Day 3: Linear Algebra

• Announcements:
  • As usual, keep an eye on the assignments page. Your first assignment will be due on Wednesday -- next time.
  • I really like the freely available pdf viewer pdf-xchange: it handles the back and forth between the book and the answers much better. Give it a try. It also allows you to highlight pdfs, add post-it notes, etc.
  • Monday is MLK, Jr. Day -- no class! It will be lonely here if you should show up....

• Today we take a look at Section One.I.2: Linear systems.Solving Linear Systems.Describing the Solution Set

• Review from last time:
  • Last time we talked about the solution method we'll use for linear systems: Gaussian elimination

  • There are three operations that we use to reduce a system where it can be solved by back-substitution.
    • Swapping distinct rows
    • Scaling rows by a non-zero constant
    • Combining distinct rows

  • These "row operations" preserve the solution set: that is, they preserve solutions of the original linear system, and do not introduce any new, spurious solutions.

  • As we discussed last time, there are three things that might happen: a linear system may have either
    • a unique solution;
    • no solution;
    • infinitely many solutions.

  • We saw examples of several systems where unique solutions existed, including one that didn't come off very well: car buying model. I want to start today with that one, to fix it up for you.

    I forgot the so-called intercept term in the model:

    

    and we'll use this model with three pieces of data to determine the parameters in our model via a system of three equations in three unknowns:

    
    
    
    

    I got this data off of the cargurus website and grabbed three different cars (I was shooting for "good values") with three different years and three different mileages. Then I solved the linear system for the parameters of the model:

    Now I can use that model to determine whether my brother-in-law's car prospect -- a 2003 Camry with 87K miles at 7.5 is a good deal:

    

    Wow! Go out and check out that car! The model says it should cost $10.2K, and it's going for $7.5K.... Got it?

• Any questions about the homework problems so far?

• Today we look at the problem of characterizing the solution of a linear system, particularly when there are infinitely many solutions.
  • Definitions in this section:
    1. free variable -- the non-leading variables in an echelon-form linear system
    2. matrix -- a rectangular grid (array) of numbers in rows and columns
    3. vector -- a matrix with only one column
    4. vector solution of a linear system (writing the solution as something other than an n-tuple)
    5. vector operations:
       1. sum
       2. scalar multiplication of a vector

  • Now: we can characterize a linear system as a matrix/vector problem, eliminating the need for the the variable names. In the car model, for instance, let's look at how the problem is presented (and solved) by Octave:

    A=[3,109,1;2,104,1;4,99,1]
    B=[8;7;10.5]
    solution=A \ B
    
    A =
    
        3  109    1
        2  104    1
        4   99    1
    
    B =
    
       8.0000
       7.0000
      10.5000
    
    solution =
    
        1.500000
       -0.100000
       14.400000
    

    There is no sign of a variable name anywhere, but we know what results: we've used the data to find the values of for our car-buying model:
    

    We write the system in "augmented matrix" form, apply elementary Gaussian row operations, and voila, we have the solution -- the unique solution -- which we write as a vector:

    solution =
    
        1.500000
       -0.100000
       14.400000
    

  • Now let's look at the solution when there are infinitely many solutions:
    • If we have just one equation, such as then we obviously have an infinite number of solutions. The solution, written as an ordered pair, is

      Notice

      1. that we have eliminated x;
      2. that we have written the solution entirely in terms of y, which is free (to take any real value we'd like);
      3. that x was a leading variable in the echelon form, but was eliminated via back-substitution of y.

      We might write that in a little different way: as vectors.

    • I took that first equation out of a more general system from our text: 2.7 Example, p. 14. Let's have a look at that one....

      Notice

      1. the advantage of the augmented matrix notation: it's so much cleaner. It eliminates what is non-essential (the choice of variable names). We realize that we're solving for three variables, but we are free to give them the names we choose.
      2. that the solution is given (p. 15) as a sum of vectors, some scaled by free variables ("scalar multiples").

Links:

• Octaveweb