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 your stuff: 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.