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Quiz 2 Solutions
February 28, 2005
Problem 1
For the ﬁrst part: the equation of the plane orthogonal to a
given
~v
∈
R
3
is
~x
·
~v
= 0. Applying this formula to our problem and writing
~x
= (
x,y,z
), we get
x
+ 2
y
+ 3
z
= 0
.
For the second part, we have a formula for the reﬂection of
~w
in a plane
P
:
ref
P
(
~w
) =
~w

2
~w
⊥
where
~w
⊥
is the component of
~w
perpendicular to
P
. But this is just
~w

2(
~w
·
ˆ
v
)ˆ
v,
where ˆ
v
is a unit vector normal to
P
. Applying this formula with ˆ
v
=
1
√
14
[123],
~w
= [456], we get
ref
P
(
~w
) =
1
7

2

25

48
Problem 2
First we ﬁnd the elementary row operations that reduce
A
to
reduced row echelon form (if possible):
1
0
1
2

1 0
0
1
1
→

2(
I
) + (
II
)
*
1
0
1
0

1

2
0
1
1
→
(
II
) + (
III
)
*
1
0
1
0

1

2
0
0

1
→

2(
III
) + (
II
)
*
,
(
III
) + (
I
)
*
1
0
0
0

1
0
0
0

1
→

(
II
)
*
,

(
III
)
*
1 0 0
0 1 0
0 0 1
1
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View Full DocumentSince
A
can be reduced to the identity matrix by elementary row operations,
A
is invertible. Applying these same row operations to the 3
×
3 identity
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 Spring '05
 CONSANI
 Linear Algebra, Algebra

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