PROBLEM 2
For the line y=ax+b to pass through the given points
(-1,1), (0,2), (1,2), (2,0) the coefficients would have
to satisfy the (inconsistent) linear system
-a + b = 1
b = 2
a + b = 2
2a + b = 0
which in matrix form is
[-1 1] [a] [1]
[ 0 1] [b] = [2]
[ 1 1] [2]
[ 2 1] [0] .
The associated normal equation for the least-squares solution is thus
[-1 0 1 2] [-1 1] [a*] [-1 0 1 2] [1]
[ 1 1 1 1] [ 0 1] [b*] = [ 1 1 1 1] [2]
[ 1 1] [2]
[ 2 1] [0]
(using the formula in 5.4.5 on page 222 of the textbook), which is to say
[6 2] [a*] [1]
[ ] [ ] = [ ]
[2 4] [b*] [5] .
Solving this (by any of the methods you've learned in class or earlier)
yields a* = -3/10, b* = 7/5 = 14/10, which matches answer (c).
[Besides checking that these values of a* and b* do satisfy
6a* + 2b* = 1 and 2a* + 4b* = 5, you can also check your work
by verifying directly that, at least among the five given choices,
answer (c) minimizes the sum of the squares of the discrepancies
-a+b-1, b-2, a+b-2, 2a+b in the inconsistent linear system for (a,b).]