Version 088 – EXAM 01 – gilbert – (53415)
1
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print-out
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have
15
questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
When
P
is the plane in
R
3
given in vector
form by
x
=
3
−
3
1
+
s
−
1
−
2
1
+
t
−
3
−
8
5
,
determine where
P
intersects the
z
-axis.
1.
z
=
−
4
2.
z
=
−
3
3.
z
=
−
6
4.
z
=
−
2
5.
z
=
−
5
correct
Explanation:
Since the
z
-axis consists of points in
R
3
with
x
=
y
= 0,
P
intersects the
z
-axis when
0
0
z
=
3
−
3
1
+
s
−
1
−
2
1
+
t
−
3
−
8
5
for some choice of
s
and
t
,
i.e.
, when
s
−
1
−
2
1
+
t
−
3
−
8
5
=
−
3
3
z
−
1
is consistent as a vector equation in
s, t
. This
will be true if and only if the associated aug-
mented matrix
A
=
−
1
−
3
−
3
−
2
−
8
3
1
5
z
−
1
is row equivalent to an echelon matrix whose
entries in the last row are all zero.
Now by row reduction in the first column of
A
, we obtain
A
∼
−
1
−
3
−
3
0
−
2
9
0
2
z
−
4
,
and after row reduction in the second column
A
∼
−
1
−
3
−
3
0
−
2
9
0
0
z
+ 5
.
Consequently,
P
intersects the
z
-axis at
z
=
−
5
.
002
10.0points
If matrices
A
and
B
are row equivalent,
they have the same reduced echelon form.
True or False?
1.
TRUE
correct
2.
FALSE
Explanation:
If
A
is row equivalent to
B
, then
A
can
be transformed by elementary row operations
first into
B
and then further transformed into
the reduced echelon form
U
of
B
.
Since the
reduced echelon form of
A
is unique, it must
be
U
.
Consequently, the statement is
TRUE
.
003
10.0points
Determine
b
when
a
=
bracketleftbigg
−
4
3
bracketrightbigg
,
b
=
bracketleftbigg
a
b
bracketrightbigg
,
c
=
bracketleftbigg
1
2
bracketrightbigg
,
and
comp
a
b
= 1
,
proj
c
b
=
−
4
5
c
.
1.
b
=
−
3

Version 088 – EXAM 01 – gilbert – (53415)
2
2.
b
=
−
1
correct
3.
b
=
−
2
4.
b
=
−
4
5.
b
=
−
5
Explanation:
Since
comp
a
b
=
a
·
b
bardbl
a
bardbl
,
proj
c
b
=
parenleftbigg
b
·
c
bardbl
c
bardbl
2
parenrightbigg
c
,
we see that
−
4
a
+ 3
b
radicalbig
(
−
4)
2
+ (3)
2
= 1
,
while
a
+ 2
b
(1)
2
+ (2)
2
=
−
4
5
.
Thus
−
4
a
+ 3
b
= 5
,
a
+ 2
b
=
−
4
.
Consequently, solving for
b
gives
b
=
−
1
.
004
10.0points
Determine the Reduced Row Echelon Form
of the matrix
A
=
1
−
1
2
3
−
2
9
2
1
13
.
1.
rref(
A
) =
1
1
5
0
1
3
0
0
0
2.
rref(
A
) =
1
0
0
0
1
0
0
0
0
3.
rref(
A
) =
1
0
0
0
1
0
0
0
1
4.
rref(
A
) =
1
0
5
0
1
3
0
0
0
correct
5.
rref(
A
) =
1
2
5
0
1
3
0
0
0
Explanation:
After performing elementary row opera-
tions downwards on the columns we see that
A
∼
1
−
1
2
0
1
3
0
3
9
∼
1
−
1
2
0
1
3
0
0
0
.
Next perform elementary row operations up-
wards. Then
A
∼
1
0
5
0
1
3
0
0
0
.
Consequently,
rref(
A
) =
1
0
5
0
1
3
0
0
0
.
005
10.0points
Under what conditions on
b
1
, b
2
does the
equation
bracketleftbigg
3
−
2
−
12
8
bracketrightbigg
x
=
bracketleftbigg
b
1
b
2
bracketrightbigg
have a solution in
R
2
?
1.
b
2
−
4
b
1
= 0
2.
b
2
+ 4
b
1
= 0
correct
3.