course mth164

??????????????assignment #015

015.

Precalculus II

04-22-2009

......!!!!!!!!...................................

22:31:06

Query 10.4.20 (5th ed 10.3.20) solve using determinants -x + 2y = 5, 4x - 8y = 6.

......!!!!!!!!...................................

RESPONSE -->

-1 2 5

4 -8 6

x=-.75

y=.625

confidence assessment: 2

The system is of the form

A x = y with

A = [[-1, 2], [4, -8]]

and y = [5, 6]`.

The determinant D is det ([[-1, 2], [4, -8]] ) = -1 * (-8) – 2 * 4= 0.

The determinant Dx is det([ [5, 2], [6, -8] ) = 5 * 6 – 2 * -8 = 46.

The determinant Dy is det( [[-1, 5], [4, 6]]) = -1 * 6 - 4 * 5 = -26.

The solutions would be

x = Dx / D = 46 / 0, which however is undefined, as is the solution for y

y = Dy / D = -26 / 0 (also undefined).

The determinant D = 0 indicates that the system is degenerate; the fact that Dx and Dy are not also 0 indicates that there is no solution.

.................................................

......!!!!!!!!...................................

22:31:25

What is your solution?

......!!!!!!!!...................................

RESPONSE -->

already given but -.75 and .625

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:31:51

Give the rows, one at a time (i.e., Enter between rows), for the determinant in the numerator of your solution for y, and give the determinant.

......!!!!!!!!...................................

RESPONSE -->

already given

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:32:02

Give the rows, one at a time (i.e., Enter between rows), for the determinant in the denominator of your solution for y, and give the determinant.

......!!!!!!!!...................................

RESPONSE -->

already given

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:33:06

Query 10.4.30 (5th ed 10.3.30) solve using determinants x - y + z = -4, 2x - 3y + 4z = -15, 5x + y - 2z = 12.

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:34:15

What is your solution?

......!!!!!!!!...................................

RESPONSE -->

the solution is 1,3,-2

confidence assessment: 2

You haven't shown the details of your solution. If you did this by setting up and evaluating the determinants, as indicated below, then you're OK. On tests, remember that all steps must be shown.

** The determinant of D = [[1,-1,1],[2,-3,4],[5,1,-2]] is

D = 1 * DET([[-3,4],[1,-2] ) - (-1 * DET ([[2,4],[5,-2]] + 1 * DET([[2,-3],[5,1]])

= 1(( -3*-2) - (4*1)) - (-1)((2*-2)- (4*5)) + 1((1*2) - (5*-3) )

= 1 (2) + 1 ( -24) + 1 ( 17)

Dx, Dy and Dz are obtained by replacing the x, y and z columns of D with the coefficients -4, -15 and 12.

Calculations similar to those above give us

Dx = DET([[-4,-1,1],[-15,-3,4],[12,1,-2]]) = -5.

Dy = DET([1, -4, 1; 2, -15, 4; 5, 12, -2]) = -15

Dz = DET([1, -1, -4; 2, -3, -15; 5, 1, 12])

So

x = Dx / D = -5 / -5 = 1

y = Dy D = -15 / -5 = 3

z = Dz / D = 10 / -5 = -2.

**

.................................................

......!!!!!!!!...................................

22:34:34

Give the rows, one at a time (i.e., Enter between rows), for the determinant in the numerator of your solution for z, and give the determinant.

......!!!!!!!!...................................

RESPONSE -->

z=-2

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:34:56

Explain in detail how you evaluated to determinant.

......!!!!!!!!...................................

RESPONSE -->

by using the minor mthod to break the matrix down

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:35:18

Give the rows, one at a time (i.e., Enter between rows), for the determinant in the denominator of your solution for z, and give the determinant.

......!!!!!!!!...................................

RESPONSE -->

-2

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:36:20

Query 10.5.6 (5th ed 10.4.6) A = [ [ 0,3,-5], [1,2,6] ]; B = [ [4,1,0], [-2,3,-2] ], C = [ [4,1], [6,2], [-2,3] ].

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:42:42

What did you obtain for the expression 2A + 4B?

......!!!!!!!!...................................

RESPONSE -->

using scalar multplication we can derive

(0,6,-10)( 2,4,12)+(16,4,0)(-8,12,-8)

confidence assessment: 2

** 2 A + 4 B = 2 * [ [ 0,3,-5], [1,2,6] ] + 4 [ [4,1,0], [-2,3,-2] ]

= [[0,6,-10], [2,4,12]] + [[16,4,0],[-8,12,-8]]

= [[ 16,10,-10], [-6,16,4]] **

.................................................

......!!!!!!!!...................................

22:42:54

Query 10.5.10 (5th ed 10.4.10) A = [ [ 0,3,-5], [1,2,6] ]; B = [ [4,1,0], [-2,3,-2] ], C = [ [4,1], [6,2], [-2,3] ].

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:50:27

What did you obtain for the expression CB?

......!!!!!!!!...................................

RESPONSE -->

i obtained an expression like this

B*C= (4,1,0)(-2,3,-2)*(4,1)(6,2)(-2,3)

confidence assessment: 2

** The correct product is [ [14, 7, -2], [20, 12, -4], [-14, 7, -6] ]. This is obtained by the following scheme:

The product of the first row [4,1] of C and the first column [4, -2] of B is 4 * 4 - 1 * 2 = 14. This product goes into the first row and first column of CB.

The product of the first row [4,1] of C and the second column [1, 3] of B is 4 * 1 + 1 * 3 = 7. This product goes into the first row and second column of CB.

The product of the first row of C and the third column of B is -2. This goes into the first row and third column of CB.

The product of the second row [6, 2] of C and the first column [4, -2] of B is 6 * 4 - 2 * 2 = 20. This product goes into the second row and first column of CB.

The rest of the second row of CB is obtained by multiplying the second row of C by the second and then the third column of B.

The third row of CB is obtained by multiplying the third row of C by each of the three columns of B. **

.................................................

......!!!!!!!!...................................

22:50:36

Query 10.5.12 (5th ed 10.4.12) A = [ [ 0,3,-5], [1,2,6] ]; B = [ [4,1,0], [-2,3,-2] ], C = [ [4,1], [6,2], [-2,3] ].

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 2

.................................................

......!!!!!!!!...................................

22:57:48

What did you obtain for the expression C(A+B)?

......!!!!!!!!...................................

RESPONSE -->

-16,18(3,-9,11)

C A+B

confidence assessment: 2

A + B = [4, 4, -5; -1, 5, 4].

C(A+B) = [ [15, 21, -16], [22, 34, -22], [-11, 7, 22] ]. **

.................................................

......!!!!!!!!...................................

23:03:37

Query 10.5.44 (5th ed 10.4.44) solve using inverse matrix x + 2z = 6, -x + 2y + 3z = -5, x - y = 6.

......!!!!!!!!...................................

RESPONSE -->

x=6

y=0

z=.33

confidence assessment: 2

.................................................

......!!!!!!!!...................................

23:03:52

What is your solution?

......!!!!!!!!...................................

RESPONSE -->

6,0,.33

confidence assessment: 2

.................................................

......!!!!!!!!...................................

23:05:05

What is your inverse matrix?

......!!!!!!!!...................................

RESPONSE -->

the inverse is

opposite of the original matrix

confidence assessment: 2

.................................................

......!!!!!!!!...................................

23:05:16

Give the rows, one at a time (i.e., Enter after each row), of the matrix you reduced to obtain the inverse.

......!!!!!!!!...................................

RESPONSE -->

used calculator

confidence assessment: 23

the calculator is not a valid means to obtain the solution

** To find the inverse matrix you set up the system as a matrix equation

[ [1, 0, 2], [ -1, 2, 3], [1, -1, 0] ] * [x, y, z] = [6, -5, 6] (write the matrices out as arrays, of course).

This system is of the for A X = Y, where A = [ [1, 0, 2], [ -1, 2, 3], [ 1, -1, 0] ], X = [x, y, z] and Y = [6, -5, 6].

A solution to A X = Y is

X = A^-1 * Y.

You find A^-1 by row-reducing the matrix [ A | I ] to obtain the form [ I | A^-1 ].

Since I = [ 1, 0, 0; 0, 1, 0; 0, 0, 1] we have

[A | I ] = [ [1, 0, 2 | 1, 0, 0 ; -1, 2, 3 | 0, 1, 0; 1, -1, 0 | 0, 0, 1]]. You'll probably need to write the matrix out in standard notation to see it well.

Adding the first row to the second and subtracting the first row from the third we get

[[ 1, 0, 2 | 1, 0, 0; 0, 2, 5 | 1, 1, 0; 0, -1, -2 | -1, 0, 1] ].

Taking half the second row we get

[[ 1, 0, 2 | 1, 0, 0], [ 0, 1, 5/2 | 1/2, 1/2, 0], [ 0, -1, -2 | -1, 0, 1] ].

Adding the second row to the third we get

[[ 1, 0, 2 | 1, 0, 0], [ 0, 1, 5/2 | 1/2, 1/2, 0], [ 0, 0, 1/2 | -1/2, 1/2, 1 ]].

Doubling the last row we get

[[ 1, 0, 2 | 1, 0, 0], [ 0, 1, 5/2 | 1/2, 1/2, 0], [ 0, 0, 1 | -1, 1, 2] ].

Subtracting 5/2 of the last row from the second and 2 times the last row from the first we get

[[1,0,0 | 3, -2, -4], [ 0, 1, 0 | 3, -2, -5], [ 0, 0, 1 | -1, 1, 2]].

Thus the inverse matrix is [[3, -2, -4], [ 3, -2, -5], [ -1, 1, 2]].

Now as before X = A^-1 * y =[ [3, -2, -4], [ 3, -2, -5], [ -1, 1, 2]] * [6, -5, 6] = [4, -2, 1].

So X = [ x, y, z ] = [4, -2, 1] and x = 4, y = -2 and z = 1.

For a 3 x 3 matrix this process is a little more trouble than determinants. For 4 x 4 and larger matrices this process is easier than determinants. **

.................................................

......!!!!!!!!...................................

23:05:25

Give the series of row operations you used to obtain your inverse matrix.

......!!!!!!!!...................................

RESPONSE -->

used calculator

confidence assessment: 2

.................................................

......!!!!!!!!...................................

23:05:47

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

......!!!!!!!!...................................

RESPONSE -->

it was challenging but very usefull in the real world

confidence assessment: 2

.............................

I think from some of your previous work that you might understand how to use determinants and Cramer's Rule.

However it does not appear that you understand how to set up a system using a matrix equation, how to find the inverse of a matrix, or how to use the inverse to get your solution.

You probably need more practice on these topics. You should work the assignments, rework the qa (as I previously advised), then redo the parts of this query that are related to these topics.