. Thus there is a unique plane passing
through
P, Q,
and
R
, namely the graph of
5
x
+ 16
y
−
6
z
= 1
,
obtained by cancelling the common factor
s
.
005
10.0 points

Version 015 – EXAM01 – gilbert – (56780)
3
3.
⎡
⎣
1 + 7
s
−
2
−
3
s
1
−
s
⎤
⎦
,
0
≤
s
≤
1
.
4.
⎡
⎣
7
−
3
s
−
3
−
3
s
−
1
−
s
⎤
⎦
,
0
≤
s
≤
1
.
5.
⎡
⎣
4 + 4
s
1
−
6
s
2
−
2
s
⎤
⎦
,
0
≤
s
≤
1
.
6.
⎡
⎣
5
−
s
−
1
−
5
s
3
−
5
s
⎤
⎦
,
0
≤
s
≤
1
.
Explanation:
The face
CDEG
of the parallelepiped lies in
the unique plane in which the vertices
D, C,
and
G
lie. Now in vector form this plane is
x
=
c
+
s
b
+
t
a
,
−∞ ≤
s, t
≤ ∞
,
and the set of all points in the parallelogram
CDEG
corresponds to
c
+
s
b
+
t
a
,
0
≤
s, t
≤
1
.
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006
10.0 points
Explanation:
If
{
x
,
y
,
z
}
is a linearly dependent set, then
c
1
x
+
c
2
y
+
c
3
z
=
0
for some choice of
c
1
, c
2
, c
3
, not all zero. Now
if
c
3
= 0, then
c
1
x
+
c
2
y
=
0
.
But if
x
,
y
are linearly independent, then
c
1
=
c
2
= 0. Thus
c
3
̸
= 0, so
z
=
−
(
c
1
/c
3
)
x
−
(
c
2
/c
3
)
y
,
i.e.
,
z
is a linear combination of
x
,
y
. Hence
z
is a vector in Span
{
x
,
y
}
when
x
,
y
are lin-
early independent and
{
x
,
y
,
z
}
is a linearly
dependent set.
Consequently, the statement is
TRUE
.

If
x
and
y
are linearly independent but
{
x
,
y
,
z
}
is a linearly dependent set, then
z
is
in Span
{
x
,
y
}
.
True or False?
1.
TRUE
correct
2.
FALSE
3.
x
1
=
−
3
4.
x
1
= 1 +
t,
t
arbitrary
5.
x
1
=
−
1 +
t,
t
arbitrary
6.
no solution
x
1
exists
correct
Explanation:

Version 015 – EXAM01 – gilbert – (56780)
4
By row reduction
B
=
⎡
⎣
1
−
1
2
−
2
0
0
−
2
−
4
0
0
1
0
⎤
⎦
∼
⎡
⎣
1
−
1
2
−
2
0
0
−
2
−
4
0
0
0
−
2
⎤
⎦
,
which is now in echelon form. But the system
x
1
−
x
2
+ 2
x
3
=
−
2
,
−
2
x
3
=
−
4
,
0
x
3
= 0
,
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
.
008
10.0 points