# Development Of The Notion Of A Matrix Engineering Essay

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Â The introduction and development of the notion of a matrix and the subject of linear algebra followed the development of determinants, which arose from the study of coefficients of systems of linear equations.Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Gauss developed Gaussian elimination around 1800 and used it to solve least squares problems in celestial computations and later in computations to measure the earth and its surface the branch of applied mathematics concerned with measuring or determining the shape of the earth. The first appearance of Gauss-Jordan elimination in print was in a handbook on geodesy written by Wilhelm Jordan.Â Â Many people incorrectly assume that the famous mathematician Camille Jordan is the Jordan in ''Gauss-Jordan'' elimination.Â

Â Â Â For matrix algebra to fruitfully develop one needed both proper notation and the proper definition of matrix multiplication.Â Â Â In 1848 in England, J.J. Sylvester first introduced the term ''matrix,' 'as a name for an array of numbers.Â Â Matrix algebra was nurtured by the work of Arthur Cayley in 1855.Â Â The famous Cayley-Hamilton theorem which asserts that a square matrix is a root of its characteristic polynomial was given by Cayley in his 1858Â Memoir on the Theory of Matrices. The use of a single letterÂ AÂ to represent a matrix was crucial to the development of matrix algebra.Â Â Early in the development the formulaÂ Â det(AB) = det(A)det(B) provided a connection between matrix algebra and determinants.Â

Introduction Of Matrices:

The term matrix was apparently coined by JAMES JOSEPH SYLVESTER about 1850, but was introduced first by ARTHUR CAYLEY in 1860. By 'matrix' we meanÂ a rectangular array of quantities or expressions set out by rows and columns; treated as a single element and manipulated according to rules. The elegant 'shorthand' representation of an array of many numbers as a single object and calculations makes matrices useful.

Matrices find many applications.Â PhysicsÂ makes use of matrices. Matrices find many applications. Physics makes use of matrices in various domains, for example in geometrical optics and matrix mechanics; the latter led to studying in more detail matrices with an infinite number of rows and columns. Graph theory uses matrices to keep track of distances between pairs of vertices in a graph. Computer graphics uses matrices to project 3-dimensional space onto a 2-dimensional screen. Matrix calculus generalizes classical analytical notions such as derivatives of functions or exponentials to matrices. The latter is a recurring need in solving ordinary differential equations. Serialism and dodecaphonism are musical movements of the 20th century that use a square mathematical matrix to determine the pattern of music intervals.

Definition Of Matrix:

A matrix is a rectangular arrangement of numbers in m rows(horizontal lines) and n columns(vertical lines). The numbers in the matrix are called its entries or its elements. To specify a matrix's size, a matrix with m rows and n columns is called an m-by-n matrix or m Ã- n matrix, while m and n are called its dimensions.it is called a rectangular matrix.

System Of Linear Equation

AÂ systemÂ has these properties:

It consists of several parts which interact and affect one another.Â

It produces an effect orÂ outputÂ as a result of some cause orÂ input.

AÂ linear equationÂ in one unknownÂ is an equation of the formÂ aÂ xÂ =Â b, whereÂ aÂ andÂ bÂ are constants andÂ xÂ is an unknown that we wish to solve for. Similarly, aÂ linear equation inÂ nÂ unknownsÂ x1,Â x2, â€¦,xnÂ is an equation of the form:

a1Â Â·Â x1Â +Â a2Â Â·Â x2Â + â€¦ +Â anÂ Â·Â xnÂ =Â b,

whereÂ a1,Â a2, â€¦,Â anÂ andÂ bÂ are constants. The name linear comes from the fact that such an equation in two unknowns or variables represents a straight line. A set of such equations is called aÂ system of linear equations. .

AÂ linear systemÂ is a system where the output is proportional to the input.

We can describe mathematically how the parts of a linear system relate to one another and to the input using aÂ system of linear equations. If a linear system hasÂ nÂ parts (whereÂ nÂ is some number), then we can describe it with a system ofÂ nÂ linear equations inÂ nÂ unknowns or variables. The unknowns in these equations are the values of the inputs the inputs. .

Here is an example of a system of two linear equations in the two unknownsÂ xÂ andÂ y:

x + y = 4

2x - 3y = 6

Here is an example of a system of three linear equations in the three unknownsÂ x,Â yÂ andÂ z

4x + 8y + 4z = 80

2x + y - 4z = 7

3x - y + 2z = 22

GAUSSIAN ELIMINATION METHOD

InÂ linear algebra,Â Gaussian eliminationÂ is an efficientÂ algorithmÂ for solvingÂ systems of linear equations, to find the rankÂ of aÂ matrix, and to calculate the inverse of anÂ invertible square matrix. It is used for solving a system of n linear equations in n unknowns, in which there are first n - 1 steps, the mth step of which consists of subtracting a multiple of the mth equation from each of the following ones so as to eliminate one variable, resulting in a triangular set of equations which can be solved by back substitution, computing the nth variable from the nth equation, the (n - 1)st variable from the (n - 1)st equation, and so forth.

Elementary row operationsÂ are used to reduce a matrix toÂ row echelon form. An extension of this algorithm, Jordan elimination, reduces the matrix further toÂ reduced row echelon form. Gaussian elimination alone is sufficient for many applications.

Algorithm Overview Of Gaussian Elimination Method

The process of Gaussian elimination has two parts. The first part (Forward Elimination) reduces a given system to either triangular or echelon form, or results in a degenerate equation with no solution, indicating the system has no solution. This is

accomplished through the use of elementary row operations. The second step uses to find the solution of the system above.

Stated equivalently for matrices, the first part reduces a matrix to row echelon form using elementary row operations while the second reduces it to reduced row echelon form, or row canonical form.

Another point of view, which turns out to be very useful to analyze the algorithm, is that Gaussian elimination computes a matrix decomposition. The three elementary row operations used in the Gaussian elimination (multiplying rows, switching rows, and adding multiples of rows to other rows) amount to multiplying the original matrix with invertible matrices from the left. The first part of the algorithm computes an LU decomposition, while the second part writes the original matrix as the product of a uniquely determined invertible matrix and a uniquely determined reduced row-echelon matrix.

Solving System Of Linear Equation By Gaussian Elimination Method

A "system" of equations is a set or collection of equations. The equation

a x + b y + c z + d w = h

where a, b, c, d, and h are known numbers, while x, y, z, and w are unknown numbers, is called a linear equation. If h =0, the linear equation is said to be homogeneous. A linear system is a set of linear equations and a homogeneous linear system is a set of homogeneous linear equations. Linear equation are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables. To solve a system means to find all values of the variables that satisfy all the equations in the system simultaneously.

Steps for solving system of linear equation by Gaussian elimination method:

1. Constructing the augmented matrix for the system;

2.Using elementary row operations to transform the augmented matrix into a triangular one;

3.Writing down the new linear system for which the triangular matrix is the associated augmented matrix;

4.Solving the new system. We may need to assign some parametric values to some unknowns, and then apply the method of back substitution to solve the new system.

Q:Â Solve the following system via Gaussian eliminationÂ

2x - 3y - z + 2w + 3v = 4

4x - 4y - z + 4w +11v = 4

2x - 5y - 2z + 2w - v = 9

2y + z +4v = -5

Ans:The augmented matrix isÂ

2 -3 -1 2 3 | 4

4 -4 -1 4 11 | 4

2 -5 -2 2 -1 | 9

0 2 1 0 4 | -5

Operating R2 changes R2 - 2R1, R3 - R1

2 -3 -1 2 3 | 4

0 2 1 0 5 | -4

0 -2 -1 0 -4 | 5

0 2 1 0 4 | -5

Next we keep the first and second row and try to have zeros in the second column. We getÂ

2 -3 -1 2 3 | 4

0 2 1 0 5 | -4

0 0 0 0 1 | -1

0 0 0 0 -1 | -1

Operating R4 changes R4 - R3

2 -3 -1 2 3 | 4

0 2 1 0 5 | -4

0 0 0 0 1 | 1

0 0 0 0 0 | 0

This is a triangular matrix. Its associated system isÂ

2x - 3y - z + 2w + 3v = 4

2y + z + 5v = -4

v = 1

Since we have 5 unkwon variables and rank[A|B] is 3 thus two variables s and t are taken. Let z =s and w =t

y = -2 -1/2z - 5/2v = -9/2 - 1/2s

The first equation implies

x = 2 + 3/2y + 1/2z - w - 3/2v

or

x = -25/4 - 1/4s - t

x -25/4 - 1/4s - t

y -9/2 - 1/2s

z = s

w t

v 1

Q: Solve the following system using Gaussian elimination:Â

2x - 2y = -6

x - y + z = 1

3y - 2z = -5

Ans: The augmented matrix is

## Â

2 -2 0 | -6

1 -1 1 | 1

0 3 -2 | -5

Operating R1 changes 1/2R1

1 -1 0 | -3

1 -1 1 | 1

0 3 -2 | -5

Operating R2 changes R2 - R1

1 -1 0 | -3

0 0 1 | 4

0 3 -2 | -5

Operating R2 interchanges to R3

1 -1 0 | -3

0 3 -2 | -5

0 0 1 | 4

This is a triangular matrix. Its associated system:

x - y + = -3

3y - 2z = -5

z = 4

From 2nd equation

y = 1

from 3rd equation

x = -2

x -2

y = 1

z 4

GAUSS JORDAN METHOD

Gauss-Jordan Elimination is a method for solving a linear system of equations. The Gauss-Jordan Elimination method works with the augmented matrix in order to solve the system of linear equation.

It is also method for finding aÂ matrix inverse . To use Gauss-Jordan Elimination, we start by representing the given system of linear equations as an augmented matrix

The goal of Gauss-Jordan elimination is to transform the matrix into reduced echelon form. consider three linear equations to be solved

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a1x+b1y+c1z=d1

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a2x+b2y+c2z=d2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a3x+b3y+c3z=d3

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Augmented matrix is given by

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [[a1,b1,c1,d1],[a2,b2,c2,d2],[a3,b3,c3,d3]]Â

has to be transformed to

Î±Î²Î³

Steps for solving gauss Jordan elimination method:

1. Representing the linear system of equations as a matrix in augmented matrix form

2. Using elementary row operations to derive a matrix in reduced echelon form

3. Writing the system of linear equations corresponding to the reduced echelon form.

Q: Solve the matrix by gauss-jordan method and back substitution:

x1 âˆ’ 2x2 + x3 = 3

2x1 + x2 âˆ’ x3 = 0

7x1 âˆ’ 4x2 + 2x3 = 31

Ans: The augmented matrix is:

1 âˆ’2 1 | 3

2 1 âˆ’1 | 0

7 âˆ’4 2 | 31

Operating R2 changes R2 - 2R1

1 âˆ’2 1 | 3

0 5 âˆ’3 | âˆ’6

0 10 âˆ’5 | 10

Operating R2 changes 1/5R2

1 âˆ’2 1 | 3

0 1 âˆ’3/5 | âˆ’6/5

0 10 âˆ’5 | 10

Operating R3 changes R3 - 10R2

1 âˆ’2 1 | 3

0 1 âˆ’3/5 | âˆ’6/5

0 0 1 | 22

This can be solved in row-echelon form.

Its linear form is:

x1 - 2x2 + x3 = 3

x2 - 3/5x3 = -6/5

x3 = 22

putting the value of x3 in 2nd equation to get x2 value

x2 = âˆ’6/5 + 3(22)/5 = 60/5 = 1

Putting the value of x3 and x2 in 1st equation for getting x1 value.

x1 = 3 + 2(12) âˆ’ 22 = 5

Therefore x1 = 5,

x2 = 12 and

x3 = 22

Q: solve the following system of linear equation by gauss Jordan method.

x + 3y - 6z = 7

2x - y + 2z = 0

x + y + 2z = -1

Ans: Arranging the matrix in the form:

1 3 -6 | 7

2 -1 2 | 0

1 1 2 | -1

Operating R2 changes R2 - 2R1

1 3 -6 | 7

0 -7 14 | -14

1 1 2 | -1

Operating R3 changes R3 - R1

1 3 -6 | 7

0 -7 14 | -14

0 -2 8 | -8

Operating R2 changes -1/7R2 and R3 changes -1/2R3

1 3 -6 | 7

0 1 -2 | 2

0 1 -4 | 4

Operating R3 changes R3 - R2

1 3 -6 | 7

0 1 -2 | 2

0 0 -2 | 2

Operating R3 changes -1/2R3

1 3 -6 | 7

0 1 -2 | 2

0 0 1 | -1

Operating R2 changes R2 + 2R3

1 3 -6 | 7

0 1 0 | 0

0 0 1 | -1

Operating R1 changes R1 + 6R3

1 3 0 | 1

0 1 0 | 0

0 0 1 | -1

Operating R1 changes R1 - 3R2

1 0 0 | 1

0 1 0 | 0

0 0 1 | -1

Thus a diagonal matrix is obtained

The solution are x= 1,

y= 0,

z= -1