Saturday, May 11, 2013

Equality Of Matrices


Equality Of Matrices

For two matrices to be equal, they must be of the same size and have all the same entries in the same places. For instance, suppose you have the following two matrices:
    A = [ [  1  3 ] [ -2 0 ] ] and B = [ [ 1  3  0 ] [-2  0  0 ] ]
These matrices cannot be the same, since they are not the same size. Even if A and B are the following two matrices:
    A = [ [  1  4] [ 2  5 ] [ 3  6 ] ] and B = [ [ 1  2  3 ] [ 4  5  6 ] ]
...they are still not the same. Yes, A and B each have six entries, and the entries are even the same numbers, but that is not enough for matrices. A is a 3 × 2 matrix and B is a 2 × 3 matrix, and, for matrices, 3 × 2 does not equal 2 × 3! It doesn't matter if A and Bhave the same number of entries or even the same numbers as entries. Unless A and B are the same size and the same shape and have the same values in exactly the same places, they are not equal.
This property of matrix equality can be turned into homework questions. You will be given two matrices, and you will be told that they are equal. You will need to use this equality to solve for the values of variables.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
  • Given that the following matrices are equal, find the values of x and y.
      A = [ [  1  2 ] [ 3  4 ] ] and B = [ [ x  2 ] [ 3  y ] ]
      x = 1,    y = 4
  • For A and B to be equal, they must have the same size and shape (which they do; they're each 2 × 2 matrices) and they must have the same values in the same spots. Then a1,1 must equal b1,1, a1,2 must equal b1,2, and so forth. The entries a1,2 anda2,1 are clearly equal, respectively, to entries b1,2 and b2,1 "by inspection" (that is, "just by looking at them"). But a1,1 = 1 is not obviously equal to b1,1 = x. For A to equal B, I must have a1,1 = b1,1, so it must be that 1 = x. Similarly, I must have a2,2 =b2,2, so then 4 must equal y. Then the solution is:
  • Given that the following matrices are equal, find the values of x, y, and z.
      A = [ [  4  0 ] [ 6 -2 ] [ 3  1 ] ] and B = [ [ x  0 ] [ 6  y+4 ] [ z/3  1 ] ]
        4 = x 
      –2 = y + 4
       
        3 = z/3
      matrices with 1,1-, 2,2-, and 3,1-entries highlighted.
      x = 4, y = –6, and z = 9.
  • To have A = B, I must have all entries equal. That is I must have a1,1 = b1,1, a1,2 = b1,2, a2,1 = b2,1, and so forth. In particular, I must have:
    ...as you can see from the highlighted matrices:
    Solving these three equations, I get:

Example


Given that the following matrices are equal, find the values of x, y and z .
Solution:
Equate the corresponding elements and solve for the variables.
x + 3 = 6 
x = 3
y = −1
z −3 = 4
z = 7

Friday, May 10, 2013

What is the Relation of Tessellations with Math


Introduction to Tessellations
              Tessellations are nothing but the arrangement of certain shape in a plane without leaving any space between them. The arrangement can be made by any polygon starting from 3 to infinity sides. If the arrangement involves with regular polygons like square, pentagon, hexagon etc., then it is called as regular tessellations. The tessellations can also be formed by semi-regular Polygons also, but the arrangement formed should be identical from each vertex point.

tessellations
Let us see Description of what is the relation of tessellations with math, Studying what is the relation of tessellations with angle measurements in math, Studying what is the relation of tessellations of different relations in math in this article.


relation of tessellations with math:


              Tessellations with math not only mean the flooring with shape, but it also involves the arrangement of shapes with particular angles of degrees. The tessellations involves the mathematical terms like vertex, angles (interior) and also the number of sides which is essential to arrange any shape. It involves the calculation of interior angles to arrange regular as well as semi regular shapes. This leads to the study of different angles of the shape. What are the relations of regular tessellations? For example, for the regular tessellation the shape involved should be exactly divisible by 360 degrees. 

types of tessellations

Studying tessellations with angle measurements in math:


Interior measure of this angles each of these polygons:
              What are the relations of interior tessellations with math measurement angles?
Shape
Angle measure in  degrees
triangle
60
square
90
pentagon
108
hexagon
120
More than six sides
More than 120 degrees
              Tessellations with math relations can plug the plane at each vertex in a regular polygons, an exact value of divisor of 360 in an interior angles. What are the relations of tessellations of the triangle, square, and hexagon?

Studying tessellations of different relations in math


Naming Conventions:
           What are the naming conventions for tessellations with math relations?
           For square is the relation of tessellations:
 relation of tessellations
           A shape of regular congruent hexagons that touch with a vertex and then count the sides of the polygons.
           Three polygons containing six sides, so this tessellations naming convention is "6.6.6".
 relation of tessellations(A)
           A triangle containing six polygons with surrounding vertex, and then each polygon has three sides: and what it’s naming convention is that "3.3.3.3.3.3".
relation of tessellations(B)

Monday, April 8, 2013

Solve Algebra Fraction

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Here we are going to study about How to solve algebra fraction and its example problems. (Source from Wikipedia)

Algebra fraction:

In algebra fraction is a simple fraction it contains algebraic expression in either numerator or denominator.

Example:
`(3x+3) / 2` = 3

Example problems:

Example: 1

Solve the following algebraic equation 

`(2x+4) / 2` + `(3x+3) / 4` = 6

Solution:

Here we have two fraction term and both have different denominator so we take Common denominator

Least common denominator = 2 * 4 = 8

Multiply first term with 4 we get

`(8x+16) / 8` + `(6x+6) / 8` = 6

Now both denominators is equal so take away

`1/ 8` {(8x+16) + (6x+6)} = 6

Multiply both sides 8 we get

8 * `1/ 8 ` {(8x+16) + (6x+6)} = 6 * 8

In left hand side 8 will be canceling

(8x+16) + (6x+6) = 48

Combine the like terms we get

14 x + 22 = 48

Add both sides -22

14x + 22 -22 = 48 -22

14x = 26

Divide both sides 14 we get

x = `26 /14`

The simplest fraction is `13 / 7`

Therefore the value of x = `13 / 7`


Example : 2


Solve the following algebraic equation

`(3x+3) / 2` = 3

Solution:

Here the denominator is 2

So multiply both sides 2 we get

2 `(3x+3) / 2` = 3 *2

In left hand side numerator 2 and denominator 2 will be canceling

3x +3 = 6

Add both sides -3 we get

3x+3-3 = 6 -3

3x = 3

Now we get without fraction equation.it is simple algebraic equation.

Divide both sides 3 we get

x =` 3 / 3`

Therefore the value of x = 1


Example: 3


Solve the following algebraic equation

`4 / (2x+3)` = 3

Solution:

Multiply both sides (2x+3)

(2x+3) * `4 / (2x+3)` = 3 (2x+3)

Numerator (2x+3) and denominator (2x+3) will be cancelling

4 = 3 (2x+3)

4 = 6x+9

Add both sides -9 we get

4-9 = 6x+9-9

-5 = 6x

Therefore x = - `(5/6)`

The value of x – `(5/6)`

Friday, April 5, 2013

Solve Second Derivative Test

Solve second derivative test involves the process differentiating the given algebraic function twice with respect to the given variable. Generally the derivative is discussed in calculus whereas it is mainly used to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution in second derivative to study for the test.


Second derivative test example problems to solve:


Example 1:

Solve the test function to find second derivative.

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

Solution:

The given equation is

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

The above function is differentiated with respect to z to find the first derivative

f '(z) =  2(6z 5)  +2 (5 z4 ) +3(4 z3) + 3

By solving above terms

f '(z) =  12z 5  +  10z4  + 12 z3 – 3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  12(5z 4 ) – 10(4z3)  + 12(3z2)

f ''(z) =  60z 4 – 40z3 +36z2  is the answer.

Example 2:
Solve the test function to find second derivative.

f(z) = 5z 2 +5z 4  + 12

Solution:

The given function is

f(z) = 5z 2 +5z 4  + 12

The above function is differentiated with respect to z to find the first derivative

f '(z) = 5(2z  )+5(4 z 3 ) + 0

By solving above terms

f '(z) = 10z +20z3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  10(1 ) +20(3z2)

f ''(z) =  10 + 60z2 is the answer.

Example 3:

Solve the test function to find second derivative.

f(z) = 4z4 +5z 5 +6z 6  + 2z

Solution:

The given function is

f(z) = 4z4 +5z 5 +6z 6  + 2z

The above function is differentiated with respect to z to find the first derivative

f '(z) = 4(4z 3 )+5(5z 4 ) +6( 6z 5) +2

By solving above terms

f '(z) = 16z 3 +25z 4 +36 z 5 + 2

The above function is again differentiated with respect to z to find second derivative

f ''(z)= 16(3z 2) +25(4z 3) +36 (5z 4)

f ''(z)= 48z 2 +100z 3 +180z 4 is the answer.


Second derivative test practice problems to solve:


1) Solve the test function to find second derivative.

f(z) = z 3 + z 4 + z 5

Answer: f ''(z) = 6z +12z2+ 20z 3

2) Solve the test function to find second derivative.

f(z) = 2z 3+3z5 + 4z 6

Answer: f ''(z) = 12z + 60z3 + 120 z 4

Wednesday, April 3, 2013

Learning Intermediate Algebra

Algebra executes most common four basic types of operations. That includes the addition, subtraction, multiplication and division operations. Algebraic problems contains the variables, constant, coefficients, exponents, terms and expressions. Balanced equations on both of the sides are considered as the fundamental concept of algebra. Important properties such as commutative, associative, identities and inverse are also involved in the algebraic problems.


Learning common term in intermediate algebra:


Variables:

Algebraic variables are use for assigning the variables. Alphabetical characters alone used as variables. There are possibilities for changes of variable values while solving the algebraic problems.  x, y and z are the three commonly used variables in algebra.

Constant:

Algebraic constants are values that never change during the process of solving the algebraic equation. Consider the algebraic form of equation 6y + 5, in which the value 5 is the constant.

Expressions:

Algebraic expressions are considered as the combinations of variables, constant, coefficients, exponents and terms.Addition, subtraction, multiplication and division expressions proves this. The example of an algebraic expression is given below

15y + 6.

Term:

Terms are used to form the algebraic expressions by some of the arithmetic operations such as addition, subtraction, multiplication and division. Consider the following example 5n^2 + 6n in which the terms 5n^2, 6n are combined to form the algebraic expression 3n^2 + 2n by using the addition operator ( + ) to perform addition operation.

Coefficient:

The coefficients are the values that are present in front of the terms of an algebraic expression. Consider the following example, 6n2 + 4n in which the coefficient of 6n2 is 6 and 4n is 4.

Equations:

Algebraic equations are the combinations of the numbers or expressions. Mostly the algebraic equations are used for evaluating the the value of the variables.


Learning Examples of intermediate algebra:


Example 1:

4x - 6 = 2x – 4

4x-6+6 =2x-4+6 (adding common value(6) in both sides)

4x      = 2x +2 (we get)

4x -2x = 2x -2x +2 (adding common value(-2x) in both sides)

2x   = 2 (we get)

2x / 2  =  2/2 (dividing common value (2) in both sides)

X  =  1 (We get x value)

Example 2:

6n +15 = 150

6n+15-15 =150-15 (adding common value(-15) in both sides)

6n = 135 (we get )

6n / 6 = 135 / 6 ( dividing common value (6) in both sides)

n = 22.5 (We get n value).

Thursday, March 14, 2013

Scale Factor in Algebra

A scale factor is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½


Explanation of Scale factor in algebra:


the following table will explain the scale factor in algebra some example given below

                            LENGTH           AREA              VOLUME

Scale Factor X      X times            X2 times              X3 times

Scale Factor 2      2 times             4 times                8 times

Scale Factor 3      3 times             9 times                27 times

Scale Factor 4      4 times             16 times              64 times

Scale Factor 5      5 times             25 times              125 times

For example in a square having 6cm all its sides. If the same squares length is reduced to 3cm then the scale factor of the square is 1/2.


Finding Scale factor in algebra:


Some examples for scale factor in algebra.

Ex : Consider the equation C= YZ, Y was the scale factor or Z, Y is also a coefficient of Z and may be called as constant of proportionality of a to Z. By using scale factor, you are creation a smaller model of something that is quite big.

In these algebra examples we discuss the scale factor

For the given problem Y = (4) (3).

4 was the scale factor of 3

4 is also a coefficient of 3

Ex :

                         squaresquare

                    square1                                            square2

In the example  square1 value of each side is 4 cm. the square2 value of each side is 8 cm. so we find the scale factor of this problem?

Sol : This example we find the scale factor of two square figures.

The second square value by first square value

Scale factor =8/4

                   =2

Ex :  The two triangle are in the same ratio. The first triangle area is 9 sq. units and  the second triangle area is 18 sq. units. Find the scale factor of the two triangle.

Sol : For the first triangle the  area is 9 sq. units.

 For the second triangle the area is 18 sq. units.

The scale factor = second triangle area /first triangle area

        = 18/9

         =2

answer for this problem = 2

Tuesday, March 12, 2013

Equation

Equation is one of the major concepts in mathematics. Taught in middle school, equation plays an important role throughout the mathematical years. Equation can be classified into algebraic equation, linear equation, quadratic equation and so on. Let’s have a basic overview of equation and formulas in this post.

Equation in simple terms defines the equality of two things. Equation includes the equal sign to define this equality. For example: X + 2 child phone for kids = 7 child phone for kids. Here, the equation is demonstrating that the things in left side are equal to the things in right side. Solving equations in mathematics is like solving a puzzle. A number is missing and one needs to find out the missing value.

For example: 2X + 6 crib toys = 12 crib toys = 2X = 12 – 6 = X = 6 / 2 = 3

Therefore, X is equal to 3.
Formula on the other hand in mathematics means a type of equation that defines the relationship between two variables. There are different formulas used in mathematics to solve equations. For example: Formula for finding volume of a box, area of a circle, radius of a circle and so on.

Example 1: Maya bought a box that includes the best flashlight available in the market. The height of the box is 4, width is 5 and length is 10. Now find the volume of the box including the best flashlight:

Applying the formula of finding the volume of a box i.e, V = h*w*l, we get
Volume of box = 4 * 5 *10 = 200.

Example 2: Arpita got a rectangular box. The length of the box is 4 and the breadth of the box is 8. Find the volume of the rectangle.

Applying the formula of finding the volume of a rectangle i.e, V = 2 (l + b),
we get Volume of rectangle = 2 (4 + 8) = 24
These are the basics about equations and formulas in mathematics.