MATRIX

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A matrix consists of a rectangular array of elements represented by single symbol. As depicted in next figure,[A] is the shorthand notation for the matrix and aij designates an individual element of the matrix.

A horizontal set of elements is called a row and a vertical set is called a column. The first subscript i designates the number of the row in which the element lies, the second subscript j designates the column.

SPECIAL TIPES OF MATRICES


MATRIX OPERATION RULES

Addition of two matrices: [A] and [B] , is accomplished by adding corresponding terms in each matrix, the elements of the resulting matrix [C] are computed:
Cij = aij + bij
The additions of two matrices are similar to the subtraction of two matrices: subtracting corresponding terms.
Cij = aij – bij
Addition and subtraction are commutative and associative.
The product of two matrices is represented [C]=[A][B], where the elements of [C] are defined as According to Chapra, where n= the column dimension of [A] and the row dimension of [B]. that is, the cij elements obtained by adding the product of individual elements from the ith row of the firs matrix, in this case [A], by jth column of the second matrix [B]. According to this definition, multiplication of two matrices can be performed only if the firs matrix has as many columns as the number of rows in the second matrix. Thus, if [A] is an n by m matrix, [B] could be an m by l matrix. For this case, the resulting [[C] matrix would have the dimension of n by l. Remember, multiplication is not generally commutative.

TECHNIQUES FOR IMPROVING SOLUTIONS

Cramer Rules
This rule applies to systems where: the equations numbers is equal at the unknown quantity, and that the determinant from the matrix be different to zero.
If the determinant from the matrix is Δ:

And be Δ1, Δ2, Δ3, …, Δn determinants getting from the substitution of the independence term in the first column, second column, third column, … and the n th column; using the Cramer rule:
X1= Δ1/ Δ x2= Δ2/ Δ x3= Δ3

Example:

X= 21/2 y= -8/2=-4 z=-11/2

GAUSS-JORDAN
This is a variation of Gauss elimination. The major difference is that when an unknown is eliminated in the Gauss-Jordan method, it’s eliminated from all other equations rather than just the subsequent ones. In addition, all rows are normalizes by dividing them by their pivot elements. Thus, the elimination step results in an identity matrix rather than a triangular matrix. Consequently, it is not necessary to employ back substitution to obtain the solution*.

Example:
3.0 X1 - 0.1 X2 - 0.2 X3 = 7.8500
0.1 X1 + 7.0 X2 - 0.3 X3 = - 19.3
0.3 X1 - 0.2 X2 + 10 X3 = 71.4000

The first line is normalized, dividing in 3 the first row to obtain:

The x1 term can be eliminated from second rows subtracting 0,1 times the first from the second row. Similarly, subtracting 0,3 times the first row from the third row willeliminated the x1 term from the third row:


Next, normalize the second row by dividing it by 7,00333:

Reduction of the x2 terms from the first and third equations gives:

The third row is them normalized by dividing it by 10,0120:

Finally, the x3 terms can be reduced from the first and the second equations to give:

LU DECOMPOSITION “Lower and Upper”
An original matrix is upset in two triangular matrixes: a lower and upper.
Methodology
1-Obtain the lower triangle matrix “L” and the upper triangle “U”.
2-Solve Ly=b
3-Save the new matrix with name “y”
4-Solve Ux=y (to find X)
5-The new matrix “x” offer the unknown quantities.



THOMAS METHOD
This method is a simplification of “LU” method upon a diagonal matrix.



If say A=LU, and apply Doolite where lii=1 to i=1 far as n, obtain:


Based in before matrix product, obtained the next expressions:
U11=b1
Ln,n-1=(an /U n-1,n-1)
Un-1,n= cn-1
Un,n= bn-Ln,n-1*Un-1,n
Where: a1=0 and cn=0
If Lux=r and Ux=d then Ld=r well them:


d1=rr1
Since k=2 as n
Dk= rk - Lk,k-1*dk-1
Finally solve Ux=d from a substitution:


Example:

U11=b1=1
For k=2
L11=(a2/U11)=(3/1)=3
U12=c1=3
U22=b2- L21*U12=1-3(3)=-8
For k=3
L32=(a3/U22)=(2/-8)=-1/4
U23=c2=2
U33=b3- L32*U23=1+(1/4)(2)=1,5
For k=4
L43=(a4/U33)=(5/1,5)=3,33
U34=c3=5
U44=b4- L43*U34=1-(10/3)(5)=-47/3
For k=5
L54=(a5/U44)=(3/(-47/3))=-9/47
U45=c4=3
U55=b5- L54*U45=1+(9/47)(3)=(74/47)

Finding the L matrix and solving L*d=r
D1=r1=5
K2---d2=r2-L21*d1 = 11-3(5)= -4
K3---d3=r3-L32*d2 = 8
K4---d4=r4-L43*d3 = -6,
K5---d5=r5-L54*d4 = 4,72



Solving U*x=d through regressive substitution:
X5= d5/U55= (3,6/(74/47))=3
To k= n-1 =4
X4=(d4-(U45*X5))/U44 =(-6,66-(3*2,27))/(-17/3)=1
To k= n-2 =3
X3=(d3-(U34*X4+U35*X5))/U33 =2
….

Cholesky Method
According to Chapra, this algorithm is based on the fact that a symmetric matrix can be decomposed as in:
[A]=[L][L]^(T)
That is, the resulting triangular factors are the transpose of each other. The terms of the last equations can be multiplied out and set equal to each other. The result can be expressed simply by recurrence relations. For the k th row*:

REFERENCES

CHAPRA, Steven C. y CANALE, Raymond P.: “Numerical Methods for engineers”. McGraw Hill, fifth edition, 2006.*
CARRILLO, Eduardo, “Lu para resolver métodos abiertos”, UIS, 2010.
Pinto F, Toledo M, Plata A, Gomez E, Budez J, Mancilla R;”Matrices y sistemas de ecuaciones”; UIS 2010.
http://personal.redestb.es/ztt/tem/t6_matrices.htm
http://docencia.udea.edu.co/GeometriaVectorial/uni2/seccion21.html