Lecture 3: How to solve a system of linear equations (Nicholson, Section 1.2)

1:00 --- defined the elementary (equation) operations. 

3:00 --- defined what it means for two linear systems to be equivalent. 

6:30 --- stated the theorem that equivalent linear systems have the same solution set. 

8:50 --- introduced the augmented matrix as a short-hand way of representing a linear system of equations.  Introduced the coefficient matrix A, the right-hand-side vector b. and the augmented matrix (A|b). 

9:30 --- Considered a specific linear system, represented it using an augmented matrix, introduced (and used) elementary row operations on the augmented matrix (which are the same thing as elementary (equation) operations) until the augmented matrix was in a simple form.  Wrote down the linear system that corresponded to the simple augmented matrix and concluded that the original linear system had no solutions. 

19:20 --- did another example in which it turned out there was exactly one solution. 

30:00 --- introduced the language “leading ones” of a matrix that’s in reduced row echelon form.  At 31:30 I gave a wrong answer to a student --- I claimed that if every student reduced the augmented matrix so that the first nonzero entry in a row is 1 then every student would have the same matrix.  This isn’t true.  What’s true is that if every student reduced the augmented matrix until it was in reduced row echelon form (the first nonzero entry in a row is 1 and above and below that 1 are zeros and the leading ones move to the right as you move down the rows and any zero rows are at the bottom of the matrix) then all students would have the same RREF matrix.

32:00 --- How to use Matlab to find the RREF of a matrix. 

34:20 --- did another example, this time there’re infinitely many solutions. How to write down the solution set.  In the example, there are 3 equations and 4 unknowns. One of the unknowns is set to be a free parameter. Does it really matter which parameter you set to the free parameter?  In this example, you could set x1 = t and find x2 = some expression of t, x3 = some expression of t.  Alternatively, you could set x2 = t and find x1 = some expression of t, x3 = some expression of t. Alternatively, you could set x3 = t and find x1 = some expression of t, x2 = some expression of t.  It happens that the structure of the RREF of the augmented matrix is such that it’s easiest to do the x3=t option.    

41:20 --- introduced the concept of a “general solution” and then verified that it’s a solution, no matter what the value of the free parameter t. 

47:00 --- I went through the exercise of writing the solution when making the choice of setting x1 = t and find x2 = some expression of t, x3 = some expression of t. Hopefully, this was a gross enough exercise to convince you of the charm of choosing x3 = t.