Notes from 2nd Class Session

Exercises are specified in the format ex(section number)(exercisenumber).  For example, to denote something from section 3.2 exercise 8, the name ex328 will be assigned.  The first exercise shown below is exercise 8 of section 1.1.

MatrixForm[
      ex118 = {{1, 1, 2, a}, {1, 0, 1, b}, {2, 1, 3, c}}]

MatrixForm[RowReduce[ex118]]

The same system can be solved using the Solve command of Mathematica.  But this produces an empty result.  You know from the computations shown above that there are solutions when c=a+b

Solve[{x + y + 2  z == a, x + z == b, 
        2  x + y + 3  z == c}, {x, y, z}]

Another Mathematica command Reduce, produces a result as shown next.

Reduce[{x + y + 2  z == a, x + z == b, 
        2  x + y + 3  z == c}, {x, y, z}]

The left-hand side or 'A' matrix of the system is used in row reduction in the next statements.

MatrixForm[ex118amat = {{1, 1, 2}, {1, 0, 1}, {2, 1, 3}}]

MatrixForm[RowReduce[ex118amat]]

Exercise 7 of 1.1 with solution for coefficients a, b, and c of the parabola passing through the three specific points (-1,2), (1,-1), and (2,3)

MatrixForm[
      ex117 = {{1, -1, 1, 2}, {1, 1, 1, -1}, {4, 2, 1, 
            3}}]

The next statement shows a=11/6, b=-3/2, and c=-4/3

MatrixForm[RowReduce[ex117]]

The parabola is plotted.

Plot[((11/6) x^2) - ((3/2) x) - (4/3), {x, -2, 
        2}]

The remaining computations are for exercise 14c of section 1.2

MatrixForm[
      ex1214c = {{1, 3, 0, 1}, {1, 4, 2, 
            0}, {0, -2, -2, -1}, {2, -4, 1, 
            1}, {1, -2, -1, 1}}]

MatrixForm[RowReduce[ex1214c]]