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Friday, August 8, 2014

Matrices

Definition of Equality of Matrices
Two Matrices A = [aij] and B = [bij] are equal if they have the same size (m x n) and aij = bij
for i and j are less than or equal to m but greater or equal to one.


Matrix Addition
If A = [aij] and B = [bij] are matrices of m x n, then their sum is the m x n matrix given by
                     A + B = [aij + bij]
The sum of two matrices of different sizes is undefined.
Examples



Scalar Multiplication
If A = [aij] is an m x n matrix and c is a scalar, then the scalar multiple of A by c is the m x n matrix given by
          cA = [caij]
Example

Matrix Multiplication

Network Analysis



Polynomial Curve Fitting





Monday, July 28, 2014

Homogeneous System of Linear Equations

A system is said to be homogeneous if all of the constant terms are zero. For a example, a homogeneous system of m equations in n variables has the form:
\begin{alignat}{7}
a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& 0 \\
a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& 0 \\
\vdots\;\;\; &&     && \vdots\;\;\; &&              && \vdots\;\;\; &&     &&& \,\vdots \\
a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& 0. \\
\end{alignat}

Example

Gauss-Jordan Elimination

Gaussian Elimination with Back Substitution

We will use Gaussian Elimination to solve the linear system
x1 2x1 3x1 +  2x2 x2 + +  3x3 x3 x3 = = = 9 8 3  .
The augmented matrix is
   1 2 3 2 1 0 3 1 1         9 8 3     

The Gaussian Elimination algorithm proceeds as follows:
   1 2 3 2 1 0 3 1 1         9 8 3     
   1 0 0 2 5 6 3 5 10         9 10 24     
(Row 1)
(Row 22Row 1)
(Row 33Row 1)
   1 0 0 2 1 6 3 1 10         9 2 24     
(Row 1)
(15Row 2)
(Row 3)
   1 0 0 2 1 0 3 1 4         9 2 12     
(Row 1)
(Row 2)
(Row 3+6Row 2)
   1 0 0 2 1 0 3 1 1         9 2 3     
(Row 1)
(Row 2)
(14Row 3)
We have brought the matrix to row-echelon form. The corresponding system
x1 + 2x2 x2 + + 3x3 x3 x3 = = = 9 2 3  

is easily solved from the bottom up:
x3=3 x2+3=2x2=1 x1+2(1)+3(3)=9x1=2  

Thus, the solution of the original system is x1=2x2=1x3=3