Lecture 2: Linear Systems: Solutions, Elementary Operations (Nicholson, Section 1.1)

Gave the definition of “a linear equation in n variables”. 

6:00 --- Defined what it means for a vector to be a solution of a linear equation.

10:15 --- Note that if you write an equation like x1 – 2 x2 + x3 = 1 then [2;1;1] is a solution in R^3.  And [2;1;1;29] is a solution in R^4.  And [2;1;1;3;-8] is a solution in R^5.   An equation doesn’t determine the R^n that a solution lives in. Certainly, the equation won’t have solutions in R^1 (what would you plug in for x2 and x3?) or solutions in R^2 (what would you plug in for x3?) but it’s perfectly reasonable to consider that equation in R^27 if needed --- it depends on what physical problem the equation is coming from. 

10:25 --- defined a “system of linear equations”. 

10:40 --- Stated that a linear system either has no solutions, has exactly one solution, or has infinitely many solutions.  (The proof will come later!)  Presented a linear system that has no solutions.  Presented a linear system that has exactly one solution.  Presented a linear system that has infinitely many solutions.

16:00 --- Presented a system of 3 linear equations in 3 unknowns and found the solution using a sequence of elementary operations.  (swapping two equations, multiplying an equation by a nonzero number, adding a multiple of one equation to another equation.)  Key in all of this is that the elementary operations don’t change the solution set.  That is, if you have S = {all solutions of the original system} and T = {all solutions of the system you get after applying one elementary operation} then S = T.   If you can prove that S = T for each of the three elementary operations then you know that the solutions of the original system are the same as the solutions of a later (easier to solve) linear system because you know that the solutions remain unchanged after each step on the way.  (The worry is, of course, that you might lose solutions or gain solutions by doing these elementary operations.  Certainly, if you multiplied one of the equations by zero you’d be at high risk of creating a new system that has more solutions than the previous system.) 

29:10 --- presented a second 3x3 system and demonstrated that there were infinitely many solutions. 

31:20 --- presented a third 3x3 system and demonstrated that there were no solutions. 

33:50 --- gave an argument based on a specific system of 2 linear equations in 2 unknowns showing how it is that the elementary operations don’t change the set of solutions.