Math 2270, Fall 2015
Instructor: Thomas Goller
1 December 2015
Quiz #9 Solutions

0
(1) Let
,
3
y= 0+


1
2
,
be data points in R2 . The goal of this problem is to find the line
1
1
1 x that best fits these data points.
(a) Write down the equation X ~ = ~y that models the problem. (1 point) Solution:
2
3
1 0 
41 1 5
1 2
0
1
2
3
3
= 4 15
1
(b) Solve the equation X T X ~ = X T ~y for ~ . (2 points)
Solution:
2
3

1 0
1 1 1 4
3 3
T
1 15 =
X X=
,
0 1 2
3 5
1 2

so
~ = (X T X) 1 X T ~y =
1
15
9

5
3
3

3
1 1 1 4 5
3
T
1 =
X ~y =
,
0 1 2
1
1

3
3

2


1 12
3
2
=
=
.
1
6
1
6
(c) Write down the equation of the line that best fits the data points. (1 point)
Solution: y = 2
x
(d) Sketch the data points and your line of best fit. (1 point)



0
1
2
Description of solution: You should plot the points
,
,
and the line
3
1
1
with equation y = 2 x, which has y-intercept 2 and x-intercept 2.
Math 2270, Fall 2015
Instructor: Thomas Goller
1 December 2015
8 2 3 2 39
6 =
< 1
(2) Compute an orthogonal basis for the subspace W = Span 425 , 4 2 5 . (2 points)
:
;
3
1
Possible solution:
2 3
1
~v1 = 425 ,
3
2
3 2 3
6
1
4 2 5 · 425 2 3 2 3
1
6
1
3
2 3 2 3 425 = 4 2 5
1
1
3
1
425 · 425
3
3
2
3
6
~v2 = 4 2 5
1
(3) Bonus problem: Compute
the

 equation of the form y =
0
⇡/2
⇡
the data points
,
,
. (1 bonus point)
0
1
1
0
2 3 2
3
1
11/2
7 4 5 4
2 =
1 5
14
3
5/2
cos x +
1
sin x that best fits
Solution: In this case
2
so the dream equation is
3 2
3
cos 0
sin 0
1 0
X = 4cos ⇡/2 sin ⇡/25 = 4 0 15 ,
cos ⇡
sin ⇡
1 0
2
3
1 0 
4 0 15
1 0
Then
XT X =

1 0
0 1
2
3

1 0
1 4
2 0
0 15 =
0
0 1
1 0
0
1
2 3
0
= 415 .
1
and X T ~y =

so the solution to X T X ~ = X T ~y is

~ = (X X) X ~y = 1/2 0
0 1
T
1
Thus the best fit equation is y =
T
1
2
cos x + sin x.

1
=
1

1 0
0 1
2 3

0
1 4 5
1
1 =
0
1
1
1/2
.
1