geyer (dag2798) – HW09 – rusin – (55735)
1
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001
10.0 points
By evaluating the determinant, express
f
(
x
) =
v
v
v
v
v
v
1
x
x
2
4
−
2
−
1
4
−
1
−
3
v
v
v
v
v
v
as a quadratic Function in
x
.
1.
f
(
x
) =
−
5
−
4
x
−
8
x
2
2.
f
(
x
) = 5 + 4
x
+ 8
x
2
3.
f
(
x
) =
−
5
−
8
x
−
4
x
2
4.
f
(
x
) = 5 + 8
x
+ 4
x
2
correct
5.
f
(
x
) = 5
−
8
x
+ 4
x
2
6.
f
(
x
) =
−
5 + 8
x
−
4
x
2
Explanation:
±or this 3
×
3 determinant, use expansion
by minors along the top row:
f
(
x
) =
v
v
v
v
−
2
−
1
−
1
−
3
v
v
v
v
−
v
v
v
v
4
−
1
4
−
3
v
v
v
v
x
+
v
v
v
v
4
−
2
4
−
1
v
v
v
v
x
2
.
Evaluating the 2
×
2 determinants, we thus
see that
f
(
x
) = 5 + 8
x
+ 4
x
2
.
keywords:
matrix, determinant, quadratic
Function, expansion by minors
002
10.0 points
±ind the value oF the determinant
D
=
v
v
v
v
v
v
v
3
−
2
−
1
−
2
1
−
3
x
y
z
v
v
v
v
v
v
v
.
1.
D
=
−
7
x
+ 11
y
−
z
2.
D
= 7
x
−
11
y
+
z
3.
D
= 7
x
+ 11
y
−
z
correct
4.
D
= 7
x
−
11
y
−
z
5.
D
=
−
7
x
+ 11
y
+
z
6.
D
=
−
7
x
−
11
y
+
z
Explanation:
±or any 3
×
3 determinant
v
v
v
v
v
v
v
A
B
C
a
1
b
1
c
1
a
2
b
2
c
2
v
v
v
v
v
v
v
=
A
v
v
v
v
v
b
1
c
1
b
2
c
2
v
v
v
v
v
−
B
v
v
v
v
v
a
1
c
1
a
2
c
2
v
v
v
v
v
+
C
v
v
v
v
v
a
1
b
1
a
2
b
2
v
v
v
v
v
.
Thus
D
=
v
v
v
v
v
v
v
3
−
2
−
1
−
2
1
−
3
x
y
z
v
v
v
v
v
v
v
= 3
v
v
v
v
v
1
−
3
y
z
v
v
v
v
v
+ 2
v
v
v
v
v
−
2
−
3
x
z
v
v
v
v
v
−
v
v
v
v
v
−
2
1
x
y
v
v
v
v
v
= 3 (1
z
+ 3
y
) + 2 (
−
2
z
+ 3
x
)
−
(
−
2
yx
)
.
Consequently,
D
= 7
x
+ 11
y
−
z
.
keywords: determinant
003
10.0 points
±ind the cross product oF the vectors
a
=
i
−
2
j
−
3
k
,
b
=
−
3
i
−
2
j
+ 3
k
.
1. a
×
b
=
−
11
i
+ 6
j
−
9
k
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geyer (dag2798) – HW09 – rusin – (55735)
2
2. a
×
b
=
−
11
i
+ 6
j
−
8
k
3. a
×
b
=
−
12
i
+ 6
j
−
8
k correct
4. a
×
b
=
−
12
i
+ 4
j
−
8
k
5. a
×
b
=
−
12
i
+ 3
j
−
9
k
6. a
×
b
=
−
11
i
+ 4
j
−
9
k
Explanation:
One way of computing the cross product
(
i
−
2
j
−
3
k
)
×
(
−
3
i
−
2
j
+ 3
k
)
is to use the fact that
i
×
j
=
k
,
j
×
k
=
i
,
k
×
i
=
j
,
while
i
×
i
= 0
,
j
×
j
= 0
,
k
×
k
= 0
.
For then
a
×
b
=
−
12
i
+ 6
j
−
8
k
.
Alternatively, we can use the de±nition
a
×
b
=
v
v
v
v
v
v
i
j
k
1
−
2
−
3
−
3
−
2
3
v
v
v
v
v
v
=
v
v
v
v
−
2
−
3
−
2
3
v
v
v
v
i
−
v
v
v
v
1
−
3
−
3
3
v
v
v
v
j
+
v
v
v
v
1
−
2
−
3
−
2
v
v
v
v
k
to determine
a
×
b
.
004
10.0 points
Which of the following statements are true for
all vectors
a
,
b
n
=
0
?
A. if
a
×
b
= 0, then
a
b
b
,
B.
|
a
·
b
|
2
− |
a
×
b
|
2
=
|
a
|
2
|
b
|
2
,
C.
a
×
b
+
b
×
a
= 0.
1.
A only
2.
C only
3.
A and C only
correct
4.
all of them
5.
B only
6.
A and B only
7.
none of them
8.
B and C only
Explanation:
A. TRUE: if
θ,
0
≤
θ
≤
π
, is the angle
between
a
, and
b
, then
|
a
×
b
|
=
|
a
||
b
|
sin
θ ,
so if
b
a
b n
= 0 and
b
b
b n
= 0, then
|
a
×
b
|
= 0
=
⇒
sin
θ
= 0
.
Thus
θ
= 0
, π .
In this case,
a
is
parallel
to
b
.

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- Spring '07
- Sadler
- Determinant, Multivariable Calculus, GEYER
-
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