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).]