Math 415 - Assignment 6 Solutions
Problems: 4.2.1, 4.2.4, 4.2.5 (c), 4.2.7 (b), 4.3.1, 4.3.3, 4.3.14 (d), 4.3.17, 4.4.3
Problem 4.2.1
First we write this quadratic function in the standard form f(x) = xTKx −2xTf+cwhere
K=
110
131
011
, f =
1
0
−3
2
, c = 2.By Gaussian elimination
110
131
011
→
110
021
011
→
1 1 0
0 2 1
0 0 1
2
=U
and since the diagonal entries of Uare all positive, Kis positive definite. This shows that the
minimizer is unique. By Theorem 4.1 the minimizer is the solution of Kx =fand we solve this by
elimination:
110
131
011
1
0
−3
2
→
110
021
011
1
−1
−3
2
→
1 1 0
0 2 1
0 0 1
2
1
−1
−1
Therefore 1
2z=−1 or z=−2,2y=−1−zor y=1
2(−1 + 2) = 1
2, x = 1 −y= 1 −1
2=1
2. Then the
minimum value is
1
2
1
2
−2
T
110
131
011
1
2
1
2
−2
−2
1
2
1
2
−2
T
1
0
−3
2
+ 2 = −3
2
Problem 4.2.4
(a) Since the (1,1) entry of Ais positive, we just need the determinant to be positive: 4 −b2>0,
i.e. −2< b < 2
(b) A=1b
b4→1b
0 4 −b2→1 0
0 4 −b2 1b
0 1
so D=1 0
0 4 −b2and L=1 0
b1
(c) This part of the question is an application of Theorem 4.4. First assume that bis such that A
is positive definite. Then the unique minimizer satisfies Ax =f=0 1 T. Thus x=A−1f=
1b
b4−10
1=−4
−4+b2
b
−4+b2
b
−4+b2−1
−4+b2 0
1=b
−4+b2
−1
−4+b2
and so the minimum is b
−4+b22
+ 2bb
−4+b2−1
−4+b2+ 4 −1
−4+b22
−2−1
−4+b2=1
−4+b2.
If b > 2 or b < −2, Ais neither definite or semidefinite, and so the minimum in this case is −∞.
If b= 2, the quadratic function reduces to p(x, y) = (x+ 2y)2−2yand so p(−2y, y) = −2y→ −∞
as y→ ∞. If b=−2, the quadratic function reduces to p(x, y)=(x−2y)2−2yand so p(2y, y) =
−2y→ −∞ as y→ ∞. Done!
Problem 4.2.5 (c)
p(x, y, z) =
x
y
z
T
3−1 1
−1 2 −1
1−1 3
x
y
z
−2
x
y
z
T
1
0
−2
−3
= 3x2−2xy + 2xz + 2y2−2yz + 3z2−2x+ 4z−3. Now let’s solve Kx =f:
3−1 1
−1 2 −1
1−1 3
1
0
−2
→
3−1 1
05
3−2
3
0−2
3
8
3
1
1
3
−7
3
→
3−1 1
0 5 −2
0−2
3
8
3
1
1
−7
3
→
3−1 1
0 5 −2
0 0 12
1
1
−11
so 12z=−11, i.e. z=−11
12 ,5y= 1 + 2z, i.e. y=1
5(1 −22
12 ) = −1
6,3x= 1 + y−z, i.e. x=
