6.
x
1
= 1 +
t, t
arbitrary
Explanation:
By row reduction
B
=
1
−
1
1
2
0
0
3
9
0
0
−
1
−
5
∼
1
−
1
1
2
0
0
3
9
0
0
0
−
2
,
which is now in echelon form. But the system
x
1
−
x
2
+
x
3
= 2
,
3
x
3
= 9
,
Explanation:
If
A
has
n
pivot positions, it has a pivot
in each of its
n
columns and in each of its
n
rows.
The reduced echelon form has a 1 in
each pivot position, so the reduced echelon
form is the
n
×
n
identity matrix.
Consequently, the statement is
TRUE
.
009
10.0points
If the augmented matrix for a system of
linear equations in variables
x
1
, x
2
, and
x
3
is
row equivalent to the matrix
B
=
1
−
1
1
2
0
0
3
9
0
0
−
1
−
5
,
determine
x
1
, if possible.
1.
x
1
=
−
2
2.
no solution
x
1
exists
correct
3.
x
1
= 0
4.
x
1
=
−
1
5.
x
1
=
−
1 +
t, t
arbitrary
0
x
3
=
−
5
,
associated with this matrix is inconsistent be
cause there is no solution
x
3
.
Consequently,
no solution
x
1
exists because the original sys
tem is
inconsistent
.
010
10.0points
The matrices
A
=
bracketleftbigg
1
0
a
1
bracketrightbigg
, B
=
bracketleftbigg
1
0
b
1
bracketrightbigg
,
have the property
AB
=
BA,
for all values of
a
and
b
.
True or False?
1.
FALSE
2.
TRUE
correct
Explanation:
When
A
=
bracketleftbigg
1
0
a
1
bracketrightbigg
, B
=
bracketleftbigg
1
0
b
1
bracketrightbigg
,
Version 023 – EXAM01 – gilbert – (57245)
5
then
AB
=
bracketleftbigg
1
0
a
1
bracketrightbigg bracketleftbigg
1
0
b
1
bracketrightbigg
=
bracketleftbigg
1
0
a
+
b
1
bracketrightbigg
,
while
BA
=
bracketleftbigg
1
0
b
1
bracketrightbigg bracketleftbigg
1
0
a
1
bracketrightbigg
=
bracketleftbigg
1
0
b
+
a
1
bracketrightbigg
.
But
a
+
b
=
b
+
a
for all
a, b
.
Consequently, the statement is
TRUE
.
011
10.0points
Determine the 3
×
3 matrix
E
such that
EA
=
B
when
A
3
R
1
+
R
3
→
R
′
3
−−−−−−−−→
A
1
4
R
2
→
R
′
2
−−−−−−→
A
2

2
R
2
+
R
1
→
R
′
1
−−−−−−−−−→
B.
1.
E
=
1
−
8
0
0
1
4
0
3
0
1
2.
E
=
1
8
0
0
4
0
−
3
0
1
3.
E
=
1
8
0
0
4
0
3
0
1
4.
E
=
1
−
8
0
0
1
4
0
−
3
0
1
5.
E
=
1
8
0
0
1
4
0
−
3
0
1
6.
E
=
1
−
8
0
0
4
0
3
0
1
correct
Explanation:
Since
A
1
=
1
0
0
0
1
0
3
0
1
A,
A
2
=
1
0
0
0
4
0
0
0
1
A
1
,
and
B
=
1
−
2
0
0
1
0
0
0
1
A
2
,
we see that
E
=
1
−
2
0
0
1
0
0
0
1
1
0
0
0
4
0
0
0
1
1
0
0
0
1
0
3
0
1
=
1
−
8
0
0
4
0
0
0
1
1
0
0
0
1
0
3
0
1
.
Consequently,
E
=
1
−
8
0
0
4
0
3
0
1
.
012
10.0points
If the columns of an
m
×
n
matrix
A
= [
a
1
a
2
...
a
n
]
span
R
m
, then the equation
A
x
=
b
is consis
tent for each
b
in
R
m
.
True or False?
1.
FALSE
2.
TRUE
correct
Explanation:
When
x
=
x
1
x
2
.
.
.
x
n
,
Version 023 – EXAM01 – gilbert – (57245)
6
then
A
x
= [
a
1
a
2
...
a
n
]
x
1
x
2
.
.
.
x
n
=
x
1
a
1
+
x
2
a
2
+
· · ·
+
x
n
a
n
.
But if
R
m
= Span
{
a
1
,
a
2
, ...,
a
n
}
,
then any vector
b
in
R
m
can be represented
by a linear combination
b
=
c
1
a
1
+
c
2
a
2
+
...
+
c
n
a
n
for suitable scalars
c
1
, c
2
, ..., c
n
. Thus
A
x
=
b
has a solution for each
b
in
R
m
with
x
j
=
c
j
.