Download Spring 2016 Math 285 Past Exam II Solutions 3-13-16

Math 285 Past Exam II Solutions
1)
3 2 2
1 1
2
a) 1
1
2  switch row 1 and row 2 and multiply by -1  3  2  2  b) Make the
5 5 4
5 5 4
(2,1) entry and (3,1) entry 0 by Row1(-3)+Row2= new Row2, Row 1 (-5)+Row 3 = new
1 1
2
Row 3: No change to the determinant.  0  5  8  c) next ‘factor out’ -5 from
0 0  14
1 1 2
8
the row 2, -14 from the row 3 (1)( 5)( 14) 0 1
 det A  (1)( 5)( 14)(1)  70
5
0 0 1
2)
a) The line of intersection is the set of solutions to the system of equations
x  2 y  3z  8
.
2 x  5 y  z  10
In a row echelon form, the augmented matrix becomes
8
1 2 3 8  1 2 3
2 5  1 10  0 1  7  6 . Z is a free variable, so let z  t . Solving the

 

second equation for y, y  7 z  6  y  7 z  6  y  7t  6 . Solving the first equation
for x, x  2 y  3z  8  x  2 y  3z  8  x  2(7t  6)  3t  8  17t  20 . Thus the line
of intersection can be given parametrically as ( x, y, z )  (17t  20,7t  6, t )
2 1 0
3 0 10
5 0 14  (1)(3(14)  5(10))
b) z 

1
2 1  5
 (1)( 3  5)
3 0
1
5 0 1
x yz a
3) Find a row echelon form of
x  y  3z  b
2 y  4z  c
a 
1  1  1 a  1  1  1
ba 
1 1 3 b   0 1 2
. This system of equation has at least one solution

 
2 
0 2 4 c  0 0 0 a  b  c 
iff the last row is entirely zero. Thus
4)
a b c  0.
a) I) T ( v  w )  A( v  w )  Av  Aw  T ( v)  T (w ) and ii) T (cv )  A(cv )  cA( v )  cT ( v ) . Thus
T is a LT
b) A basis for the range of a linear transformation can be found by computing the
column space of the matrix: The row echelon form of
1 1 1 
2 3 1
1 1 1 is 0 1 1 . The




first column and the second column each contains a leading 1. Thus by going back
to the original matrix, a basis for the range is {(2,1),(3,1)}
5)
c)
det( S 1 AS )  det( s 1 ) det( A) det( S ) 
1
det( A) det( S )  det( A)
det( S )
d) Since 2 can be factored out from each row, 2 becomes 8 outside the determinant
function: det( 2 A
2
)  8 det( A2 )  8 det( A) det( A)  128
6)
a)
b)
c)
d)
e)
f)
g)
Its sign is -1 since there is one inversion.
rank=n-# of free variables=3-2=1.
False: they have the same dimension, but they are not equal.
True.
False: A matrix is invertible iff its determinant is not 0.
True. The number of solutions of a system of homogeneous equations is 1 or infinite.
True
7) T1 (ax  b)  3ax  (a  b)  T (1)  T (0 x  1)  1  -
1
(2) and
2
3
1
T ( x)  T ( x  0)  3x  1  (5 x)  (1) . Thus placing the coefficients into the columns,
5
2
 1 1
 2 2 
C
T B 
3
 0

5

8)
a) First show the zero vector is in the set:
0 0 
0
 is a diagonal matrix since all the
0 0 
off diagonal entries are 0. Thus it contains the zero vector.
b) Next show it is closed under addition:
Let
0 
 a 0
c 0 
a  c
A
and B  
be diagonal matrices. Then A  B  
is still


b  d 
0 d 
 0
0 b
a diagonal matrix with real entries. It is closed under addition.
c) Let
 a 0
ka 0 
, k be a real number. Then kA  
A
 is still a diagonal matrix with

 0 kb
0 b
real entries. It is closed under scalar multiplication.
Therefore, it is a subspace.
9)
a) Recall that
f   f , f  . Thus
1
2
1
2
f  (  (sin 2 x)(sin 2 x)dx) 2  (  (sin 2 2 x)dx) 2  ( 
0
0
2
0
1  cos 4 x
dx) 2  
2
1
b) Recall that two vectors are orthogonal iff their inner product is zero.
2
 sin 2 x, cos 2 x   (sin 2 x)(cos 2 x)dx 
0
sin 2 2 x
4
2
0
 0 (use u-sub with u=sin2x to
integrate). Thus they are orthogonal.
10)
a)
( x1 , y1 )  ( x2 , y2 )  (2x1 x2 , y1  y2 )
,
( x2 , y2 )  ( x1, y1 )  (2x2 x1, y2  y1 )  (2 x1 x2 , y1  y2 ) .
It
is commutative.
b) Looking for an element (a,b) such that ( x, y )  (a, b)  ( x, y ) for all x and y.
1
( x, y )  (a, b)  (2 xa, y  b)  ( x, y )  2 xa  x, y  b  b  a  , b  0 . The zero vector is
2
1
( ,0 )
2
1
2
c) Looking for an element (a,b) such that (1,2)  ( a, b)  ( ,0) , the zero vector for all x
1
2
1
2
and y. (1,2)  (a, b)  ( ,0)  (2a,2  b)  ( ,0)  a  4, b  2 . Thus  (1,2)  ( 4,2)
d) 1v  1( x, y )  (1x, y )  ( x, y )  v . Yes, 1v  v
11)
i)
ii)
Pick two first degree polynomials ax  b and cx  d . Then
p(( ax  b)  (cx  d ))  p(( a  c) x  (b  d )) 
3(a  c) x  2(b  d )  (3ax  2b)  (3cx  2d )  p(ax  b)  p(cx  d )
Let k be a real number. Then p(k (ax  b))  p(kax  kb)  3kax  2kb
 k (3ax  2b)  kp(ax  b)
Thus it is a linear transformation.
a) The matrix with respect to B  {x  1, 2 x  1} and {3,2 x  5} : Express the image of each
vector in B using the vectors in C. Then their coefficients are the columns of the
matrix.
p( x  1)  3x  2  c1 (3)  c2 (2x  5)  2c2 x  3c1  5c2
as partial fractions),
2c2  3, 3c1  5c2  2  c1  
Equating their coefficients (same technique
11
3
, c2  (they are placed in the first
6
2
column).
Similarly,
p(2x  1)  6x  2  c1 (3)  c2 (2x  5)  2c2 x  3c1  5c2
(same technique as partial fractions),
in the second column).
Equating their coefficients
2c2  6, 3c1  5c2  2  c1  
17
, c2  3 (they are placed
3
17 
 11
 6  3 
Thus [T]  

3

3 
 2

C
B
12)
a) LI? Let
c1v1  c2v2  0  c1 (1,0,2)  c2 (2,4,5)  (0,0,0)  (c1  2c2 ,4c2 ,2c1  5c2 )  (0,0,0) .
Write the system of equations in a matrix form , we get
1 2 0 1 2 0
0 4 0  0 1 0 . Since there are no free variables, c  c  0 . Thus the
1
2

 

2 5 0 0 0 0
vectors are LI.
b) The subspace spanned by the vectors: Find (x,y,z) that
c1 (1,0,2)  c2 (2,4,5)  ( x, y, z)
has a solution: we use the row echelon form:
c1 (1,0,2)  c2 (2,4,5)  ( x, y, z)  (c1  2c2 ,4c2 ,2c1  5c2 )  ( x, y, z) .
1 2
0 4

2 5
In augmented matrix,



x
x  1 2


y
y   0 1

4


z 
0 0  2 x  y  z 

4

This system has a solution iff the last row is entirely 0. Thus the subspace spanned is
given by the equation
 2x 
y
z0
4
13)
1 0 0
3 1  2


a) 1 2
0   switch Row 1 and Row 3, multiply by -1.  1 2 0 .

1 0 0 
3 1 2
1 0
to Row 3, Row1 (-1) to Row 2. No change in determinant.   0 2
0 1
Add Row1 (-3)
0
0 ‘Factor out”
2
1 0 0
1 0 0
2 from Row 2.  (2) 0 1
0 . Add Row 2 (-1) to Row 3:  (2) 0 1 0 . Thus the
0 1 2
0 0 2
determinant is (-2) times the product of the diagonal elements.  det( A)  4
b) The (2,3) entry of the adjoint matrix is the determinant of the 2x2 obtained by
crossing off the third row and second column , times
entry, use (3,2)). That is,
(1) 23
3 2
 2
1 0
 (1)i j
(note that for (2,3)
14) To find the composition, it is best to first express each linear transformation using
matrices. Here any bases can be used: so simply use the standard bases. Then the
composition is the product of their matrices. Recall that the matrix of a LT can be
constructed as
T (e1 ),T (en ) .
Since we can write
1  (1,0)  e1, x  (0,1)  e2 , T (e1 ) and
T (e2 ) are given. So you can simply place the given vectors xfhxfbinto columns.
1  1
0 1 
0 1  1  1 1 2 
. Then the matrix of T2T1  
T1  
, T2  




 . Thus
1 2 
3  1
3  1 1 2  2  5
1 2  1  7 
T2T1 (3x  1)  
   
  13x  7
2  5 3  13
2nd Solution:
Alternately, this problem can be solved without using matrices:
T2T1 (3x  1)  T2 (3T1 ( x)  1T1 (1))  T2 (3(2x  1)  1( x  1))  T2 (7 x  2)
 7T2 ( x)  2T2 (1)  7( x  1)  2(3x)  13x  7
15) A linear transformation is invertible iff its matrix is invertible. Thus first write the LT in
the matrix form. Then compute the determinant of the matrix to see if it is 0 or not. In
the matrix form, we can write 1  (1,0,0)  e1 , x  (0,1,0)  e2 , x  (0,0,1)  e3
2
T (ax 2  bx  c)  (2a  c) x 2  bx  T (1)  T (0 x 2  0 x  1)   x 2 , T ( x)  T (0 x 2  1x  0)  x, T ( x 2 )  2 x 2
 0 0 0


Thus the matrix of T is T (e1 ), T (e2 ), T (e3 )  0 1 0 . The determinant of this matrix is not


 1 0 2
0 since it has a row consisting of zeros. Therefore, the linear transformation is not
invertible.
1 2 1 0 1
2
1 0



16) First find the row echelon form of 0
1  1 2  0 1  1 2 

0  2 1 5 0 0 1  5
All the nonzero rows of the row echelon form form a basis for the row space: thus a basis
for the row space is {(1,0,1,2), (0,1,01,2), (0,0,1,5)}
For a basis of the column space, take the columns of the original matrix with leading 1 in
the row echelon form: take the first 3 columns. Thus a basis for the column space is
{(1,0,0), (0,1,2), (1,1,1)}
17)
a1  b1
a2  b2
a3  b3
a1  b1
a2  b2
a3  b3
a1  b1
a2  b2
a3  b3
c1 a1
c2  a2
c3 a3
c1 b1
c2  b2
c3 b3
a1
affecting the determinant)  a2
a3
a1
a2
a3
c1 a1
c2  a2
c3 a3
a1
separated the columns twice). But a2
a3
a1
a2
a3
 b1
 b2
 b3
b1
b2
b3
a1
a2
a3
c1
a1
c2  (1) a2
c3
a3
c1
a1 b1
c2   a2 b2
c3
a3 b3
b1
b2
b3
c1
c2
c3
c1
c2
c3
a1  b1
a2  b2
a3  b3
a1
a2
a3
c1
c2 ( a column can be separated without
c3
 b1
 b2
 b3
c1 b1
c2  b2
c3 b3
a1
a2
a3
c1 b1
c2  b2
c3 b3
 b1
 b2
 b3
c1
c2 (I
c3
c1
c2  0 , since it has two identical rows.
c3
since -1 factors out from the column 2.
since two columns are switched, and
 b1
 b2
 b3
b1
b2
b3
c1
c2  0 since two
c3
columns are the same after factoring out -1 from the second column. Therefore,
a1
 a2
a3
a1
a2
a3
a1
 2 a2
a3
c1 a1
c2  a2
c3 a3
b1
b2
b3
 b1
 b2
 b3
c1 b1
c2  b2
c3 b3
a1
a2
a3
 b1
 b2
 b3
c1 b1
c2  b2
c3 b3
c1
a1
c2  0  a2
c3
a3
b1
b2
b3
c1 a1
c2  a2
c3 a3
b1
b2
b3
c1
c2  0
c3
c1
c2
c3
18) A)
i)
First show the zero vector is in the space: (0,0,0) satisfies 0  0  2(0)  0 . Thus it
contains the zero vector.
( x1 , x2 , x3 )  S and ( y1 , y2 , y3 )  S (note that since
( y1 , y2 , y3 )  S , x1  x2  2 x3  0 and y1  y2  2 y3  0 ). Then ( x1 , x2 , x3 ) 
ii) Closed under addition: Let
( x1 , x2 , x3 )  S
and
( y1 , y2 , y3 )  ( x1  y1 , x2  y2 , x3  y3 ) and
( x1  y1 )  ( x2  y2 )  2( x3  y3 )  ( x1  x2  2 x3 )  ( y1  y2  2 y3 )  0 .
Thus it is closed under
addition.
iii) Under scalar multiplication: let k be a real number,
k ( x1 , x2 , x3 )  (kx1 , kx2 , kx3 )
and
( x1 , x2 , x3 )  S .
Then
kx1  kx2  2kx3  k ( x1  x2  2 x3 )  0 .
Thus it is closed under
scalar multiplication.
Therefore it is a subspace.
B) To find a basis:
( x1 , x2 , x3 )  S  x1  x2  2 x3  0 .
row echelon form has rank 1. Let
x2  s, x3  t .
Thus there are two free variables since the
Then
( x1 , x2 , x3 )  S  x1   x2  2 x3  s  2t .
 x1 
x  
 2
 x3 
 s  2t 
 s 


 t 
 1  2
s  1   t  0  . Thus ( x1 , x2 , x3 )  S  it is a linear combination of (1,1,0) and
 0   1 
(2,0,1) . They are clearly LI and spans the subspace. Therefore, {( 1,1,0), (2,0,1)} is a basis
for S.
19)
1 2  1 4 0
1 2  1 4 0


Find the row echelon form of the augmented matrix 3 5
1 2 0  0 1  4 10 0 .

2 4  2 8 0
0 0 0 0 0
Solve this: notice that x3 , x4 are free variables: x3  s, x4  t . Use the second equation and
solve for
x2 : x2  4 x3  10 x4  4s  10t . Take the first and solve for x1 :
x1  2 x2  x3  4 x4  7s  16t .
Thus
 x1 
x 
 2 
 x3 
 
 x4 
 7 s  16t 
 4 s  10t 




s


t


 7  16 
 4   10
.
s   t 
1  0 
  

0  1 
Thus
 7   16 
 4   10
}
{ , 
1  0 
  

0  1 
spans the null space. Since they are clearly LI, it is a basis. Thus the dimension of the null space
is 2.
20)
a) First show it contains the zero vector:
0 0 
A
 is in S since its second row
0 0 
entries are 0. Thus it contains the zero vector.
b) Next show it is closed under addition:
i)
Let
a b
c d 
a  c b  d 
and B  
be in S. Then A  B  
is still in S
A


0 
0 0 
 0
 0 0
since the second row is 0. It is closed under addition.
ii)
Let
a b
ka kb
A
, k be a real number. Then kA  
 is still in S since the

0 0
 0 0
second row is 0. It is closed under scalar multiplication.
c)
i)
ii)
1 0
 0 1  0 0 
c1 
 c2 


 . Then
0 0 
 0 0  0 0 
c c  0 0
A   1 2  
  c1  c2  0 . Therefore they are LI.
 0 0  0 0
a b a b 
1 0
0 1 
a 
 b
Spans: for any A  
, 


 , a linear combination

0 0 
0 0 
 0 0  0 0
1 0 0 1
of 
, 
 ( a is like c1 , b is like c2 ) . Thus it spans S. Therefore, it
0 0 0 0
Show LI: Suppose
is a subspace.
21) The subspace spanned by the vectors {(1,0,2), (2,1,4)} : Find all (x,y,z) that
c1 (1,0,2)  c2 (2,1,4)  ( x, y, z) has a solution:
we use the row echelon form:
c1 (3,1,2)  c2 (2,1,4)  ( x, y, z)  (3c1  2c2 , c2  c2 ,2c1  4c2 )  ( x, y, z) .
x 
 1  2 x  1  2
0


1 y   0 1
y 

 2 4 z  0 0  2 x  z 
In augmented matrix,
This system has a solution iff the last row is entirely 0. That is, the subspace spanned is a
plane given by the equation  2 x  z  0 (notice that two LI vectors in R3 spans a plane)
iii)
(1,-4,2) is in the span of {(1,0,-2),(-2,1,4)}. The vectors in the span of S must
satisfy the equation  2 x  z  0 : here  2(1)  2  0 .
22) It is clearly LD since there are three vectors in a 2 dimensional space. We use the row
echelon form to find dependency relation. Find
c1 , c2 , c3 satisfying
c1 (1,3)  c2 (3,1)  c3 (0,4)  (0,0) . Now
c1 (1,3)  c2 (3,1)  c3 (0,4)  (0,0)  (c1  3c2 ,3c1  c2  4c3 )  (0,0)
1 3 0 0
1 3 0 0
In the augmented matrix form, 
. In row echelon form, 


3  1 4 0
3  1 4 0
0
0
1 3
0 1  2 / 5 0 . c3 is a free variable: let c3  s . Solving the last equation for c2 , we


have c2  2 / 5c3  2 / 5s . Solving the first equation for c1 , c1  3c2  6 / 5s . Thus
6
2
(c1 , c2 , c3 )  ( s, s, s ) . Since we are looking for a relationship (one of many possible
5
5
relationships), we can pick s  5 (just a random number). This choice of s makes
(c1 , c2 , c3 )  (6,2,5) . Therefore,
c1 (1,3)  c2 (3,1)  c3 (0,4)  (0,0)  6(1,3)  2(3,1)  5(0,4)  (0,0)
23)
a) To show the product defined is an inner product, we must verity the three conditions:
i)
(positive definiteness)
1
1
0
0
 f , f   f ( x) f ( x)dx   f ( x) 2 dx . Since the integrand is
never negative,  f , f  0 and  f , f  0 iff f (x ) is a zero polynomial.
ii)
(Symmetry)
iii)
(Linearity)
1
1
0
1
0
 f , g   f ( x) g ( x)dx   g ( x) f ( x)dx  g , f 
1
 af , g   af ( x) g ( x)dx  a  f ( x) g ( x)dx  a  f , g  and
0
1
0
1
 f  g , h   ( f ( x)  g ( x))h( x)dx   ( f ( x)h( x)  g ( x)h( x))dx
0
0
1
1
0
0
  f ( x)h( x)dx   g ( x)h( x)dx  f , h    g , h 
Therefore
1
 f , g   f ( x) g ( x)dx is an inner product.
0
b) Next use Gram-Schdmit process to construct an orthogonal basis:
v1  u1  1
v2  u 2 
 v2 , u1 
u1
2
1
u1 ,  v2 , u1  x,1   x(1)dx 
0
1
1
0
0
u1  u1 , u1  1,1   1(1)dx   1dx 
2
Therefore,
v2  u2 
 v2 , u1 
u1
2
1
,
2
1
2
1
u1  x  2 (1)  x  1 . {1,x-1} is an orthogonal basis.
1
2
24) Since S  {(1,3,2), (4,2,4), (0,7,0)} contains 3 vectors R3 , it spans R3 if they are LI. We
1 4 0
1  4
can compute their determinant to see if they are LI. 3
2 7  (1) 23 7 
 12 ,
2 4 

2 4 0
which is not 0. Thus S spans R3
25) Since they are all differentiable functions, it is best to use Wronskian.
f1
f1 '
f2
f2 '
f1"
f2"
f3 1  x 2  x  x 2 1  x
f3 '  1
2x  1
 1  2((1  x)  (1  x))  4 , which is not the zero vector.
f3 '
0
2
0
Therefore, the polynomials are LI. Thus this set forms a basis since there are 3 LI
polynomials in a 3 dimensional space.
26) Subspace is obvious . Next show it is a basis.
 1 0  0 1   0 0 
{
, 
, 
} is a basis for all 2 x 2 upper triangular matrices.
 0 0  0 0   0 1 
1 0
0 1 
 0 0  0 0 
 c2 
 c3 
Show LI: Suppose c1 



 . Then
0 0 
0 0 
 0 1  0 0 
Claim:
c c  0 0
A   1 2  
  c1  c2  c3  0 . Therefore they areLI.
 0 c3  0 0
a b a b 
1 0
0 1
0 0
a 
 b
c 
Spans: for any A  
, 


 , a linear combination of


0 0 
0 1
0 c  0 c 
0 0
1 0 0 1 0 0
0 0, 0 0 , 0 1  ( a is like c1 , b is like c2 ) . Thus it spans S. Therefore, it is a

 
 

subspace.
27)
To verify that it is a linear transformation, Show a)
T (( x1 , x2 )  ( y1 , y2 ))  T ( x1 , x2 )  T ( y1 , y2 ), T (k ( x1 , x2 ))  kT ( x1 , x2 )
T (( x1, x2 )  ( y1, y2 ))  T ( x1  y1, x2  y2 )  ( x1  y1  x2  y2 , x1  y1  x2  y2 ) 
( x1  x2 , x1  x2 )  ( y1  y2 , y1  y2 )  T ( x1, x2 )  T ( y1 , y2 )
T (k ( x1, x2 ))  T (kx1, kx2 )  (kx1  kx2 , kx1  kx2 )  k ( x1  x2 , x1  x2 )  kT( x1, x2 ) .
i)
ii)
Thus it is a linear transformation.
For the kernel, first find a matrix of T. Then compute the null space by finding its row
T ( x1, x2 )  ( x1  x2 , x1  x2 )  T (1,0)  (1,1), T (0,1)  (1,1) . Thus placing them
1 1 
into columns, the matrix of T is 
 . In row echelon form in augmented matrix, ,
1 1
echelon form.
1 1

1  1
0 1 1 0
. You can see that the null space is trivial since there are no free

0 0 1 0
variables. No basis for the null space. Using the rank-nullity theorem, the dimension of
the range is 2-0=2.
28) Compute the determinant . Recall that a matrix is invertible iff det( A)  0 . Use the
cofactor expansion along the third column.
2 0
 1
 k
  det( A)  0  2(k  2(4k  1))  0  k   2
0
2


7
4k  1 k 0
Thus A is invertible iff
k
2
7
29)
a)
(a, b)  (c, d )  (2a  2c,2b  2d ), (c, d )  (a, b)  (2c  2a,2d  2b)  (2a  2c,2b  2d ) .
It is
commutative.
b) Zero element is (c, d ) such that (a, b)  (c, d )  (a, b) for all (a, b) . But
a
b
(a, b)  (c, d )  (2a  2c,2b  2d )  (a, b)  2a  2c  a,2b  2d  b  c   , d   . But
2
2
this choice of (c, d ) depends on (a, b) . Thus it is impossible to find (c, d ) that works
for all (a, b) . Thus there is no zero element.
c)
30) Doing row echelon form is better than using the determinant: a dependency relation
can be found if it is not LI. The determinant does not give you a relationship. Suppose
c1 (1,3,1)  c2 (1,3,7)  c3 (2,3,2)  (0,0,0) . Then
(c1  c2  2c3 ,3c1  3c2  3c3 , c1  7c2  2c3 )  (0,0,0)
1  1  2 0 1  1  2 0

1

 
Using augmented matrix, 3 3  3 0  0 1
0 Since there is a free variable


2
1 7
2 0 0 0
0 0
(thus there is non-trivial solutions) , the set is dependent. To find a relation, find
1
2
c1 , c2 , c3 .
1
2
c3 is a free variable, so let c3  s . Use the second equation to find c2 : c2   c3   s .
the first equation to find
1
2
3
2
c1 : c1  c2  2c3   s  2s  s .
3
2
Thus (c1 , c2 , c3 )  ( s,
Use
1
s, s) . Since
2
we are finding a relation (there are infinitely many of them), we can assign a number to s.
s  2 . Then (c1 , c2 , c3 )  (3  1,2) Thus a relation is
c1 (1,3,1)  c2 (1,3,7)  c3 (2,3,2)  (0,0,0)  3(1,3,1)  (1)( 1,3,7)  2(2,3,2)  (0,0,0)
So let
31)
0 2 2 1 0 0
 2 2 4 0 1 0




A  2 2 4 0 1 0 ( R1  R 2)  0 2 2 1 0 0 ( R1  2, R 2  2) 
0 3 1 0 0 1 
0 3 1 0 0 1 




1 1 2 0 1 / 2 0
1 0 1  1 / 2 1 / 2 0




1 0
0 1 1 1 / 2 0 0 ( R 2(1)  R1, R 2(3)  R3)  0 1 1 1 / 2
0 3 1 0
0 0  2  3 / 2 0 1 
0 1



1 0 1  1 / 2 1 / 2 0
1 0 0  5 / 4 1 / 2  1




1 0  0 1 0  1 / 4 1  1
0 1 1 1 / 2
0 0  2  3 / 2 0 1 
0 0 1 3 / 4
0
1 



Thus
 5 / 4 1 / 2  1
A    1 / 4 1  1
 3 / 4
0
1 
1
32)
a) (1,3) and (1,1) is a basis for R2 since
1 1
 2  0 . Two LI vectors in R2 is a basis.
3 1
b) First express (1,0) as a linear combination of (1,3) and (1,1). Then place the
coefficients
c1 ,c2
in the first column. Next express (0,1) as a linear combination of
(1,3) and (1,1). Then place
c1 ,c2
in the second column.
(1,0)  c1 (3,1)  c2 (1,1)  (1,0)  (3c1  c2 , c1  c2 )  (1,0)
1
1
 3c1  c2  1, c1  c2  0  c1  , c2   .
2
2
(0,1)  c1 (3,1)  c2 (1,1)  (0,1)  (3c1  c2 , c1  c2 )  (0,1)
1
3
 3c1  c2  0, c1  c2  1  c1   , c2 
2
2
 1

Thus the matrix is  2
1

 2
1
 
2
3 

2 
33) Write each of the element in
Since
1  3 2 0  1  1 0 0
{
, 
, 
, 
} using the standard basis .
0 1  1 2  1  1 0 1
M 2 (R) is a 4 dimensional vector space over R, the change of basis matrix is 4 x 4.
1  3 1 0 0 1 0 0 0 0
0 1   10 0  30 0  01 0  10 1 . Then write the coefficient 1, -3, 0, 1 in

 
 
 
 

2 0  1  1 0 0
the first column. Repeat this process for 
, 
, 
 and we obtain
1 2  1  1 0 1
 1 2 1 0
 3 0  1 0


 0 1  1 0


 1 2  1 1
First ,
34)
a) True. If AB is invertible then det( AB)  0  det( A) det( B)  0  det( A) and det( B ) are
both not zero.
b) False. If
1 0
0 0 
1 0
A
and B  
, A B  
 is invertible but A, B are not


0 1
0 0 
0 1 
invertible.
c) True. The rank is at most 2, and there are 3 variables. Thus at least one of the
variables is free.
d) True. R satisfies all the vector space axioms.
e) False. A system of homogeneous equations always have trivial solution.
35)
f)
0 is an unique element (a,b) such that
( x1 , y )  (a, b)  ( x, y), ( x, y).
( x, y )  (a, b)  ( x  a, yb)  ( x, y )  x  a  x, yb  y  a  0, b  1. Thus (0,1) is the zero
element.
g)  (3,4) is an element (a,b) such that (3,4)  ( a, b)  (0,1),
(3,4)  (a, b)  (3  a,4b)  (0,1)  3  a  0, 4b  1  a  3, b 
1
.
4
Thus the additive
1
4
inverse of (3,4) is (3, )
36)
To verify that it is a linear transformation, Show a)
T (( x1 , x2 , x3 )  ( y1 , y2 , y3 ))  T ( x1 , x2 , x3 )  T ( y1 , y2 , y3 ), T (k ( x1 , x2 , x3 ))  kT ( x1 , x2 , x3 )
iii)
T (( x1 , x2 , x3 )  ( y1 , y2 , y3 ))  T ( x1  y1 , x2  y2 , x3  y3 ) 
( x1  y1  2( x3  y3 ), x2  y2 )  ( x1  2 x3 , x2 )  ( y1  2 y3 , y2 )  T ( x1 , x2 , x3 )  T ( y1 , y2 , y3 )
iv)
T (k ( x1 , x2 , x3 ))  T (kx1 , kx2 , kx3 )  (kx1  2kx3kx2 )  k ( x1  2 x3 , x2 )  kT ( x1 , x2 , x3 ) .
Thus it is a linear transformation.
For the kernel, first find the matrix of T. Then compute the null space by finding its row
echelon form.
T ( x1 , x2 , x3 )  ( x1  2 x3 , x2 )  T (1,0,0)  (1,0,0), T (0,1,0)  (0,1,0), T (0,0,1)  (2,0,0) .
1 0  2

Thus placing them into columns, the matrix of T is 0 1
0  . This is already in a row

0 0 0 
echelon form. You can see that x3 is a free variable since there is no leading 1. Using the
first and second equations, we get
x1  0, x2  0
the null space is (0,0, s )  s (0,0,1) . Thus a
basis for the null space is (0,0,1).
37)
3a  3b 3b 3c
ab
 d  e  e  f  (1)(3) d  e
gh
h
i
gh
a b
the second row)  (3)( d e
g h
b
e
h
c
f ( ‘factor out’ 3 from the first row, (-1) from
i
c b b
f e e
i h h
c
f ) (A column can be separated without
i
b b
changing the determinant). But e e
h h
Therefore,
3a  3b 3b 3c
a
 d  e  e  f  (1)(3) d
gh
h
i
g
38) To find the adjoint matrix of
c
f  0 , since it has two identical columns.
i
b
e
h
c
f  (3)(5)  15
i
1 1 1
0 2 0  :


0 0 2
a) First compute the determinant: Since A is an upper triangular, it is the product of its
diagonal elements: det(A)=1(2)(2)=4
b) Next find the cofactor matrix:
 4 0 0
 2 2 0


 2 0 2
c)
d)
 4  2  2

Take the transpose of the cofactor matrix. 0
2
0 

0 0
2 
1
1

1  2  2 


1
1
0 
Divide each entry by the determinant: A  0
2


1 
0 0

2 
39)
Since
{v1 , v2 , v3} , c1 , c2 , c3 , not all 0 such that c1v1  c2v2  c3v3  0 .
LD, observe that
0. Thus
To show
{v1 , v2 , v3 , v4 }
is
c1v1  c2v2  c3v3  0  c1v1  c2v2  c3v3  0v4  0 and not all coefficients are
{v1 , v2 , v3 , v4 }
is LD
40) First find the row echelon form of
1 3 0 0 1 3 0 0 1 3 0 0
1 4 1 1 : 1 4 1 1  0 1 1 1

 
 

1 3 0 0 0

 . c3 and c4 are free variables:
0 1 1 1 0
c3  s, c4  t . Solving the second equation for c2 , c2  s  t . Using the first equation,
For the null space, solve the augmented matrix
 c1  3s  3t 
3 3
c    s  t 
 1  1
2



.
Thus
c1  3c2  3s  3t

 s    t   . Thus a basis for the null space is
c3   s 
1 0
  

   
0 1
c4   t 
3 3
 1  1
{ ,  }
1 0
   
0 1
For the column space, take the first column and the second column of the original matrix
since the column1 and column 2 have the leading 1 in row echelon form. A basis for the
column space is
{(1,1), (3,4)}
41)
a) To show it is a subspace (observe that the requirement to be in S is the polynomial is
degree 2 or less with 0 constant term),
i)
Show 0 is in S: It is in s since 0  0 x  0 x
2
(ax  bx 2 )  (cx  dx 2 )  (a  c) x  (b  d ) x 2  S
iii)
To show it is closed under addition,
iv)
To show it is closed under scalar multiplication,
k (ax  bx 2 )  kax  kbx 2  S
ax  bx 2  a( x)  b( x 2 ) . Thus {x, x 2 } is a basis. The dimension is 2.
0  1
Claim: A  
 is a basis for the set of 2 x 2 skew symmetric matrices.
1 0 
b)
42)
\
a) Since there is only one vector, it is LI
b) Span: Let A be a skew symmetric matrix. Then
0  a 
0  1
A
 a


a 0 
1 0 
Thus it is a 1 dimensional subspace.
First express (3,4) in terms of v1, v2 : c1 (1,2)  c2 (2,3)  (3,4)  c1
43)
(3,4)  v1  2v2 .
Thus
 1, c2  2 .
Thus
T (3,4)  T (v1  2v2 )  Tv1  2Tv2  (1,3)  2(0,2)  (1,1)
44)
a) False: the number of solutions is 0, 1 or infinite. This is same as in elementary
algebra.
b) True. Since det( AB)  det( A) det( B) , A and B must both have non zero determinant.
A  AT T AT  ( AT )T AT  A
) 

B
2
2
2
1 0
0 0 
1 0
d) False: If A  
, B
, det A  det B  0 , but A  B  
 so det( A  B)  1


0 1
0 0 
0 1 
c) True:
BT  (
e) True: since there are two rows, its row echelon form has at most two leading 1s.
This leaves at least two free variables (recall that the variables without the leading 1
become free variables)
f)
 0 1  2  0 0 0 
1 0 0  , 0 0 0 are some examples of 3 x 3 skew symmetric matrices.


 
2 0 0  0 0 0
g) False: if a matrix A is invertible then det( A)  0
45)
a)
det( A)  4, , det( B)  2, det(2A -1B2 )  24
a1  b1
b) a2  b2
a3  b3
3b1
3b2
3b3
c1 a1
c2  a2
c3 a3
3b1
3b2
3b3
c1 b1
c2  b2
c3 b3
a1
affecting the determinant)  3 a2
a3
1
1
det( B) 2  16 (2)  8
det( A)
4
3b1 c1
3b2
3b3
b1
b2
b3
c2 ( a column can be separated without
c3
c1
b1
c2  3 b2
c3
b3
b1
b2
b3
c1
c2 ( 3 can be factored out). But
c3
the second determinant is 0. K=3
c) This is already in a row echelon form. There are two free variables: y and z.
y  s, z  t  x  2s  3t The solutions are ( x, y, z )  (2s  3t , s, t )  s(2,1,0)  t (3,0,1)
46)
a)
T ( f ( x)  g ( x))  ( f ( x)  g ( x))'  f ' ( x)  g ' ( x)  T ( f ( x))  T ( g ( x))
T (kf ( x))  (kf ( x))'  k ( f ' ( x))  kT ( f ( x))
Thus it is a LT.
b) The kernel is all functions whose derivative is 0. Using common sense, it is all
constant functions.
c) No, since the kernel is not trivial.
47) The operation is clearly commutative, thus you need only to check one direction.
i)
( x, y )  (0,1)  ( x, y ) for all x, y. Thus 0  (0,1)
ii)
( x, y )  (a, b)  (0,1)  ( x  a, by )  (0,1)  a   x, b 
 u does not exist.
rs ( x, y)  (rs, (rs )2 y)  (rs, r 2 s 2 y)
1
.
y
Thus
1
 u  ( x, ) if y  0 .
y
If
y  0,
then
iii)
and
r (s( x, y))  r (sx, s 2 y)  (rsx, r 2 s 2 y) .
Thus it is true;
48)
First compute
1 0 1 0
, 
} .
0 1 0 2
T (1,1) and T (0,1) (using basis vectors) and express them using C  {
1 0
1 0
1 0
T (1,1)  
  c1 0 1  c2 0 2  c1  1, c2  0 (use common sense) .
0 1




Place the values in the
first column of the matrix.
0 0
1 0
1 0
 c1 
 c2 
T (0,1)  


  c1  1, c2  1(use common sense) Place the values in the
0 1
0 1
0 2
second column of the matrix.
1b) Thus the matrix of T is
basis
1
T (1,1)  
0
1
1
Suppose the matrix of
1 1 
B  C  {(1,1), (0,1)} is given as 

1 1
. Find
T : R 2  R 2 with respect to the
T (2,3)
Solution: You need to interpret the given matrix correctly.
1 1 
1 1


From the matrix,
Then
T (1,1)  1(1,1)  1(0,1)  (1,0)
T (0,1)  1(1,1)  (1)(0,1)  (1,2)
T (2,3)  T (2(1,1)  5(0,1))  2T (1,1)  5T (0,1)  2(1,0)  5(1,2)  (7,10)
49)
a)
5(0)  0v1  0v2  0 .
Since there is a nontrivial way to make the zero vector, it is LD.
(you can use any nonzero number for the coefficient of 0)
c1 (w1  w2 )  c2 (w1  w2 )  0 . (need to show c1  c2  0 )
c1 (w1  w2 )  c2 (w1  w2 )  0  (c1  c2 )w1  (c1  c2 )w2  0 Since {w1, w2} is LI,
c1  c2  0, c1  c2  0 . Solving this using Math 51, c1  c2  0 . Thus they are LI.
b) Suppose
50)
a)
T ((a1x  b1 )  (a2 x  b2 ))  T ((a1  a2 ) x  (b1  b2 ))
 (2(a1  a2 )  3(b1  b2 ))  (4(a1  a2 )  5(b1  b2 ) x)  (16(a1  a2 )  9(b1  b2 )) x 2
 (2a1  3b1 )  (4a1  5b1 ) x  (16a1  9b1 ) x 2  (2a2  3b2 )  (4a2  5b2 ) x  (16a2  9b2 ) x 2
 T (a1x  b1 )  T (a2 x  b2 )
And
T (c(ax  b))  T (cax  cb)  (2ca  3cb)  (4ca  5cb) x  (16ca  9cb) x 2
 c[(2a  3b)  (4a  5b) x  (16a  9b) x 2 ]  cT (ax  b)
Thus T is a LT
b) Method1: Find a matrix representation M of T. Then the range of T is the column
space of M. Here we use the ordered bases
T (1)  T (0 x  1)  3  5x  9 x
representation. Its REF is
2

1
0

0 

.
B  {1, x}, C  {1, x, x 2}
T ( x)  T (1x  0)  2  4 x  16 x
2
. Thus
3 2
5 4 is a matrix


9 6
2
3
1  , so the first column and the second column both have a

0

leading 1. Going back to the original matrix, a basis for the range is
{3  5x  9 x 2 ,2  4 x  16 x 2}
Alternative solution:
T (ax  b)  [(2a  3b)  (4a  5b) x  (16a  9b) x 2 ]  a(2  4 x  16 x 2 )  b(3  5x  9 x 2 ) . Thus
2
2
the range is spanned by {3  5 x  9 x ,2  4 x  16 x } . The polynomials are LI since one is not
2
2
a multiple of the other. Thus {3  5 x  9 x ,2  4 x  16 x } is a basis for the range.
c) The rank-nullity theorem says dim( P1 ) =dim(Kernel)+dim(Range). dim( P1 ) =2 and
dim(range)=2, so dim(kernel)=0. Thus T is 1-1 since its kernel is trivial.
51)
a) False: They span a line or a plane if the vectors are nonzero.
b) True. A scalar may be pulled out.
c) False: v does not even make sense.
d) False: 1 is a real number, so it is always the usual 1.
2
52)
a) To find a matrix, find T (1,0) and T (0,1) .
T (1,0)  (0,1)
T (0,1)  (1,0)
and
Thus its matrix is
b)
 0  1
 1 0 


dim( R 4 )  dim(ker T )  dim( RngT )
and
dim( R 4 )  4, dim(ker T )  1  dim( RngT )  3 . Since
dim( R3 )  dim( RngT ) , T is onto.
53)
1 0

x


y
0 1
  3x  2 y  z  0 when the system is consistent.
2
0 0 3x  2 y  z 


b) Make sure (2,1,z) is not in the plane 3 x  2 y  z  0 . So 3(2)  2(1)  z  0  z  8 . Thus you
must avoid z  8 .
0 x
1

2 y   REF
a)
0
 3  4 z 
54)
a) Recall that Span of a set is all the linear combinations of the vectors in the set.
Span {v1 , v1  v2}  c1v1  c2 (v1  v2 ) for all
c1  c2 can also be
b)
c)
any number,
c1,c2  R . But c1v1  c2 (v1  v2 )  (c1  c2 )v1  c2v2
c1v1  c2 (v1  v2 )  (c1  c2 )v1  c2v2 =span{ {v1 , v2}
Since
T2  T1 (v1  v2 )  T2 (T1 (v1  v2 ))  T2 (T1 (v1 )  T1 (v2 ))  T2 (T1 (v1))  T2 (T1(v2 ))  T2  T1 (v1 ) 
T2  T1 (v2 )
And T2  T1 (cv)  T2 (T1 (cv))  T2 (cT1 (v))  cT2 (T1v)  cT2  T1 (v) Thus T2  T1 is a LT.
Suppose c1T (v1 )  c2T (v2 )  0 . NTS c1  c2  0 . Since T is a LT, c1T (v1 )  c2T (v2 )  0 
T (c1v1  c2v2 )  0 . Since T is 1-1, its kernel is trivial. Thus c1v1  c2v2  0 . But this
implies c1  c2  0 since {v1 , v2 } is LI. Therefore {T (v1 ), T (v2 )} is LI.