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Ch 5
4.  Find the volume of the parallelepiped formed by the three vectors:

A = i-2j
B = 2i+j
C = i-2j+2k

Finding this volume is a fairly routine procedure.  The formula I will
use is V = A*(BxC).  The area of the parallelogram formed by the
two vectors, B & C, is the absolute value of their cross product.
Then, dotting that result with A, will give the total volume of the
solid.  The calculations follow:

IDL> A=[1,-2,0]
IDL> B=[2,1,0]
IDL> C=[1,-2,2]

IDL> area=C#B
IDL> vol=total(A*area)
IDL> print,area
2          -4           4
1          -2           2
0           0           0
IDL> print,vol
10.0000

You may have noticed that the order on the cross product has been
reversed.  This due to an unknown problem with the IDL program.

10.  Solve the following simultaneous equations using Cramer's rule:

x - y + z = 2
x + 2y + z = 4
2x + 3y - z = 1

Cramer's rule allows us to calculate the of the variables by using
the value of the matrix determinant.  However in this case, we will
first form the constant matrix and then multiply that by the inverse
of the matrix of coefficients.  The calulations follow:

IDL> a = [[1,-1,1],[1,2,1],[2,3,-1]]  (This is the matrix of coefficients)
IDL> print,a
1      -1       1
1       2       1
2       3      -1

IDL> b = invert (a)                  (This is the inverse of matrix a)
IDL> print,b
0.555556    -0.222222     0.333333
-0.333333     0.333333      0.00000
0.111111     0.555556    -0.333333

IDL> c = [2,4,1]                     (This is constants matrix)
IDL> print,c
2       4       1

IDL> sol = c#b                 (This will give us the answers we want)
IDL> print,sol
x = 0.555555
y = 0.666667
z = 2.11111

14.  Show that the following three vectors are linearly independent:

a1 = (1,1,1)
a2 = (-1,1,0)
a3 = (2,0,1)

Linear independence can be demonstrated by showing that the
determinant of the matrix formed by the three vectors is equal
to zero.  In general the method of cofactors could be used to
solve this problem.  However, with the IDL program it is a very
easy matter.  The calculations follow:

IDL> A = [[1.,1.,1.], [-1.,1.,0.], [2.,0.,1.]]
IDL> print, A
1.00000      1.00000      1.00000
-1.00000      1.00000      0.00000
2.00000      0.00000      1.00000

IDL> detA = determ (A)
IDL> print, detA
0

So, as you can see the deteminant is equal to zero therefore the
three vectors are linearly independent.

32.  Find the inverse of A.

Once again the IDL program allows us to take the inverse directly.
The calculations follow:

IDL> A = [[2,1,0], [1,1,1], [2,0,2]]
IDL> print,A
2       1       0
1       1       1
2       0       2

IDL> b = invert (A)
IDL> print,b
0.500000    -0.500000     0.250000
-0.00000      1.00000    -0.500000
-0.500000     0.500000     0.250000

41.  Determine whether the matrix is a proper or an improper
transformation.

A matrix transformation is proper if the determinant is equal
to 1 and improper if the determinant is equal to -1.  Again we
will use the IDL program to save ourselves some work.  IDL do

IDL> A = [[-4./5.,0,-3./5.], [0.,1.,0.], [3./5.,0.,-4./5.]]
IDL> print,A
-0.800000      0.00000    -0.600000
0.00000      1.00000      0.00000
0.600000      0.00000    -0.800000
IDL> detA = determ (A)
IDL> print,detA
1.00000

As you can see, the value of the determinant is equal to 1 and
this proves that the transformation is proper.

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