But this will contain cross-product terms un-
less
P
T
AP
is a
diagonal
matrix,
i.e.
, unless
A
is orthogonally diagonalized by
P
.
Consequently, the statement is
FALSE
.
007
10.0 points
If
A
is a 2
×
2 symmetric matrix, then the set
of
x
such that
x
T
A
x
=
c
, for some constant
c
,
corresponds to either a circle, an ellipse, or a
hyperbola.
True or False?
1.
TRUE

jones (dzj62) – HW14 – gilbert – (53525)
3
2.
FALSE
correct
Explanation:
There are also “degenerate” cases where
the solution set of
x
T
A
x
=
c
can be a single
point, two intersecting lines, or no points at
all.
Consequently, the statement is
FALSE
.
008
10.0 points
A quadratic form can always be written as
Q
(
x
) =
x
T
A
x
with
A
a symmetric matrix.
True or False?
1.
TRUE
correct
2.
FALSE
Explanation:
By definition a quadratic form has the form
Q
(
x
) =
a
11
x
2
1
+
. . .
+
a
nn
x
2
n
+ 2(
a
12
x
1
x
2
+
. . .
+
a
n
−
1
,n
x
n
−
1
x
n
)
.
Thus
Q
(
x
) =
x
T
A
x
with
x
=
⎡
⎢
⎢
⎣
x
1
x
2
.
.
.
x
n
⎤
⎥
⎥
⎦
,
A
=
⎡
⎢
⎢
⎣
a
11
a
12
. . .
a
1
n
a
12
a
22
. . .
a
2
n
.
.
.
.
.
.
.
.
.
.
.
.
a
1
n
a
2
n
. . .
a
nn
⎤
⎥
⎥
⎦
,
and
A
T
=
A
.
Consequently, the statement is
TRUE
.
009
10.0 points
The graph of
x
T
A
x
=
−
x
2
1
−
12
x
1
x
2
+ 15
x
2
2
=
−
1
is a conic section.
Find an orthogonal matrix
P
so that the
transformation
x
=
P
y
diagonalizes
A
, and
then use this to identify the conic section
given that
λ
1
≥
λ
2
.
1.
ellipse 17
y
2
1
−
3
y
2
2
=
−
1
2.
ellipse 17
y
2
1
+ 3
y
2
2
= 1
3.
hyperbola 17
y
2
1
−
3
y
2
2
=
−
1
correct
4.
hyperbola 17
y
2
1
+ 3
y
2
2
= 1
5.
empty set
6.
point (0
,
0)
7.
straight lines 17
y
2
1
−
3
y
2
2
= 0
Explanation:
The quadratic form can be written as
−
x
2
1
−
12
x
1
x
2
+ 15
x
2
2
= [
x
1
x
2
]
−
1
−
6
−
6
15
x
1
x
2
.
The eigenvalues of
A
are the solutions of
det
−
1
−
λ
−
6
−
6
15
−
λ
=
λ
2
−
14
λ
−
51
= (
λ
−
17)(
λ
+ 3) = 0
,
i.e.
,
λ
1
= 17 and
λ
2
=
−
3. Associated eigen-
vectors are
u
1
=
−
1
3
,
u
2
=
3
1
;
these are orthogonal. The normalized eigen-
vectors
v
1
=
1
√
10
−
1
3
,
v
2
=
1
√
10
3
1
,
are orthonormal, and
P
= [
v
1
v
2
] =
1
√
10
−
1
3
3
1

jones (dzj62) – HW14 – gilbert – (53525)
4
is an orthogonal matrix such that
A
=
P
17
0
0
−
3
P
−
1
=
PDP
T
is an orthogonal diagonalization of
A
.
Now
set
x
=
P
y
. Then
x
T
A
x
= (
P
y
)
T
(
PDP
T
)
P
y
=
y
T
(
P
T
P
)
D
(
P
T
P
)
y
=
y
T
17
0
0
−
3
y
= 17
y
2
1
−
3
y
2
2
=
−
1
.
Consequently, the conic section is a
hyperbola
.
010
10.0 points
The graph of
x
T
A
x
= 3
x
2
1
+ 12
x
1
x
2
+ 19
x
2
2
= 5
is a conic section.
Find an orthogonal matrix
P
so that the
transformation
x
=
P
y
diagonalizes
A
, and
then use this to identify the conic section
given that
λ
1
≥
λ
2
.
1.
straight lines 21
y
2
1
+
y
2
2
= 0
2.
hyperbola 21
y
2
1
−
y
2
2
=
−
5
3.
ellipse 21
y
2
1
+
y
2
2
= 5
correct
4.
ellipse 21
y
2
1
−
y
2
2
=
−
5
5.
empty set
6.
hyperbola 21
y
2
1
+
y
2
2
= 5
7.
point (0
,
0)
Explanation:
The quadratic form can be written as
3
x
2
1
+ 12
x
1
x
2
+ 19
x
2
2
= [
x
1
x
2
]
3
6
6
19
x
1
x
2
.
The eigenvalues of
A
are the solutions of
det
3
−
λ
6
6
19
−
λ
=
λ
2
−
22
λ
+ 21
= (
λ
−
21)(
λ
−
1) = 0
,
i.e.
,
λ
1
= 21 and
λ
2
= 1.
Associated eigen-
vectors are
u
1
=
1
3
,
u
2
=
−
3
1
;
these are orthogonal. The normalized eigen-
vectors
v
1
=
1
√
10
1
3
,
v
2
=
1
√
10
−
3
1
,
are orthonormal, and
P
= [
v
1
v
2
] =
1
√
10
1
−
3
3
1