Problem Set 3 Individual

Problem I3-1

Kreyszig 6.7, 2,4,6,8

2.

\(Inverse[
		{
		{1, 2, 
            5}, 
		{0, \(-1\), 
            2}, 
		{2, 4, 11}
		}] // 
      MatrixForm\)

4.

\(Inverse[
		{
		{\(-7\), 0, 
            0}, 
		{0, 8, 13}, 
		{0, 
            3, 5}
		}] // MatrixForm\)

6.

\(Inverse[
		{
		{0, 
            0,  .3}, 
		{0, \(- .2\), 
            0}, 
		{ .4, 0, 0}
		}] // 
      MatrixForm\)

8.

\(Inverse[
		{
		{1, 2, 
          3}, 
		{4, 5, 6}, 
		{7, 8, 
          9}
		}]\)

Problem I3-2

Kreyszig 6.7: 10,11,12,13,14: Note, the work that follows does not include solutions for all of the matrices in I3-1, but what follows is the model for solving each problem.

10.

For these problems, we can use logic statements or we can compute the left and right hand sides of the equations.  The matrix I've used here is from Problem 1 in the text.

\(\(A = {
	{2, 0, \(-1\)}, 
	{5, 1, 
            0}, 
	{0, 1, 
            3}
};\)\)

Inverse[A . A] == Inverse[A] . Inverse[A]

11.

Same as 10.

Inverse[A . A] == Inverse[A] . Inverse[A]

12.

Transpose[Inverse[A]] == Inverse[Transpose[A]]

13.

First just make up a symmetric matrix.

\(\(SymMat\  = \ {
		{1, 2, 
            3}, 
		{2, 
            5, \(-3\)}, 
		{3, \(-3\), 
            4}
		};\)\)

Inverse[SymMat] // MatrixForm

This is symmetric!

14.

Inverse[Inverse[A]] == A

Problem I3-3

Kreyszig 6.8: 19,20,22,23,24

19.

For two vectors to be orthogonal, their angle of separation must be 90 degrees, meaning their dot product (called "inner product" in the text) must be 0.  By definition, the dot product of two vectors a & b is a scalar of the value: Norm(a) * Norm(b) * Cos(θ), where θ is the angle between the two vectors.  This principle allows the following problems to be solved efficiently

Vec14  =  
        Transpose[{{2, 0, 3, 0, 8}}];
Vec14 // MatrixForm

Vec17  =  {{3, 2, -2, 4, 0}};
Vec17 // MatrixForm

Vec17 . Vec14 // MatrixForm

20.

Here we are to find all unit vectors in terms of v1, v2 that are orthoganal to another vector.

VecInQues  =  {{3, -4}};

VecV  =  Transpose[{{v1, v2}}];

UnitVector = 
      Solve[{VecInQues . VecV == 0,  v1^2 + v2^2 == 1}, {v1, 
          v2}]

The Schwaz & triangle inequalities as well as the parallelogram equality are shown on page 361.

22.

Vec13  =  
        Transpose[{{ .4, 1.3, -2.2}}];
Vec16  =  Transpose[{{2, 3, -5}}];

Abs[Flatten[Transpose[Vec13] . Vec16]][[1]] ≤ 
      Norm[Vec13]*Norm[Vec16]

Abs[Flatten[Transpose[Vec13] . Vec16]][[1]]

Norm[Vec13]*Norm[Vec16]

23.

Norm[Vec13 + Vec16] ≤ Norm[Vec13] + Norm[Vec16]

Norm[Vec13 + Vec16]

Norm[Vec13] + Norm[Vec16]

24.

Vec21  =  Transpose[{{2, 1}}];
Vec13  =  Transpose[{{1, 3}}];

(Norm[Vec21 + Vec13])^2 + (Norm[Vec21 - Vec13])^2 == 
      2 (Norm[Vec21]^2 + Norm[Vec13]^2)

ListPlot[{{2, 1}, {3, 4}, {1, 3}, {0, 0}, {2, 1}, {1, 3}, {3, 4}, {0, 
            0}}, PlotJoined -> True];

The diagonals represent a-b and a+b (note a+b > a-b here).  The parallelogram equality says that the sum of the squares of the 4 sides is equal to the sum of the squares of the diagonals.  Simple trigonometric identities would allow this to be proven on paper.