If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
SOLUTIONS/COMMENTARY ON QUERY 15
**** 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] ]. ** Check out that problem and get the statement right. **
MATRIX IS [ [ 1 -3 -1 | 2], [ 2 -5 2 | 6 ] , [ -3 -6 4 | 6 ]
ROW OPERATIONS ARE R2=-2 r1 + r3, R3 = 3r1+r3
Your solution:
Confidence Assessment:
Given Solution:
** 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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**** Query problem 10.3.30 (5th ed 10.2.30) reduced echelon form of
[ [1, 0, 0, 0, 1], [0, 1, 0, 0, 2], [0, 0, 1, 2, 3] ]
REDUCED ECHELON FORM IS GIVEN. PROBLEM ASKS TO WRITE THE EQUATIONS; MATRIX IS ALSO CORRECTED ABOVE
Your solution:
Confidence Assessment:
Given Solution:
For the matrix [ [1, 0, 0, 0, 1], [0, 1, 0, 2, 2], [0, 0, 1, 2, 3] ]:
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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**** Query problem 10.3.48 (5th ed 10.2.54) system 2x + y - 3z = 0,
-2x + 2y + z = -7, 3x - 4y - 3z = 7.
Your solution:
Confidence Assessment:
Given Solution:
MATRIX [2, -3, -1, 0; -1, 2, 1, 5; 3, -4, -1, 1] FOR #48 YIELDS [1, 0, 1, 0; 0, 1, 1, 0; 0, 0, 0, 1]
SOLUTION TO system 2x + y - 3z = 0, -2x + 2y + z = -7, 3x - 4y - 3z = 7:
** 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]. **
DER
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15:32:15
** 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. **
(checks out)
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**** 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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15:46:43
Your solution:
Confidence Assessment:
Given Solution:
&&&&&& Correct matrix is
[ [0, -2, 0, 0, -4], [1, 0, 0, 5, 8], [1, 0, 3, 0, 4], [-1, 0, 1, 1, 0]] with solution
„ 44 †
¦ 1 0 0 0 ———— ¦
¦ 23 ¦
¦ ¦
¦ 0 1 0 0 2 ¦
¦ ¦
¦ 16 ¦
¦ 0 0 1 0 ———— ¦
¦ 23 ¦
¦ ¦
¦ 28 ¦
¦ 0 0 0 1 ———— ¦
… 23 ‡ &&&&&&&&&&&
** 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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Query 10.4.20 (5th ed 10.3.20) solve using determinants -x + 2y =
5, 4x - 8y = 6.
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Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION WITH INSTRUCTOR RESPOSE:
if we set up the matrix for the value of D we find
(-1,2)
(4,-8)
and since the value of D is ad-bc and in this case ad-bc is (-1)(-8)-(2)(4)
which is equal to 0 and if D equals 0 then cramers rule cannot be used to
solve the equation.it is not applicable.
** This shows that the system is degenerate. The lines defined by the two equations are parallel. Either they coincide or there is no solution. You can check to see which is the case. **
** In this case the slope-intercept forms of the lines are y = x/2 + 5/2 and y = x/2 - 3/4 (just solve both equations for y). The lines have the same slope (the 0 determinant told us that) but different y intercepts.
There is therefore no solution.
Had the lines been the same the solution would have been the entire line. **
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.
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**** Query 10.4.30 (5th ed 10.3.30) solve using determinants x - y + z = -4, 2x - 3y + 4z = -15, 5x + y - 2z = 12.
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13:36:00
Your solution:
Confidence Assessment:
Given Solution:
** 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.
**
**** 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] ].
What did you obtain for the expression 2A + 4B?
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Your solution:
Confidence Assessment:
Given Solution:
** 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]] **
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**** 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] ].
What did you obtain for the expression CB?
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Your solution:
Confidence Assessment:
Given Solution:
** 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. **
DERIVE CHECKS
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14:11:43
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**** 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] ].
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14:16:01
Your solution:
Confidence Assessment:
Given Solution:
** A + B = [4, 4, -5; -1, 5, 4].
C(A+B) = [ [15, 21, -16], [22, 34, -22], [-11, 7, 22] ]. **
** Common error: [[ 50 -3 ], [ 18 21 ]].
Note that it is true that (A+B) * C = [[ 50 -3 ], [ 18 21 ]]
However the requested product was C ( A + B), which is [[15, 21, -16]. [22, 34, -22],[-11, 7, 22]]. **
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Query 10.5.44 (5th ed 10.4.44) solve using inverse matrix x + 2z = 6, -x + 2y + 3z = -5, x - y = 6.
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Your solution:
Confidence Assessment:
Given 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. **
COMMON ERROR: USING CRAMER'S RULE WHERE INVERSE MATRIX WAS REQUESTED
we can find the value of D using Cramers rule. we find that D=1 so we can
find x by saying D sub x= 6(3)-0+2(-7) so x=4 we can then find the value of
y since we know x in the equation x-y=6 we can say 4-y=6 so y=-2 we can then
also find the value of z by saying 4+2z=6 2z=2 so z=1
** Cramer's Rule is not a method for finding the inverse matrix. **
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