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Using Mathematica for linear algebra Solving simultaneous equations Mathematica will solve simulataneous equations for us, e.g. (Note that one must use == in the relations, not =) Solve@82x + 5y - z 2, x + y + 2z 1, x + 5z 3<, 8x, y, z<D ::x fi- 9 2 ,y 5 2 ,z 3 2 >> This is how Mathematica deals with a case without a solution: it gives an empty solution vector. Solve@8x + y 2, x + y 5<, 8x, y<D 8< Here’s what happens if the equations have an infinite number of solutions: Solve@8x + y 2,2x + 2y 4<, 8x, y<D Solve::svars : Equations may not give solutions for all "solve" variables. 88y 2 - x<< We can also solve our original problem using the augmented matrix eeaug = 882, 5, - 1, 2<, 81, 1, 2, 1<, 81, 0, 5, 3<<; eeaug MatrixForm 25 - 12 1121 1053 We ask Mathematica to “row reduce” the augmented matrix (see Boas 3.2) and can read off the answer RowReduce@eeaugD MatrixForm 100 - 9 2 010 5 2 001 3 2 Finally, we can solve by writing the problem in matrix form, E r=D, with E and D as follows. ee = 882, 5, - 1<, 81, 1, 2<, 81, 0, 5<<; ff = 82, 1, 3<; MatrixForm@eeD 25 - 1 112 105 MatrixForm@ffD 2 1 3 Here's the solution:

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Page 1: Solving simultaneous equations - University of Washingtoncourses.washington.edu/ph227814/227/A13/notes/Linear1.nb.pdf · Solving simultaneous equations Mathematica will solve simulataneous

Using Mathematica for linear algebra

Solving simultaneous equations

Mathematica will solve simulataneous equations for us, e.g.

(Note that one must use == in the relations, not =)

Solve@82 x + 5 y - z � 2, x + y + 2 z � 1, x + 5 z � 3<, 8x, y, z<D

::x ® -9

2

, y ®5

2

, z ®3

2

>>

This is how Mathematica deals with a case without a solution:

it gives an empty solution vector.

Solve@8x + y � 2, x + y � 5<, 8x, y<D8<

Here’s what happens if the equations have an infinite number of solutions:

Solve@8x + y � 2, 2 x + 2 y � 4<, 8x, y<D

Solve::svars : Equations may not give solutions for all "solve" variables. �

88y ® 2 - x<<

We can also solve our original problem using the augmented matrix

eeaug = 882, 5, -1, 2<, 81, 1, 2, 1<, 81, 0, 5, 3<<; eeaug �� MatrixForm

2 5 -1 2

1 1 2 1

1 0 5 3

We ask Mathematica to “row reduce” the augmented matrix (see Boas 3.2) and can read off the answer

RowReduce@eeaugD �� MatrixForm

1 0 0 -9

2

0 1 05

2

0 0 13

2

Finally, we can solve by writing the problem in matrix form, E r=D, with E and D as follows.

ee = 882, 5, -1<, 81, 1, 2<, 81, 0, 5<<; ff = 82, 1, 3<;

MatrixForm@eeD2 5 -1

1 1 2

1 0 5

MatrixForm@ffD2

1

3

Here's the solution:

Page 2: Solving simultaneous equations - University of Washingtoncourses.washington.edu/ph227814/227/A13/notes/Linear1.nb.pdf · Solving simultaneous equations Mathematica will solve simulataneous

rr = [email protected]

:-9

2

,

5

2

,

3

2

>

Vectors & operations

Vectors are entered in row form:

v1 = 81, -1, 3<81, -1, 3<

v2 = 83, 1, 2<83, 1, 2<

Dot product:

Dot@v1, v2D8

It can also be written as just a Dot:

v1.v2

8

Cross product:

v1cr2 = Cross@v1, v2D8-5, 7, 4<

is antisymmetric, as it should be,

Cross@v2, v1D85, -7, -4<

and is orthogonal to v1 and v2:

8Dot@v1cr2, v1D, Dot@v1cr2, v2D<80, 0<

You can pick out individual components of a vector as follows:

(Note the double square parentheses.)

v1@@2DD-1

Complex vectors

v3 = 81 + I, 2, -3 I<; v4 = 86 + I, -3 I, 4<;

To get the complex inner product, need to explicitly put in the complex conjugation:

2 Linear1.nb

Page 3: Solving simultaneous equations - University of Washingtoncourses.washington.edu/ph227814/227/A13/notes/Linear1.nb.pdf · Solving simultaneous equations Mathematica will solve simulataneous

[email protected] + ä

Reversing order gives complex conjugate:

[email protected] - ä

The norm or length of v3 is

Sqrt@[email protected]

15

Mathematica has a built in Norm function (which takes lots of types of argument), but in this case gives

the standard norm:

Norm@v3D

15

Cross product works for complex vectors too:

Cross@v3, v4D817, -1 - 22 ä, -9 - 5 ä<

Matrices

Functions of Matrices

Eigenvalues and Eigenvectors

Linear1.nb 3