Math 285 Final Part II 06-13-13
1)
(5 points each) Let
T : P1  P1 be a function defined by T (ax  b)  (a  b) x  b .
a) Show that T is a linear transformation.
b) Find
T 1 (ax  b)
a)
T ((ax1  b)  (ax2  b))  T ((a1  a2 ) x  (b1  b2 ))  (a1  a2  b1  b2 ) x  (b1  b2 )
 (a1  b1 ) x  b1  (a2  b2 ) x  b2  T (a1 x  b)  T (a2 x  b)
T (c(ax  b))  T (cax  cb)  (cax  cb)  cb  c[( ax  b)  b)  cT (ax  b)
ii)
First find a matrix of T with respect to the standard bases from {1, x} to {{1, x} . Then compute its
i)
b)
inverse. Finally write the matrix in the LT form.
T (1)  T (0 x  1)  x  1  (1)  1  1  x, T ( x)  T (1x  0)  x  0  1  1  x .
 1 0
 1 1 .


 1 0    1 0
1
 1
   1 1 . Since the first column T (1)  1  x , the second

1

1

 

1
1
T ( x)  x , T (ax  b)  aT 1 ( x)  bT 1 (1)  ax  b(1  x)  (a  b) x  b , which is
Its inverse is
column
Thus T can be represented as
the same as the original function.
2) Let V= R 3 and S  {( x1 , x2 , x3 )  R 3 : x1  x2  x3  0}
a) Show S is a subspace of R .
b) Find a basis for S.
3
a)
0  (0,0,0)
Show
ii)
Show S is closed under addition: pick two typical elements in S:
( y1 , y2 , y3 )
is in the set: since
0  0  0  0 , 0  (0,0,0) is in S.
i)
be in S. Since they are in S,
x1  2  x3  0 and
Let
( x1 , x2 , x3 ) and
y1  y2  y3 )  0 . We need to show
( x1 , x2 , x)  ( y1 , y2 , y3 )  ( x1  y1 , x2  y2 , x3  y3 ) is in S. But
( x1  y1 )  ( x2  y2 )  ( x3 , y3 )  ( x1  x2  x3 )  ( y1  y2  y3 )  0  0  0
Show S is closed under scalar multiplication: Let ( x1 , x2 .x3 ) be in S, c be in F.
c( x1 , x2 , x3 )  (cx1 , cx2 , cx3 ) and cx1  cx2  cx3  c( x1  x2  x3 )  c  0  0
iii)
a.
Find a basis for S: First observe that if
x1  x2  x3  0 .
1 1  1
0 0 0 


0 0 0 
column vectors,
( x1 , x2 , x3 ) is in S, then ( x1 , x2 , x3 ) satisfies
That is, S is the set of solutions to
. Thus
Then
x1  x2  x3  0 .
In row-echelon form, it is
x2 , x3 are free variables. x3  t , x2  s, x1   x2  x3  s  t .
 x1   s  t 
 1
 x    s   s  1  t
 2 

 
 x3   t 
 0 
1
0 .
 
1
We now claim that
 1
{ 1,
 0 
Thus in
1 
0} is a basis
 
1
for S:
a) First they are LI since there are only two vectors and one is not a multiple of the other.
b) They span S since a typical vector can be expressed as
 x1   1
 x   s  1  t
 2  
 x3   0 
1
0 
 
1
.
3)
Let S be the subset of M3(R) consisting of all diagonal matrices. Show that S is a subspace of M3(R) and
find a basis for S (you are now required to verify that the set found is indeed a basis.
Subspace:
I)
II)
Show S is nonempty:
0 0 0
0  0 0 0
0 0 0
is a diagonal matrix since its off-diagonal elements are 0.
Next choose two typical elements in S:
x  y 1
 0

 0
0
x2  y 2
0
 x1
u   0
 0
0
x2
0
0
0 
x3 
and
 y1
v   0
 0
0
y2
0
0
0  .
y3 
Then
uv 

0 
x3  y3 
0
Is still a diagonal matrix.
III)
Let
c be in F,
 x1

v  0
 0
0
x2
0
0
0  .
x3 
Then
 x1

To find a basis. A typical diagonal matrix 0

 0
0 0 0 
1 0 0 0 0 0


 x3 0 0 0 . Then {1 0 0 0 , 0 1 0

 

0 0 1 
0 0 0 0 0 0
cx1

cv   0
 0
0
cx2
0
0 
0 
cx3 
is again a diagonal matrix.
0
1 0 0
0 0 0 
x2 0  can be expressed as x1 0 0 0  x2 0 1 0
0 0 0
0 0 0
0 x3 
0 0 0 
, 0 0 0} is a basis.
0 0 1
0
4)
(5 points each)
a) If
A
and
B are 3 3 matrices with det( A)  3
nn
Recall that for an
Thus
det( 2 A1 B 2 )  23
b)
Suppose
matrix,
,
det( B )  4 , compute det( 2 A1 B 2 ) .
det( aA)  a n det( A), det( A1 ) 
1
, det( AB)  det( A) det( B)
det( A)
1
1
128
det(b) det( B)  8 (4 2 ) 
det( A)
3
3
a2  b2
3b2
c2
a1
b1
c1
a1  b1
3b1
c1  k a2
b2
c2 . Find k without evaluating the determinants.
a3  b3
3b3
c3
b3
c3
a3
a2  b2
3b2
c2
a1  b1
3b1
c1
a1  b1
3b1
c1  switch row1 and row2  a2  b2
3b2
c 2  separate column 1 (no change in det)
a3  b3
3b3
c3
a3  b3
3b3
c3
a1
3b1
c1
 b1
3b1
c1
 ( a2
3b2
c12   b2
3b2
c12 )  factor out 3 from column 2 of the first matrix, -1 from column 3
a3
3b3
c3
 b3
3b3
c3
a1
b1
c1
b1
b1
c1
from column 2 of the second matrix  3 a2
b2
b2
c 2 . But the determinant of the second
a3
b3
c 2  3 b2
c3
b3
b3 c3
matrix is 0 since they have identical columns. Thus k  3
5)
(5 points each) On
i.
ii.
a) Is + associative?
i)
R 2 define operations of addition and multiplication by a real number as follows:
(a, b)  (c, d )  (a  d ,2b  2c)
k (a, b)  (2ka,2kb)
b) Does  0 V s.t. v  0  v, 0  v  v v V ?
(( a, b)  (c, d ))  (e, f )  (a  d ,2b  2c)  (e, f )  (a  d  f ,4b  4c  2e)
(a, b)  ((c, d )  (e, f ))  (a  b)  (c  f ,2d  2e)  a  2d  2e  2bb  2c  2 f
. Since they are not
always equal, + is not associative,
j)
Suppose
0 exists.
Let
0  ( x , y ) . 0  ( c, d )  ( c, d )
for all
(c, d ) .
1
( x , y )  ( c , d )  ( c , d )  ( x  d , 2 y  2 d )  ( c, d )  x  d  c , 2 y  2 d  d  x  d  c , y   d :
2
depend on c and d. Thus it is impossible to find (x,y) that works for all v. 0 does not exist.
x and y
6) A mass weighing 64 pounds stretches the spring 32 ft. The mass is initially released from a
point 1 ft above the equilibrium position with an upward velocity of 1 ft/sec. Assume there is
no damping.
a) Find the equation of the motion.
b) Find the amplitude, period and frequency.
c) If an external force of the form
F (t )  cos t
is applied to the system, find the value of
lim y .
t 
Justify your answer.
F  kx  64  32k  k  2 . mg  64  m  2
d y c dy k
F (t )
dy 2
We obtain a DE

 y

 y  0  yc  c1 cos t  c2 sin t .
2
dt
m dt m
m
dt 2

y(0)  1, y' (0)  1  c1  1, c2  1 . Thus yc  cos t  sin t  2 cos(t  )
4
a) First find k:
2
b) Thus the amplitude is
c) If
Thus
2 , the period is 2
F (t )  cos t , the DE becomes
, the frequency is
Using the initial values of
1
2
y" y  cos t  y p is of the form At cos t  Bt sin t , A, B are not both 0.
y  yc  y p  cos t  sin t  At cos t  Bt sin t , A, B not both 0 and lim y  diverges to , (the system
resonates).
t 