1.7:3,12,19,32; 1.8:8,13,14,16,37; 1.9:11,12
DUE: Wednesday, FEb. 14, Beginning of Class
1.7:3) Solve the associated system Ax = b. Row reduce the augmented matrix
[ 1 0 -1 0 ] [ 1 0 -1 0 ] [ 1 0 -1 0 ]
[ 3 1 -5 -5 ] ~ [ 0 1 -2 -5 ] ~ [ 0 1 -2 -5 ]
[ -4 2 1 -6 ] [ 0 2 -3 -6 ] [ 0 0 1 4 ]
[ 1 0 0 4 ] Thus x = [ 4 ] is the unique x satisfying Tx = b
~ [ 0 1 0 3 ] [ 3 ]
[ 0 0 1 4 ] [ 4 ]
1.7:12) b is in the range of the linear transformation if and only if
Ax = b is consistent (has a solution). Row reduce the augmented matrix [A b]
to an echelon form to determine consistency.
[ 1 2 -7 5 9 ] [ 1 2 -7 5 9 ]
[ 0 1 -4 0 5 ] ~ [ 0 1 -4 0 5 ] ~
[ 1 0 1 6 0 ] [ 0 -2 8 1 -9 ]
[ 2 -1 6 8 -9 ] [ 0 -5 20 -2 -27]
[ 1 2 -7 5 9 ] [ 1 2 -7 5 9 ]
[ 0 1 -4 0 5 ] ~ [ 0 1 -4 0 5 ]
[ 1 0 0 1 1 ] [ 0 0 0 1 1 ]
[ 0 0 0 -2 -2 ] [ 0 0 0 0 0 ]
The system is consistent so b is in the range of the transformation x -> Ax.
(In fact, we see that there is a free variable x_3 so infinitely many
solutions, but computation of the solutions is not necessary for the
conclusion that b is in the range of the linear transformation).
1.7:19) Utilize the linearity of T.
T(2u) = 2 T(u) = 2 [ 2 ] = [ 4 ]
[ 0 ] [ 0 ]
T(3v) = 3 T(v) = 3 [ 1 ] = [ 3 ]
[-4 ] [-12]
Using above result,
T(2u + 3v) = T(2u) + T(3v) = [ 4 ] + [ 3 ] = [ 7 ]
[ 0 ] [-12] [-12]
1.7:32) T(x_1,x_2) = ( 4x_1 - 2x_2, 3|x_2| )
To show that T is not linear, you need to find any counterexample,
that is, either
1) Find vectors u,v in R^2 such that T(u+v) does not equal T(u)+T(v)
or 2) Find a real number c and a vector u such that T(cu) not equal cT(u).
We will show 2):
Let c = -1, and u be any vector (x_1,x_2) with x_2 > 0, e.g.
u = (0,1). Then
T(-u) = T{(0,-1)} = (2,3)
However,
-T(u) = -T{(0,1)} = -(-2,3) = (2,-3)
As these are not equal, T is not linear.
1.8:8) To determine the standard matrix representation A of transformation T,
we need to determine what T does to
each of the unit vectors e_1 = [ 1 ], e_2 = [ 0 ]
[ 0 ] [ 1 ]
T(e_1) = e_1 = [ 1 ], T(e_2) = e_2 - 3e_1 = [-3 ]
[ 0 ] [ 1 ]
The standard matrix has column 1 equal to T(e_1) and column 2 equal to T(e_2)
A = [ 1 -3 ]
[ 0 1 ]
1.8:13) In this and the next problem, you should use plane geometry to
visualize what T does to the vectors e_1 and e_2.
T is a composition of linear transformations: T = VU , where
U rotates points counterclockwise pi/4 radians:
U(e_1) = [ 1/sqrt{2} ] = (1/sqrt{2}) (e_1 + e_2)
[ 1/sqrt{2} ]
U(e_2) = [-1/sqrt{2} ] = -(1/sqrt{2}) e_1 + (1/sqrt{2}) e_2
[ 1/sqrt{2} ]
V reflects a point about x_2 axis, that is V(x_1,x_2) = (-x_1, x_2).Hence
V(e_1) = -e_1 , V(e_2) = e_2
The rotation U followed by the reflection V is the composition T = VU.
Thus T(e_1) = V{U(e_1)} = (1/sqrt{2}){V(e_1) + V(e_2)} = (1/sqrt{2}) [-1 ]
[ 1 ]
and T(e_2) = V{U(e_2)} = (1/sqrt{2}){-V(e_1) + V(e_2)} = (1/sqrt{2}) [ 1 ]
[ 1 ]
Therefore the standard matrix is
A = [ T(e_1) T(e_2) ] = (1/sqrt{2}) [ -1 1 ]
[ 1 1 ]
Equivalently, A can be found by the multiplication of, respectively, the
standard matrix of V and the standard matrix of U:
[-1 0 ] 1/sqrt{2} [ 1 -1 ] = 1/sqrt{2} [ -1 1 ]
[ 0 1 ] [ 1 1 ] [ 1 1 ]
1.8:14) T is a composition of linear transformations; T = VU, where
U reflects points through the origin, that is U(x_1, x_2) = (-x_1, -x_2)
(identical to reflection about x_1 axis AND x_2 axis), so
U(e_1) = -e_1 , U(e_2) = -e_2
V rotates vectors pi/2 clockwise, so
V(e_1) = [ 0 ] = -e_2
[-1 ]
V(e_2) = [ 1 ] = e_1
[ 0 ]
Therefore T(e_1) = V{U(e_1)} = V(-e_1) = -V(e_1) = -(-e_2) = e_2
T(e_2) = V{U(e_2)} = V(-e_2) = -V(e_2) = -e_1
Thus the standard matrix of T has column 1 equal to e_2 and column 2 = -e_1:
A = [ 0 -1 ]
[ 1 0 ]
1.8:16) [ 5x_1 - 6x_2 ] = x_1 [ 5 ] + x_2 [-6 ] = [ 5 -6 ] [ x_1 ]
[ x_2 ] [ 0 ] [ 1 ] [ 0 1 ] [ x_2 ]
[ 2x_1 - 2x_2 ] [ 2 ] [-2 ] [ 2 -2 ]
1.8:37) Determine if the columns of the standard matrix are linearly
independent (implies that T is 1-1).
Row reduce to an echelon form. Columns are lin. independent iff
every column is a pivot column; equivalently for this square matrix check if
pivot position in every row.
[ -5 10 -5 4 ] [ -5 10 -5 4 ] [-5 10 -5 4 ]
[ 8 3 -4 7 ] ~ [ 0 65 -60 67 ] ~ [ 0 65 -60 67 ]
[ 4 -9 5 -3 ] [ 0 -10 10 2 ] [ 0 -10 10 2 ]
[ -3 -2 5 4 ] [ 0 40 -40 -8 ] [ 0 0 0 0 ]
No further reduction necessary.
No pivot position in row 4 so columns not linearly independent so T is not 1-1.
1.9:11)
a) Let x_0 be the initial vector:
x_0 = [ Population In CA in 1990 ] = [ 29,716,000 ]
[ Population Out of CA in 1990 ] [ 218,994,000 ]
After 1 year:
Proportion In CA who stay In = (29,716,000 - 509,500)/29,716,000 = .98285
Proportion In CA who move Out = 509,500/29,716,000 = 1 - .98285 = .01715
Proportion Out of CA who move In = 564,100/218,994,000 = .00258
Proportion Out of CA who stay Out =(218,994,000 - 564,100)/218,994,000 = .99742
Therefore the migration matrix M is
In Out
M = [ .98285 .002576 ] In
[ .01715 .99742 ] Out
b) Utilize the relation x_(k+1) = M x_k , k = 0,1,2,...
Year 2000 is 10 years hence from 1990, so we find x_10:
x_10 = [ Population In CA in 2000 ] = M^10 x_0 = [ 30,215,000 ]
[ Population Out of CA in 2000 ] [ 218,495,000 ]
(to the nearest 1,000)
Use a scientific calculator/computer to calculate M^10 above, where M^10 is the
matrix M multiplied by itself 10 times:
M^10 = [ .84295 .02359 ]
[ .15705 .97641 ]
c)The proportion of total population In CA in 2000 is, from (b),
30,215,000/(218,495,000 + 30,215,000) = .1215, i.e. 12.15%.
1.9:12) Let x_0 be the initial status vector of the cars. We are given
x_0 = [ # of cars at airport on Mon. ] = [ 304 ]
[ # of cars at eastside on Mon. ] [ 48 ]
[ # of cars at westside on Mon. ] [ 98 ]
The migration matrix M is given in text.
M = [ .97 .05 .10 ]
[ .00 .90 .05 ]
[ .03 .05 .85 ]
Wednesday is 2 days hence from Mon, 1 day hence from Tues, so we wish to find
x_2 = M (M x_0) = M (x_1) = M [ 307 ] = [ 309.69 ]
[ 48 ] [ 47.95 ]
[ 95 ] [ 92.36 ]
( Tues ) ( Wed )
Round entries to the nearest integer (# cars in whole numbers) to get [ 310 ]
[ 48 ]
[ 92 ]
the state vector for Wednesday.