Page 351 - 35Linear Algebra
P. 351

351


                          (a) Write down a linear system of equations you could use to find the slope
                              m and constant term b.
                          (b) Arrange the unknowns (m, b) in a column vector X and write your
                              answer to (a) as a matrix equation

                                                          MX = V .

                              Be sure to give explicit expressions for the matrix M and column vector
                              V .
                          (c) For a generic data set, would you expect your system of equations to
                              have a solution? Briefly explain your answer.

                                          T
                                                      T
                          (d) Calculate M M and (M M)     −1  (for the latter computation, state the
                              condition required for the inverse to exist).
                          (e) Compute the least squares solution for m and b.
                          (f) The least squares method determines a vector X that minimizes the
                              length of the vector V − MX. Draw a rough sketch of the three data
                              points in the (x, y)-plane as well as their least squares fit. Indicate how
                              the components of V − MX could be obtained from your picture.


                   Solutions

                      1. You can find the definitions for all these terms by consulting the index of
                         this book.

                      2. Both junctions give the same equation for the currents

                                                     I + J + 13 = 0 .

                         There are three voltage loops (one on the left, one on the right and one going
                         around the outside of the circuit). Respectively, they give the equations
                                                  60 − I − 80 − 3I = 0

                                                  80 + 2J − V + 3J = 0

                                             60 − I + 2J − V + 3J − 3I = 0   .              (F.1)

                         The above equations are easily solved (either using an augmented matrix
                         and row reducing, or by substitution). The result is I = −5 Amps, J = −8
                         Amps, V = 40 Volts.

                      3.  (a) m.


                                                                  351
   346   347   348   349   350   351   352   353   354   355   356