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).]