HW09-solutions - geyer(dag2798 HW09 rusin(55735 This...

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

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 .