Linear Algebra

When performing the cross product of two vectors (vector A and vector B), one of the properites of the resultant vector C is that it is perpendicular to both vectors A & B. In two dimensional space, this is not possible, because the resultant vector will be perpendicular to the plane that A & B reside in. Using the (i,j,k) unit vector notation, you could add a 0*k to each vector when doing the cross product, and the resultant vector will have zeros for the i & jcomponents, and only have k components.

Two vectors define a plane, and their cross product is always a vector along the normal to that plane, so the three vectors cannot lie in a 2D space which is a plane.

The answer depends on the context. Some examples:

• z can represent the length of a side of a polygon, for example, a triangle with sides of lengths x, y and z;
  
• z can represent the vertical axis in 3-dimensional coordinate geometry (where x and y are used for the base plane);
• z can represent a variable in the complex plane (z = x + yi);
• z can represent the probability that a random observation from a Normal distribution is at least as extreme as the one seen.

#include

#include

#define MAX1 3

#define MAX2 3

struct sparse

{

int *sp ;

int row ;

} ;

void initsparse ( struct sparse * ) ;

void create_array ( struct sparse * ) ;

void display ( struct sparse ) ;

int count ( struct sparse ) ;

void create_tuple ( struct sparse *, struct sparse ) ;

void display_tuple ( struct sparse ) ;

void transpose ( struct sparse *, struct sparse ) ;

void display_transpose ( struct sparse ) ;

void delsparse ( struct sparse * ) ;

void main( )

{

struct sparse s[3] ;

int c, i ;

for ( i = 0 ; i <= 2 ; i++ )

initsparse ( &s[i] ) ;

clrscr( ) ;

create_array ( &s[0] ) ;

printf ( "\nElements in Sparse Matrix: " ) ;

display ( s[0] ) ;

c = count ( s[0] ) ;

printf ( "\n\nNumber of non-zero elements: %d", c ) ;

create_tuple ( &s[1], s[0] ) ;

printf ( "\n\nArray of non-zero elements: " ) ;

display_tuple ( s[1] ) ;

transpose ( &s[2], s[1] ) ;

printf ( "\n\nTranspose of array: " ) ;

display_transpose ( s[2] ) ;

for ( i = 0 ; i <= 2 ; i++ )

delsparse ( &s[i] ) ;

getch( ) ;

}

/* initialises data members */

void initsparse ( struct sparse *p )

{

p -> sp = NULL ;

}

/* dynamically creates the matrix of size MAX1 x MAX2 */

void create_array ( struct sparse *p )

{

int n, i ;

p -> sp = ( int * ) malloc ( MAX1 * MAX2 * sizeof ( int ) ) ;

for ( i = 0 ; i < MAX1 * MAX2 ; i++ )

{

printf ( "Enter element no. %d:", i ) ;

scanf ( "%d", &n ) ;

* ( p -> sp + i ) = n ;

}

}

/* displays the contents of the matrix */

void display ( struct sparse s )

{

int i ;

/* traverses the entire matrix */

for ( i = 0 ; i < MAX1 * MAX2 ; i++ )

{

/* positions the cursor to the new line for every new row */

if ( i % MAX2 0 )

printf ( "\n" ) ;

printf ( "%d\t", * ( p.sp + i ) ) ;

}

}

/* deallocates memory */

void delsparse ( struct sparse *p )

{

free ( p -> sp ) ;

}

First, we need to recall that a linear equation does not involve any products or roots of variables. All variables occur only to the first power and do not appear as arguments for trigonometric, logarithmic, or exponential functions. For example, x + √y = 4, y = sin x, and 2x + y - z + yz = 5 are not linear.

To solve a system of equations such as

3x + y = 5

2x - y = 3

all information required for the solution is emboded in the augmented matrix (imagine that I put those information into a rectangular arrays)

3 1 5

2 -1 3

and that the solution can be obtained by performing appropriate operations on this matrix.

The matrix of this system linear equations is a square matrix A such as

3 1

2 -1

Think this matrix as

a b

c d

To find an inverse of this square matrix A (2 x 2), we need to find a matrix B of the same size such that AB = I and BA = I, then A is said to be invertible and B is called the inverse of A. If no such a matrix can be found, then A is said to be singular.

An invertible matrix has exactly one inverse.

A square matrix A is invertible if ad - bc ≠ 0 (where ad - bc is the determinant)

The formula of finding the inverse of a square matrix A is

A-1 = [1/(ad - bc)][d -b the second row -c a](I'm sorry, I can't draw the arrays)

So let's find the inverse of our example.

A-1 = [1/(-3 -2)][-1 -1 the second row -2 3] = [-1/-5 -1/-5 the sec. row -2/-5 3/-5] =

1/5 1/5

2/5 -3/5

A n x m matrix cannot have an inverse. A n x n matrix may or may not have an inverse.

To find the inverse of a n x n matrix we should to adjoin the identity matrix to the right side of A, thereby producing a matrix of the form [A | I]. Then we should apply row opperations to this matrix until the left side is reduced to I. This opperations will convert the right side to A-1, so the final matrix will have the form [I |A-1 ].

(There are many other methods how to find the inverse of a n x n matrix, but I can't show them by examples. I am so sorry that I can't be so much useful to you).

In N dimensions i.e. 3 space dimensions and 1 time dimension (our perceived universe) there is a way of collapsing all of the contain information (any nontrivial function i.e. x^4 + y^4 +Z^4 = any real number other than zero (Eigen function) at time being discrete towards 0 gives an informational representation of 3 dimensions in 4 dimensions after the first derivation and a 2 dimensional information representation of 4 dimensional space. After deriving any Eigen function the Eigen Values in a domain and range can be represented by a singularity. (i.e. the singularity that expanded "in the beginning) which is commonly known as the Big Bang theory).

In real life this can be useful in reducing large amounts of information into a "simplest" symbol. i.e. a 2 dimensional string (string theory) can be described by an Eigen Function. All the information of said string can be described in binary code i.e. 00111010101010110011101010 can be reduced by defining a string of binary code by a symbol say: q (where there can be an infinite amount of symbols)
Therefor a string of two dimensional information represented by symbols can then be reduced again by another symbol.

This can be done until all of the information in any space time of n dimensions is represented by a symbol defined by all of the information in said space time dimensions up to time t with complete regression assuming that information CAN be regressed with no error.

A real life example would be huge amounts of binary code is reduced and then transported to a television in which the simplified information is then expanded back to the original information which is then transformed by the television into a moving picture.

A more controversial assessment of a real life example in the field of Mathematics and Physics is that all of the information in our current space time (4-D) up to t=present can be represented by binary code at a snapshot t=plank time reduced to a symbol. The amount of plank times from the Big Bang to now can be reduced back down to the singularity from whence The Big Bang occurred. (This assumes that there is no error in the possible irreducible holonomies. The decomposition and classification of information can be represented by a singularity)