Math 285 Past Exam II Solutions
1)
3 2 2
a)
1
1
5
5
1
1
2
2  switch row 1 and row 2 and multiply by -1  3  2  2  b) Make the
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
Row 3: No change to the determinant.
1
 0 5
0
0
2
 8  c) next ‘factor out’ -5 from
 14
1 1 2
8
 det A  (1)( 5)( 14)(1)  70
the row 2, -14 from the row 3 ( 1)( 5)( 14) 0 1
5
0 0 1
2)
a) 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  s . Solving the

 

second equation for y, y  7 z  6  y  7 z  6  y  7 s  6 . Solving the first equation
for x, x  2 y  3z  8  x  2 y  3z  8  x  2(7 s  6)  3s  8  17 s  20 . Thus
( x, y, z )  (17 s  20,7 s  6, s )
2 1 0
3
b)
z
0
10
5 0 14  (1)(3(14)  5(10))

1
2 1  5
 (1)( 3  5)
3 0
1
5
0
1
3) Find a row echelon form of
x yz a
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 rank A = rank A# iff the last row is entirely zero. Thus
4) A) To show T is a LT,
i) T(x+y)=A(x+y)=Ax+Ay=Tx+Ty
ii) T(kx)=A(kx)=k(AX)=kTx
1
a b c  0 .
c) 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)
1
1
a) det( S AS )  det( s ) det( A) det( S ) 
1
det( A) det( S )  det( A)
det( S )
b) Since 2 comes out from each of the three rows, 2 becomes 8 outside the determinant
function: det( 2 A
2
d)
a)
b)
c)
d)
e)
f)
g)
e)
)  8 det( A2 )  8 det( A) det( A)  128
It 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
T1 (ax  b)  3ax  (a  b)  T (1)  1 (a  0 b  1) , and T ( x)  3x  1 (a  1, b  0) . Thus
 1 1
placing them into columns, T  

 0 3
f)
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
and B  
be diagonal matrices. Then A  B  
is still
A


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.
g)
a) Recall that
2
f   f , f  . Thus
1
2
2
1
2
f  (  (sin 2 x)(sin 2 x)dx)  (  (sin 2 x)dx)  ( 
0
2
0
2
2
0
1  cos 4 x
dx) 2  
2
1
b) Recall that two vectors are orthogonal iff their inner product is zero.
 sin 2 x, cos 2 x  ( 
2
0
1
2
sin 2 2 x
(sin 2 x)(cos 2 x)dx) 
4
2
0
 0 (use u-sub to integrate). Thus
they are orthogonal.
10)
c)
( x1, y1 )  ( x2 , y2 )  (2x1 x2 , y1  y2 ) , ( x2 , y2 )  ( x1, y1 )  (2x2 x1, y2  y1 )  (2x1 x2 , y1  y2 ) . It is
commutative.
d) 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 ) element is
2
1
( ,0 )
2
11)
a)
i)
ii)
Pick two first degree polynomials ax  b, 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.
b) 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 Equating their coefficients (same technique
11
3
as partial fractions), 2c2  3, 3c1  5c2  2  c1   , c2  (they are placed in the first
6
2
column).
p(2x  1)  6x  2  c1 (3)  c2 (2x  5)  2c2 x  3c1  5c2 Equating their coefficients
17
, c2  3 (they are placed
(same technique as partial fractions), 2c2  6, 3c1  5c2  2  c1  
3
Similarly,
in the second column).
17 
 11

 

C
3
Thus [T]B   6

3

3 
 2

3
12)
c1v1  c2v2  0  c1 (1,0,2)  c2 (2,4,5)  (0,0,0)  (c1  2c2 ,4c2 ,2c1  5c2 )  (0,0,0) . By
setting each component 0, we obtain c1  2c2  0, 4c2  0, 2c1  5c2  0 . The second
equation gives c2  0 . Then substituting c2  0 into the first equation, we get c1  0 .
a) LI? Let
Thus the vectors are linearly independent.
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) . In augmented matrix,
1 2
0 4

2 5




1
2
x
x


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. That is, the subspace spanned is
 2x 
y
z0
4
13)
1 0 0
3 1  2


0   switch Row 1 and Row 3, multiply by -1.  1 2 0 . Add Row1 (-3)
a) 1 2

3 1 2
1 0 0 
1 0 0
to Row 3, Row1 (-1) to Row 2. No change in determinant.
 0 2
0 ‘Factor out”
0 1 2
1 0
2 from Row 2.
 (2) 0 1
0
1 0
0 . Add Row 2 (-1) to Row 3:  (2) 0 1
0 1 2
0
0 . Thus the
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,
14)
4
(1) 23
3 2
1
0
 2
 (1)i j (note that for (2,3)
To find the composition, it is best to first express each linear transformation using a
matrix. Then the composition is the product of their matrices. Recall that the
standard matrix of a LT can be constructed as
T (e1 ),T (en ) (other bases can be
used, but this is the easiest). 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 into 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)  
   
  12 x  7
2  5 3  12
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)   x 2 , T ( x)  x, T ( x 2 )  2 x 2 . Thus the matrix of T is
 0 0 0
T (e1 ), T (e2 ), T (e3 )   0 1 0 . The determinant of this matrix is not 0 since it has a row
 1 0 2
consisting of zeros. Therefore, the linear transformation is not invertible.
16)
2
1 2  1 0 1
1 0

1  1 2  0 1  1 2 
a) First find the row echelon form of 0

0  2 1 5 0 0 1  5
All the nonzero rows of the row echelon 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)}
b) Yes, it is true. The dimension of the row space and the dimension of the column
space are both determined by the number of leading 1s in a REF.
17)
5
a1  b1
a1  b1
c1
a1
a1  b1
c1
b1
a1  b1
c1
a2  b2
a2  b2
c2  a2
a2  b2
c2  b2
a2  b2
c2 ( a column can be separated without
a3  b3
a3  b3
c3
a3  b3
c3
b3
a3  b3
c3
a1
a1
c1
a1
 b1
c1
b1
a1
c1
b1
 b1
c1
 a2
a2
c2  a2
 b2
c2  b2
a2
c2  b2
 b2
c2 (I
a3
a3
c3
 b3
c3
a3
c3
 b3
c3
a3
affecting the determinant)
separated the columns twice). But
a1
 b1
c1
a1
b1
c1
a2
 b2
b2
c2
a3
 b3
c2  (1) a2
a3
c3
b3
c3
b1
a1
c1
b1
c1
b2
a2
c2   a2 b2
c2
b3
a3
c3
c3
a1
a3
b3
a3
b3
b3
a1
a1
c1
a2
a2
c2  0 , since it has two identical rows.
a3
a3
c3
since -1 factors out from the column 2.
since two columns are switched, and
b1
 b1
c1
b2
 b2
c2  0 since two
b3
 b3
c3
columns are the same after factoring out -1 from the second column. Therefore,
a1
a1
c1
a1
 b1
c1
b1
a1
c1
b1
 b1
c1
a1
b1
c1
a1
b1
c1
 a2
a2
c2  a2
 b2
c2  b2
a2
c2  b2
 b2
b2
c2  a 2
b2
c2  0
a3
a3
c3
 b3
c3
a3
c3
 b3
c2  0  a2
a3
c3
b3
c3
b3
c3
a3
a1
b1
c1
 2 a2
b2
c2
a3
b3
c3
b3
b3
a3
18)
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
( x1 , x2 , x3 )  S and ( y1 , y2 , y3 )  S , x1  x2  2 x3  0 and y1  y2  2 y3  0 ). Then ( x1 , x2 , x3 ) 
( 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
ii) Closed under addition: Let
addition.
( x1 , x2 , x3 )  S . Then
k ( x1 , x2 , x3 )  (kx1 , kx2 , kx3 ) and kx1  kx2  2kx3  k ( x1  x2  2 x3 )  0 . Thus it is closed udenr
iii) Under scalar multiplication: let k be a real number,
scalar multiplication.
Therefore it is a subspace.
B) To find a basis:
( x1 , x2 , x3 )  S  x1  x2  2 x3  0 . Thus there are two free variables since the
row echelon form has rank 1. Let
6
x2  s, x3  t . Then ( x1 , x2 , x3 )  S  x1   x2  2 x3  s  2t .
 x1   s  2t 
 1  2
x   

   
 2   s   s  1   t  0  . Thus ( x1 , x2 , x3 )  S  it is a linear combination of (1,1,0) and
 x3   t 
 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. It
A


0 
0 0 
 0
 0 0
is closed under addition.
ii)
Let
a b 
ka kb
, k be a real number. Then kA  
A
 is still in S. It is closed

0 0
 0 0
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 areLI.
 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.
7
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) . In augmented matrix,
x 
 1  2 x  1  2
0
1 y   0 1
y 

 2 4 z  0 0  2 x  z 
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)}. In a) we found that any vector
satisfying  2 x  z  0 is in the span of {(1,0,-2),(-2,1,4)} and  2(1)  2  0 .
22) It is LD since there are three vectors in a 2 dimensional space. We use the row echelon
form to find dependency relation. Find
Now
c1 , c2 , c3 satisfying c1 (1,3)  c2 (3,1)  c3 (0,4)  (0,0) .
c1 (1,3)  c2 (3,1)  c3 (0,4)  (0,0)  (c1  3c2 ,3c1  c2  4c3 )  (0,0)
In the augmented matrix form,
1 3 0 0
1 3 0 0
3  1 4 0 . In row echelon form, 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
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)
relationships), we can pick
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)
iii)
(Symmetry)
(Linearity)
1
1
0
1
0
1
 f , g   f ( x) g ( x)dx   f ( x) g ( x)dx   g ( x) f ( x)dx  g , f 
0
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 
8
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 
u1
2
 v2 , u1 
u1
2
1
u1 ,  v2 , u1  x,1   x(1)dx 
0
1
1
0
0
 u1 , u1  1,1   1(1)dx   1dx 
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
can compute their determinant to see if they are LI.
which is not 0. Thus S spans R
3
2
2
4
1  4 
7  (1) 23 7 
 12 ,
2 4 

0
3
Since S  {(1,3,2), (4,2,4), (0,7,0)} is LI and spans R3, it is a basis for R3.
25) Since S consists of 3 polynomials in a 3 dimensional vector space, you only need to
show S is LI.
Since they are all differentiable functions, it is best to use Wronskian.
1  x 2  x  x2 1  x
f1
f2
f3
f1 '
f2 '
f3 '  1
f1"
f2"
f3 '
0
2x  1
2
 1  2((1  x)  (1  x))  4 , which is not always 0.
0
Therefore, the polynomials are LI and this set forms a basis.
26)
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 
c c  0 0
A   1 2  
  c1  c2  c3  0 . Therefore they are LI.
 0 c3  0 0
Claim:
9
a
A
0
0 1 0
0 0 , 0

 
Spans: First pick a typical upper triangular matrix . For
0 1 
0 0
1 0
, a linear combination of 
 b
c 

,

0 1
0 0
0 0 
like c2 ) . Thus it spans S. Therefore, it is a basis.
b a b 
1 0
, 
a 



c  0 c 
0 0 
0
 ( a is like c1 , b is
1
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 )
i)
ii)
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 ) .
Thus it is a linear transformation.
For the kernel, it is best to find a matrix of T first. Then compute the null space by
T ( x1, x2 )  ( x1  x2 , x1  x2 )  T (1,0)  (1,1), T (0,1)  (1,1) . Thus
1 1 
placing them into columns, the matrix of T is 
 . In row echelon form in augmented
1 1
finding its row echelon form.
matrix, ,
1 1

1  1
0 1 1 0
. You can see that the null space is trivial since there is

0 0 1 0
no free variables. No basis for the null space. Using the rank-nullity theorem, the
dimension of the range is 2-0=2.
28) Recall that a matrix is invertible iff det( A)  0 . Thus first compute the determinant as a
function of k . Use cofactor expansion along the third column.
2 0
 1
2
 k
0 2  det( A)  0  2(k  2(4k  1))  0  k  

7
4k  1 k 0
Thus A is invertible iff
10
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 this
2
2
choice of (c, d ) depends on (a, b) . Thus it is impossible to find one vector (c, d ) that
works for all (a, b) . Thus there is no zero element.
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

 
0 Since there is a free variable
Using augmented matrix, 3 3  3 0  0 1


2
1 7
2 0 0 0
0 0
(thus there is non-trivial solutions) , the set is dependent. To find a relation, find c1 , c2 , c3 .
1
1
c3 is a free variable, so let c3  s . Use the second equation to find c2 : c2   c3   s . Use
2
2
1
3
3
1
the first equation to find c1 : c1  c2  2c3   s  2s  s . Thus (c1 , c2 , c3 )  ( s, s, s ) . Since
2
2
2
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




0 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



11
1 0 1  1 / 2 1 / 2 0
1 0



1
0 0 (R3( ))  0 1
0 1 1 1 / 2
2
0 0  2  3 / 2 0 1 
0 0



1 0 0  1 / 2

 R3(1)  R 2, R3(1)  R1  0 1 0  1 / 4
0 0 1 3 / 4

Thus
1 1/ 2 1/ 2
1 1/ 2
1
1 3/ 4
0
1/ 2
0
0
0 

0 
 1 / 2
1/ 2 

1/ 2 
 1 / 2
 1 / 2 1 / 2 1 / 2 
A   1 / 4 0
1 / 2 
 3 / 4
0  1 / 2
1
32)
a) (1,3) and (1,1) is a basis for R2 since
1 1
3 1
 2  0 . Two LI vectors in R2 is a basis.
b) Next express (1,0) as a linear combination of (1,3) and (1,1). Then place
c1 , c2 in the
first column. Next express (0,1) as a linear combination of (1,3) and (1,1). Then
c1 , c2 in the second column. (1,0)  c1 (3,1)  c2 (1,1)  (1,0)  (3c1  c2 , c  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 , c  c2 )  (0,1)
1
3
 3c1  c2  0, c1  c2  1  c1   , c2 
2
2
place
1
1
 1

Thus the matrix is  2
1

 2
1
 
2
3 

2 
33) Write each of the element in
Since
12
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
and B  
, A B  
A
 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 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.
13
b)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), T (0,1,0)  (0,1), T (0,0,1)  (2,0) . Thus
1 0  2
 . This is already in a row echelon
0 1 0 
placing them into columns, the matrix of T is 
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).
c)The dimension of the range is 3-1=2 (use the rank-nullity theorem)
37)
3a  3b
3b
ab
b
c
d e
 e  f  (1)(3) d  e
e
f ( ‘factor out’ 3 from the first row, (-1) from
gh
h
i
gh
h
3c
i
a b
the second row)
 (3)( d
e
g h
c
c
f e e
f ) (A column can be separated without
i
i
changing the determinant). But
Therefore,
b b
h h
b b
c
e
f  0 , since it has two identical columns.
e
h h
i
3a  3b
3b
a
b
c
d e
 e  f  (1)(3) d
e
f  (3)(5)  15
h
i
gh
h
3c
i
38) To find the adjoint matrix of
g
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
14
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
Divide each entry by the determinant: A  0
0 
2


1 
0 0

2 
39)
{v1 , v2 , v3} , c1 , c2 , c3 , not all 0 such that c1v1  c2v2  c3v3  0 . To show {v1 , v2 , v3 , v4 } is
LD, observe that c1v1  c2v2  c3v3  0  c1v1  c2v2  c3v3  0v4  0 and not all coefficients are
0. Thus {v1 , v2 , v3 , v4 } is LD
Since
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 
   
2
  s  1  t  1 . Thus a basis for the null space is
c1  3c3  3s  3t . Thus    
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)
iii)
15
Show 0 is in S: It is in s since 0  0 x  0 x
To show it is closed under addition,
2
(ax  bx 2 )  (cx  dx 2 )  (a  c) x  (b  d ) x 2  S
iv)
b)
To show it is closed under scalar multiplication,
k (ax  bx 2 )  kax  kbx 2  S
ax  bx 2  a( x)  b( x 2 ) . Claim: {x, x 2 } is a basis. A) for LI, suppose c1 x  c2 x 2  0 . But
2
2
2
zero means zero polynomial, so c1 x  c2 x  0  c1 x  c2 x  0 x  0 x  c1  c2  0 B) for
2
2
2
span, for a typical polynomial ax  bx  a( x)  b( x ) , a linear combination of x and x .
Thus it spans S. Thus
42) Claim:
0  1
A
 is a basis for the set of 2 x 2 skew symmetric matrices. \
1 0 
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.
43)
First express (3,4) in terms of the given basis v1 , v2 :
c1 (1,2)  c2 (2,3)  (3,4)  c1  1, c2  2 . Thus (3,4)  v1  2v2 . 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: B  (
T
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)
 0 1  2  0 0 0 
1 0 0  

f)

 , 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 B )  2
-1
16
2
4
1
1
det( B) 2  16 (2)  8
det( A)
4
b)
a1  b1
3b1
c1
a1
3b1
c1
b1
3b1
c1
a2  b2
3b2
c2  a2
3b2
c2  b2
3b2
c2 ( a column can be separated without
a3  b3
3b3
c3
3b3
c3
3b3
c3
a3
affecting the determinant)
b3
a1
b1
c1
b1
b1
c1
 3 a2
b2
c2  3 b2
b2
c2 ( 3 can be factored out). But
a3
b3
c3
b3
c3
b3
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) We know from calculus that that derivative can be separated for addition and a
constant can be pulled out,
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 the set of
all constant functions.
c) No, since the kernel is not trivial.
17