This preview shows pages 1–3. Sign up to view the full content.
Solutions to Review Problems One
1. Represent the following system of equations by an augmented matrix, reduce
this matrix to upper triangular form by Gaussian Elimination and then use back
substitution to ﬁnd the solution.
x
+ 2
y
+ 4
z
= 0
3
x
+ 8
y
+ 8
z
= 2
2
x
+ 6
y
+ 9
z
= 7
(1)
1 2 4

0
3 8 8

2
2 6 9

7
may be reduced by the operations
R
2

3
R
1
,
R
3

2
R
1
,
to upper triangular form
1 2
4

0
0 2

4

2
0 0
5

5
. Backsubstitution then gives the
solution
z
= 1,
y
=
4
z
+2
2
= 3 and
x
=

2
y

4
z
=

10.
2. Put the matrix
A
=
1 2 0 0 3
2 4 1 3 0
2 4 2 3 1
into reduced row echelon form and
then give the general solution to the equation
A
→
x
=
→
0 and express in parametric
form.
(2) The operations
R
2

2
R
1
,
R
3

2
R
1
,
R
3

2
R
2
,

1
3
R
3
reduce
A
to row
echelon form
1 2 0 0
3
0 0 1 3

6
0 0 0 1

7
3
. Finally the operation
R
2

3
R
3
gives reduced
row echelon form
1 2 0 0
3
0 0 1 0
1
0 0 0 1

7
3
. Then
x
1
,
x
3
and
x
4
are the dependent
variables,
x
2
=
s
and
x
5
=
t
are the independent variables and the solution is
(

2
s

3
t,s,

t,
7
3
t,t
).
3. Let A be an
m
×
n
matrix, where
m < n
. Explain why the homogeneous
system of equation
A
→
x
=
→
0 has a nontrivial solution.
(3) There are three possible outcomes for any system of equations: no solution,
a unique solution, or inﬁnitely many solutions. A homogeneous equation always
has solution
→
x
=
,
→
0 , so the ﬁrst outcome is impossible. Since
m < n
, the system
A
→
x
= 0 is underdetermined, which implies that the solution is not unique. Thus
we must be in the third outcome where there are inﬁnitely many solutions and
therefore at least one nontrivial solution.
4. Let
A
be an
m
×
n
matrix with entries
a
ij
, let
B
be an
n
×
r
matrix with entries
b
ij
, and let
C
be an
n
×
r
matrix with entries
c
ij
. Prove that
A
(
B
+
C
) =
AB
+
AC
.
(4)
B
+
C
has entries
b
jk
+
c
jk
, so that
A
(
B
+
C
) has entries
∑
n
j
=1
a
ij
(
b
jk
+
c
jk
),
which equals
∑
n
j
=1
(
a
ij
b
jk
+
a
i
jc
jk
), which equals
∑
n
j
=1
a
ij
b
jk
+
∑
n
j
=1
a
ij
c
jk
), which
equals the entries of
AB
+
AC
, since
AB
has entries
∑
n
j
=1
a
ij
b
jk
and
AC
has entries
∑
n
j
=1
a
ij
c
jk
).
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 2
5. Prove that if
A
is invertible, then
A
T
is also invertible.
(5) Suppose that
A
is invertible and let
B
be the inverse of
A
, so that
AB
=
I
=
BA
. Then we have
B
T
A
T
= (
AB
)
T
=
I
T
=
I
and
A
T
B
T
= (
BA
)
T
=
I
T
=
I
, so
that
A
T
has inverse
B
T
.
6. Suppose that
→
u
is a solution of
A
→
x
=
→
0 and that
→
v
is a solution of
A
→
x
=
→
c
. Explain carefully why 2
→
u
+
→
v
is a solution of
A
→
x
=
→
c
.
(6) We are given
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 12/27/2011 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

Click to edit the document details