assignement 14

course mth 164

?????????y?????assignment #014

014.

Precalculus II

04-20-2009

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22:23:27

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

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RESPONSE -->

3 -1 I 7

9 -3 I 21

which results in

x=2.3

y=.7

approx.

confidence assessment: 2

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22:23:46

What is your solution to the equation?

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RESPONSE -->

the solution is 2.3 and .7 approx

confidence assessment: 2

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22:24:12

Explain in detail how you obtained your solution.

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RESPONSE -->

by using minors to find the x and y values and then checking them to the problems

confidence assessment: 2

If you substitute x = 2.3 and y = .7 into the equations, you do not get identities. So these values are not solutions.

** The two equations in fact coincide. 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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22:44:20

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 -->

has a det of 0 for matrix a so it is not possible to perform

confidence assessment: 1

You answer is correct, as is your reasoning, but it's important to understand how row reduction leads to this conclusion:

** 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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22:44:28

What is your solution to the equation?

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RESPONSE -->

?

confidence assessment: 1

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22:44:35

Explain in detail how you obtained your solution.

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RESPONSE -->

?

confidence assessment: 0

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22:52:50

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

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RESPONSE -->

ok

confidence assessment: 2

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22:55:59

How many adults and how many seniors bought tickets?

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RESPONSE -->

the senoirs bought 215 tickets and the adults bought 110 tickets

confidence assessment: 2

Your answers are correct. It's not clear how you obtained them.

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.

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22:56:20

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

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RESPONSE -->

the matrices and the minor method to solve the equations

confidence assessment: 2

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22:58:18

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 -->

ok

confidence assessment: 2

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23:01:12

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

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RESPONSE -->

-2,0 =r2

3,0 =r3

6, 0(0,-14,5)= r3

confidence assessment: 2

** 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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23:03:27

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

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RESPONSE -->

0,-14,5

I think this is what is asked for?

confidence assessment: 1

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23:03:38

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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23:04:20

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 -->

ok

confidence assessment: 2

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23:10:07

What is the reduced echelon form of the matrix?

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RESPONSE -->

it is 10000

00100

00010

confidence assessment: 2

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23:10:17

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

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RESPONSE -->

many

confidence assessment: 2

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23:11:20

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 -->

ok

confidence assessment: 2

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23:13:11

What is your solution to the system?

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RESPONSE -->

4.3

-.5

2.7 approx and x y z

confidence assessment: 2

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23:13:23

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

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RESPONSE -->

used calculator

confidence assessment: 2

The calculator is not a valid method of solution in this course.

The matrix row-reduces to

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

These solutions agree with yours up to roundoff error, but your solutions need to be done using the specified methods. In this case you need to use row reduction.

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23:13:31

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

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RESPONSE -->

used calculator

confidence assessment: 2

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23:13:39

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

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RESPONSE -->

used calculator

confidence assessment: 2

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23:17:22

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: 2

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23:18:45

What are your four currents?

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RESPONSE -->

the foru values given to usi in the matrix were given in different form but are all still usable when put into matrix form

confidence assessment: 2

** 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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23:20:13

Give the matrix you solved to obtain your solution.

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RESPONSE -->

i=the sum of all matrices minus their prior values

confidence assessment: 2

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23:20:54

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

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RESPONSE -->

many and always

confidence assessment: 3

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You don't appear to be using row reduction to solve these problems. Note that the calculator is not a valid method of solution.

Be sure to study the notes I've inserted and let me know if you have questions about anything.