B U Department of Mathematics
Math 201 Matrix Theory
Fall 2003 Final Exam
This archive is a property of Bo˘
gazi¸
ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions.
This archive is a nonprofit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of
it, without nonprofit purpose may result in severe civil and criminal penalties.
1.)
Suppose
A
is a 4
×
3 matrix, and the complete solution to
Ax
=
1
4
1
1
is
x
=
0
1
1
+
c
0
2
1
, c
∈
R
(a) Find the second and the third columns of
A
.
(b) Determine the ranks of the coefficient matrix and the augmented matrix. Give all the known
information about the first column of
A
.
Solution:
(a)
Let
b
T
= [1 4 4 1]. From the particular solution when
c
= 0 it follows that
column
2
+ column
3
=
b.
The homogenous solution says that
2column
2
+ column
3
= 0
.
From these we obtain
column
2
=

b,
column
3
=
b
(b)
We have dim Null(
A
) = 1, which means rank(
A
) = 2. Since the system is consistent, we
have rank([
A

b
]) = rank(
A
) = 2. Since the matrix must have two linearly independent
columns, and all the remaining columns are multiples of
b
we can infer that the first
column of
A
is not a multiple of
b
.
2.)
Let
v
1
=
1
1
,
v
1
=
1
4
,
w
1
=
2
5
,
w
2
=
2
8
and
A
=

2
3
4
5
.
(a) Show that
β
I
=
{
v
1
, v
2
}
is a basis for
R
2
.
Solution:
Since dim
R
2
= 2, any pair of linearly independent vectors form a basis for
R
2
.
The
vectors
v
1
and
v
2
are linearly independent because
1
1
1
4
= 3
and therefore
β
I
is a basis for
R
2
.
(b) Suppose
A
= [
T
]
β
I
is the matrix representation of the linear transform
T
:
R
2
→
R
2
relative
to the basis
β
I
. Find the matrix representation of the same linear transformation
B
= [
T
]
β
II
with
respect to the basis
β
II
=
{
w
1
, w
2
}
.
Solution:
If we can find a matrix
M
such that [
w
1
w
2
] = [
v
1
v
2
]
M
then
B
is related to
A
by
B
=
M

1
AM
.
We note that
w
1
=
v
1
+
v
2
and
w
2
= 2
v
2
, so
M
=
1
0
1
2
is the desired matrix.
M

1
=
1
0

1
2
1
2
B
=
M

1
AM
=
1
6
4
2
.
3.)
Let
A
=
cos
θ
sin
θ
sin
θ
0
.
(a) Using the GramSchmidt process, find the
A
=
QR
factorization of
A
.
(b) Find the projection matrix which projects onto the column space of
A
.
Solution:
(a)
Let
a
=
cos
θ
sin
θ
and
b
=
sin
θ
0
.
Since
a
= 1 we set
q
1
=
a
.
Then we have
b
=
b

(
q
1
T
b
)
q
1
and
q
2
=
b / b
. We compute
b
= sin
2
θ
sin
θ

cos
θ
,
q
2
=
sin
θ

cos
θ
.
So
Q
=
cos
θ
sin
θ
sin
θ
cos
θ
and
R
=
q
T
1
a
q
T
1
b
0
q
T
2
b
=
1
cos
θ
sin
θ
0
sin
2
θ
(b) If
θ
is not an integer multiple of
π
, then
R
has linearly independent columns, hence
R
T
R
is invertible, and
P
=
A
(
A
T
A
)

1
A
T
=
Q
T
Q
=
I
=
1
0
0
1
.
If
θ
is an integer multiple of
π
, then the second column of
A
is zero, and the first column
is
a
= [1 0]
T
, so the column space is the one dimensional space spanned by
a
.
The
projection matrix onto this space is:
P
=
aa
T
a
T
a
=
1
0
1
0
=
1
0
0
0
4.)
(a) Using the cofactor matrix, find the inverse
A

1
of
A
=
1
1
1
1
2
2
1
2
3
.
You've reached the end of your free preview.
Want to read all 42 pages?
 Spring '20