Inverting Matrices

The inverse of a square matrix A, sometimes called a reciprocal matrix, is a matrix   such that         

Where I is the identity matrix.

A square matrix A has an inverse if the determinant . A matrix possessing an inverse is called nonsingular, or invertible.

The matrix inverse of a square matrix m may be taken in mathematics using the function inverse[m].

Inverse of 2x2 Matrix

For a 2X2  matrix

A= `[[a,b],[c,d]]`

The matrix inverse is

                  

             =  

                   =    

Inverse of 3x3 Matrix

For a 3X3 matrix

          

The matrix inverse is

A general matrix can be inverted using methods such as the gauss Jordan elimination, Gaussian elimination.

Prism Solve for h

Prism is one of the three dimensional shape. Normally the prisms faces are flat surface. Here we have two bases, one is at the top and another is at the bottom. In prisms we have two kinds of prisms; they are rectangular prism and triangular prism. When a prism bottom is rectangular shape then it’s denoted as rectangular prism. Cube and cuboids are some examples are rectangular prism. When a prism bottom is triangular shape then its said to be triangular prism. 

The diagram given below the structure of prism:

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Description to prism solve for h:

The following types are used in the prism shape they are

Volume of the regular prism is = BH cubic units,

Where B represents the base area of shape, and H represents height.

The Lateral surface area of the regular prism is given by,

              L.S.A = PH square units.

Where P represents the perimeter of shape and H represents the height of prism

Prism based on rectangular:

The area of the base is calculated by l*w

The perimeter of the base can be calculated by 2(l+w)

Therefore the total surface area is calculated by 2lw+2(l+w)H

Volume of the rectangular prism is V= L * W * H

Triangular based prism:

The area is measured as 1/2 bh

The perimeter of the base is measured by a+b+c.

So the surface are is measured by  bh+ (a+b+c) H

Circular based prism:

The area of the base is calculated as `pi` r²

The perimeter of the base is calculated by 2`pi` r

Therefore the surface area of the prism is 2`pi` r²+2`pi` rH

Let we see some basic problems regarding prisms.

Example 1:

 The dimensions of the rectangular prism is given below ,they are length is 10cm, height is 4 cm, and width is 14 cm.

Solution:

Given dimensions length = 10 cm, height = 4 cm, width = 14 cm.

Formula:

  Surface area = 2lw+2(l+w)H

apply  the values in the above formula, we get

                       = 2 *10*14+2(10+14)4

                       = 240 + 2(24)4

                      = 240 + 48*4

                     = 240 +192

                     = 432 cm²

      Volume = l * w * h  

                      = 10 * 4 * 14 cm³

                      = 560 cm³

Answer:

Surface area = 432 cm²

Volume = 560 cm³

Examples problems to solve h in prism

Example 1:

solve the h value whose prism volume is 1210 cubic units and base is 21 cm.

Solution:

The volume of the prism is V = 1210cubic units.

Base of the prism is = 21cm

                                   V= BH

apply the values in the formula we get

                              1210 = 121*H

                                     h = 1210 /121

                                     h = 10cm.

Example2:

The volume of the  rectangular prism is 360 cm³ and its length and widths are 10 ,4 cm , solve the h value

Solution:

The volume of the rectangular prism is V = 360cm³

Length of the rectangular prism is l = 10cm

Width of the rectangular prism is W = 4 cm

Apply these values in formula we get

                                     360 = 10*4*h

                                      360 = 40 *h

                                     h = 360 / 40

                                    h = 9 cm

Sample problems:

1) The volume of a regular prism is  120cm³.The base of the regular prism is 10 cm , solve its height h?

solution : height h = 12 cm.

2) The volume of the rectangular prism is 420cm3 and its length = 7 cm and width =6 cm and solve its height h?

solution: height h = 10 cm.

Conic Tutorial

The sections of a cone are  ellipses, Circles, parabolas and hyperbolas. Because these diagrams can be obtained by passing a plain through vertex of the double-napped cone. In this conic tutorial we are going to see the sections of the cone.

Circle - conic tutorial:
The circle in the section of a cone is like ellipse. When two foci are coincided to each other the circle is formed. This conic section can be formed when a circular cone is intersected with a perpendicular plane.

Sections of a cone - Parabolas - conic tutorial:

One of the section of the cone is the parabola. The points will not be on the same line but in the plain are equal from a fixed line and in a fixed point.

Solve Maxima and Minima:
The local minimum and the local maximum value must be known for finding the maxima and minima values.

Ellipse - conic tutorial:
The section of a cone is an ellipse. It is from a constant two fixed point with the sum of its focus of all the points. The concept of an ellipse makes different terminologies like focus of ellipse, loci of ellipse, etc.

Equation of ellipse:
Equation of ellipse is `(x - h)^2 / a^2 ` + `(y - k)^2 / b^2` = 1 . here real numbers are h, k, a and b.

Sections of a cone-General Equation of an Ellipse

The points of locus will be F ( p, q ) and the equation of directrix ax + by + c = 0. This is the general equation of ellipse

Theorem on Line contact with an Ellipse:
The condition for the line y = m x + c to touch the ellipse `(x^2/a^2)+(y^2/b^2) ` = 1 is that c = ±  `sqrt(a^2 . m^2 + b^2)` .

Finding the point of contact for Ellipse: 
The tangent y = m x + `sqrt(a^2 . m^2 + h^2)`  touches the ellipse `(x^2/a^2)+(y^2/b^2) `= 1, at the point `[(-a^2m/sqrt(a^2m^2+b^2) , b^2/sqrt(a^2m^2+b^2)]`

Hyperbola - conic tutorial:

A hyperbola is the section of a cone. A conic which an eccentricity greater than 1. It has two branches extending up to infinity, two foci, two directions, two vertices and two axes. ( i . e . )  transverse and conjugate and has a center which is the midpoint of the line of two foci and perpendicular conjugate axis is the line through the center and to the transverse axis.

Two Non Intersecting Lines

The two lines are concurrent means then the pencil is formed by the set of lines passing in their common point: if they are parallel, by the set of parallels to both, in the same direction means, they are non-intersecting lines. In fact, if the lines meet in the ordinary sense of the term or are parallel, the point can be at once obtained. These are called non-intersecting lines.

Example problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
     Y = x – 5 and y = x + 5.
     Two non intersecting lines graph for the following equation:
     Y = x - 5
Solution:
   Now plug x values for 0, 1, 2, 3, 4
   When x = 0                                                 
   Y = (0) -5
   Y = 0-5
   Y = -5

   The co ordinate’s point is (0,-5).
   When we plug x = 1
   We get
   Y = 1 -5
   Y = -4

   The co ordinate’s point is (1, -4).
   When we plug x = 2
   We get
   Y = 2 -5
   Y = 2-5
   Y = -3

   The co ordinate’s point is (2,-3).
    When we plug x = 3
    We get
    Y = 3 -5
    Y = -2

   The co ordinate’s point is (3, -2).
    When we plug x = 4
    We get
    Y = 4-5
    Y = -1
The co ordinate’s point is (4, -1).

X	0	1	2	3	4
Y	-5	-4	-3	-2	-1

Two non intersecting lines graph for the following another equation:
   Y = x + 5
Solution:
    Now plug x values for 0, 1, 2, 3, 4
    When x = 0                                                 
    Y = 0+5
    Y = 0+5
    Y = 5

    The co ordinate’s point is (0, 5).
    When we plug x = 1
    We get
     Y = 1 +5
     Y = 6

    The co ordinate’s point is (1, 6).
     When we plug x = 2
     We get
     Y = 2 +5
     Y = 7

     The co ordinate’s point is (2,7).
     When we plug x = 3
     We get
     Y = 3 +5
     Y = 8

     The co ordinate’s point is (3, 8).
     When we plug x = 4
     We get
     Y = 4+5
     Y = 9

     The co ordinate’s point is (4, 9).

X	0	1	2	3	4
Y	5	6	7	8	9

The non intersecting lines represent in graph is,

Practice problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
    Y = x – 4 and y = x + 4.
2. Find the equation for two non intersecting lines that passes through the two points:
    Y = x + 8 and y = x - 8.
3. Find the equation for two non intersecting lines that passes through the two points:

    Y = x + 2 and y = x - 2.

Trinomial Notes

Trinomial notes is a polynomial with three monomial terms. In trinomial notes, sum of three monomials is said to be  trinomial. For example, 2x5 – 3x2 + 3 is a trinomial . An Algebraic expression of the form axn is called a monomial in x, The sum of two monomials is called a binomial and the sum of three monomials is called a trinomial.  sum of the finite number of a monomials in term x is called a polynomial of x.

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How To Factor trinomial Notes:

Example 1:
Factor   trinomial notes x² + 3x − 10.

Solution.
The binomial factors will have this form:
(x   a)(x   b)
What are the factors of 10?  Let us take that they are 2 and 5:

x² + 3x − 10 = (x   2)(x   5).

We have to know how to choose the signs so that,  coefficient for middle term  will be sum of the outers plus the inners -- will be +3.                                  Choose factors , +5 and −2.

=  x² + 3x − 10

= (x − 2)(x + 5)
When 1 is the coefficient for x², the order of  factors it does not matter.

(x − 2)(x + 5) = (x + 5) (x − 2) are same factors.

Example Problems In trinomial notes:

Example 1:
Find factors for trinomial notes x² − x − 12.

Solution:
We must finding factors for 12 whose algebraic sum will be  coefficient of x is −1.
Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3). are facors

Example 2:

Find factors for trinomial notes    x2 + 5xy + 4y2
Solution:
x2+5xy+4y2

=x2+xy+4xy+4y2

=x(x+y)+4y(x+y)

=(x+y)(x+4y)

Example 3:

Find factors for trinomial notes    2a2 - 7ab + 6b2
Solution:
2a2-7ab+6b2
=2a2-3ab-4ab+6b2

=a(2a-3b)-2b(2a+3b)

=(2a-3b)(a-2b)

Example 4:

Find factors for trinomial notes 4x2 - 13xy + 3y2
Solution:
4x2-13xy+3y2
=4x2-xy-12xy+3y2

=x(4x-y)-3y(4x-y)

=(4x-y)(x-3y)

Acute Angles Pictures

Here we are going to see about the introduction to angles. The angle is referred as a figure which is formed by distribution of  two rays with a common point. This common point is referred as end point. The word angle is from the Latin word angulus which defines the point in corner. There are many types of angles in this we are going to see about the acute angles.

The acute angle is the type of angle which measures the angle between 0 to 90 degree and less than the 90 degree. The normal picture of the acute angle is given as,

Pictures of acute angle:
            The acute angles are referred by using the pictures which measures the angle less then 90 degree. The pictures of the acute angles are given below as examples.
Picture 1:

This is the picture of acute angle which defines that the angle measured in this picture is 48⁰ which is less than 90 degree.

Picture 2:

This is other picture of acute angle which defines that the angle measured in this picture is 28⁰ which is less than 90 degree.

Picture 3:

This is the picture of acute angle which defines that the angle measured in this picture is 37⁰ which is less than 90 degree.

Picture 4:

This is also a picture of acute angle which defines that the angle measured in this picture is 63⁰ which is less than 90 degree.

Picture 5:

This is also a picture of acute angle which defines that the angle measured in this picture is 49⁰ which is less than 90 degree.

Picture 6:

This is also a picture of acute angle which defines that the angle measured in this picture is 72⁰ which is less than 90 degree.

Example problems

Problem 1:
When the angle a-180⁰ is an acute angle. what is the value of a?
Options:
a) 240⁰
b) 300⁰
c) 290⁰
d) 320⁰

Solution:
Acute angle is the angle which measures less than 90 degree or 90 degree.
When 240⁰-180⁰ = 60⁰.
Hence the value of a is 60⁰.

Problem 2:
When the angle y - 90⁰ is an acute angle. what is the maximum value of y?
Options:
a) 170⁰
b) 200⁰
c) 100⁰
d) 540⁰

Solution:
Since the acute angle measures less than 90 degree. The maximum value of y is 170⁰.
Where 200⁰- 90⁰ = 110⁰, since it is greater than 90 degree this is not an acute angle.  
100⁰- 90⁰ = 10⁰, since it is less than 90 degree this is an acute angle but it is the minimum value.

540⁰- 90⁰ = 450⁰, since it is greater than 90 degree this is not an acute angle.  
170⁰- 90⁰ = 80⁰, since it is less than 90 degree this is an acute angle.    

Finding Volume in Math

Finding volume in math article deals with the definition of volume and the formula for the various shapes and the model problems related to volume of various shapes.

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Definition of Volume:

Volume is always measured in cube.units. It is defined as the space that is occupied by the entire three-dimensional shape.

Volume formula for various shapes in math.

Volume of the cube is a3 cubic units

Here “a” is the side of the cube.

Volume of the rectangular prism is L*B*H cubic units

Here L is length, B is breadth and H is the height.

Volume of the cylinder is BH cubic units

Here B is the base area and H is the height

Volume of the pyramid is `(1/3)` BH cubic units.

Here B is the base area and H is the height

Volume of the cone is `(1/3) ` BH cubic units.

Here B is the base area and H is the height

Model problem for finding volume in math.

Problem:

1.  Finding volume of the cube when the side measures about 5 cm in length.

Solution:

Since it is a cube, the formula

Volume of the cube is a3 cubic units

Here “a” is the side of the cube.

Here a= 5cm

The volume of the cube =53

=125 cm3

The volume of the cube is125cm3

2. Finding volume of the rectangular prism when the length = 8cm, breadth = 5cm and the height is 4cm.

Solution:

Since it is a rectangular prism, the formula

Volume of the rectangular prism is L*B*H cubic units

Here L is length, B is breadth and H is the height
L= 8cm

B= 5cm

H = 4cm

Volume of the rectangular prism = (8* 5*4) cubic units.

= 160 cm3

Volume of the trectangular prism is 160cm3

3. Find the volume of the cylinder whose base area is 46m^2 and the height of the cylinder is 10m

Solution:

Base area of the cylinder is 46m2

Height of the cylinder is 10m

Formula:

Volume of the cylinder is BH cubic units

Here B is the base area and H is the height

= 46*10

= 460 m3

the volume of the cylinder is 460m3