of variables

confidence assessment: 2

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I think you have the right basic ideas; can't tell whether you've had enough practice to perform well on a test.

I've inserted a number of notes. Let me know if you have questions.