Thursday, July 15, 1999;
2:00 pm
Answers to the even number problems in the current week Homework
Assignments (Sections 4.6, 4.7, 4.8, 5.1(1))
4.6:
#4: {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}
4.7:
#2: (3,2,-1)
#8: (3,1,3)
#22: (1,1,-1)
4.8:
#6: {(1/sqrt(5), 0, 2/sqrt(5)), (-4/(3sqrt(5)), 5/(3sqrt(5)),
2/(3sqrt(5))
5.1:
#2: lambda^2 - 5lambda + 7
#4: f(lambda) = lambda^3; lambda1 = lambda2 = lambda3 = 0; x1 =
(1,0,0), x2 = (2,0,0), x3 = (3,0,0)
#6: f(lambda) = lambda(lambda - 2); lambda1 = 0, lambda2 = 2; x1 =
(1,-1), x2 = (1,1)
Thursday, July 8, 1999;
4:30 pm
Answers to the even number problems in the current week Homework
Assignments (Sections 4.3, 4.4, 4.5)
4.3:
#2d: yes
#10a: no
#10b: (2, 6, 5) = (4, 2, 1) - 2(1, -2, 3)
4.4:
#2a: no
#2b: no
4.5:
#4: {[4, 3, -2, 1, 0], [1, 1, -2, 0, 1]}
Wednesday, June 30, 1999;
4:30 pm
Answers to the even number problems in the current week Homework
Assignments (Sections 3.3, 3.6, 4.1, 4.2)
3.3:
#4b: yes
#4c: yes
#30: (a) 160 61 123 47 43 17 102 40; (b) OF COURSE
3.6:
#2b: -x + 1 = 0
#2c: y + 4 = 0
#10a: x - z + 2 = 0
#10b: 3x + y - 14z + 47 = 0
4.2:
#10:
Let w1 = a1u + b1v and
w2 = a2u + b2v be two vectors in W.
Then:
w1 + w2= (a1u + b1v) +
(a2u + b2v) = (a1 + a2)u +
(b1 + b2)v
is in W. Also, if c is a scalar, then
cw1 = c(a1u + b1v) =
(ca1)u + (cb1)v
is in W. Therefore, W is a subspace of R3
(3-space).
#22: (a) and (b)
Thursday, June 24, 1999;
12:30 pm
Answers to the even number problems in the current week Homework
Assignments (Sections 2.1, 2.2, 3.1, 3.2)
2.1:
#2a: even
#2b: odd
#2c: even
#6a: 2
#6b: 24
2.2:
#6b: -20
#6c: -20
#14b: nonsingular
#14c: singular
#22: x = 22/5, y = - 26/5, z = 12/5
3.1:
#6a: (1,7); (-3,-1); (-2,6); (-7,1)
#6b: (1,-1); (-9,-5); (-8,-6); (-22,-13)
#6c: (1,2); (5,2); (6,4); (13,6)
#8a: x = -2, y = -9
#8b: x = -6, y = 8
#22b: -1/(sqrt(2)sqrt(41))
#22c: -4/(sqrt(5)sqrt(13))
#24b: u1 and u5; u4 and u6
#24c: u1 and u3; u3 and u5
3.2:
#10a: sqrt(14)
#10b: sqrt(30)
#12b: sqrt(6)
#12c: sqrt(13)
Wednesday, June 23, 1999;
6:00 pm
In order to introduce the concept of Linear Transformations better,
consider the following example of an application of LTs.
Cartoonist's Question
A modern cartoonist uses computers and Linear Algebra to transform images
she draws. Suppose she is interested in conveying the sensation of motion by
gradually tilting and stretching (horizontally) the image of Fig.1 (a) to get
that of Fig.1 (b).
Fig.1
If the necessary gradual stretching along the x-axis is 50%,
how can she model this situation and have a computer draw the tilted
image? The method should be independent of the initial image (frame) so
it can be applied to other frames.
As you see from the course material, the answer to this question involves a
simple matrix-vector multiplication. In fact, what we need to do is multiply
the coordinate vector of any plane point that we want to transform on the left
(Fig.1, (a)) by some matrix.
Wednesday, June 23, 1999;
9:50 PM
Answers to the even number problems in the current week Homework
Assignments (Sections 1.4, 1.5, 1.6):
1.4:
#2: A(BC) = {a11 = -2, a12 = 34, a21 = 24,
a22 = -9}
#14a: a11 = -3, a12 = -2, a21 = 4, a22 = 1
#14b: a11 = -24, a12 = -30, a21 = 60, a22 =
36
1.5:
#2a: a11 = 1, a12 = 0, a13 = 3, a21 = 5,
a22 = -1, a23 = 5, a31 = 4, a32 = 2,
a33 = 2, a41 = -3, a42 = 1, a43 = 4
#2b: a11 = 1, a12 = 0, a13 = 3, a21 = -3,
a22 = 1, a23 = 4, a31 = 12, a32 = 6,
a33 = 6, a41 = 5, a42 = -1, a43 = 5
#2b: a11 = 1, a12 = 0, a13 = 3, a21 = -3,
a22 = 1, a23 = 4, a31 = 4, a32 = 2,
a33 = 2, a41 = 2, a42 = -1, a43 = -4
#8c: x = 1, y = 1, z = 0
#10b: x = 1 - r, y = 3 + r, z = 2 -
r, w = r
#10c: No solution
1.6:
#4: Singular
#6a: Singular
#6b: a11 = 1, a12 = -1, a13 = 0, a21 = 3/2,
a22 = 1/2, a23 = -3/2, a31 = -1, a32 = 0,
a33 = 1
#10a: a11 = 3/5, a12 = -3/5, a13 = -2/5, a21 = 2/5,
a22 = 3/5, a23 = -4/5, a31 = -1/5, a32 = 2/5,
a33 = 2/5
#10b: Singular
#14: a11 = -1, a12 = -4, a21 = 1, a22 = 3
Thursday, June 10, 1999;
3:30 pm
Answers to the even number problems in the current week Homework
Assignments (Sections 1.1, 1.2, 1.3):
1.1:
#8: No solution
#12: No solution
1.2:
#6: (a) (A)T = [aTij]; a11 = 1, a12 = 2,
a21 = 2, a22 = 1, a31 = 3, a32 = 4;
((A)T)T = [aTij]; a11 = 1, a12 = 2,
a13 = 3, a21 = 2, a22 = 1, a23 = 4
(b) A = [aij]; a11 = 5, a12 = 4,
a13 = 5, a21 = -5, a22 = 2, a23 = 3,
a31 = 8, a32 = 9, a33 = 4
(c) A = [aij]; a11 = -6, a12 = 10, a21 = 11,
a22 = 17
(d) A = [aij]; a11 = 0, a12 = -4, a21 = 4,
a22 = 0
(e) A = [aij]; a11 = 3, a12 = 4, a21 = 6,
a22 = 3, a31 = 9, a32 = 10
(f) A = [aij]; a11 = 17, a12 = 2, a21 = -16,
a22 = 6
1.3:
#2: (a) 4; (b) 0; (c) 1; (d) 1
#14: col1(AB) = 1[1 2 3]T + 3[-2 4 0]T + 2[-1 3 -2];
col2(AB) = -1[1 2 3]T + 2[-2 4 0]T + 4[-1 3 -2]
#16:
-2x - y + 4w = 5
-3x - 2y + +7z + 8w = 3
x + 2w = 4
3x + z + 3w = 6
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