Learning Elementary Linear Algebra

Introduction:

Linear algebra is one of the subtopics in algebra. Algebra is classified into many subtopics like linear, non linear, quadratic, inequalities, system of equations. Linear algebra is one of the important topics in algebra. Linear algebra deals with variables and numbers. Simple linear equation consists of one variables and a constant. In elementary level, simple linear algebra is followed. Linear equation means, the x axis varies linearly with respect to the y – axis.

Elementary Linear Algebra Larson Pdf:

Ron Larson, mathematician in the Penn State Erie, The Behrend College, Pennsylvania. He wrote lot of books in math for intermediate level and also for the college level. He is well known for his article publishing and text books. His books mainly deal with algebra and pre algebra. Larson pdf in elementary linear algebra is an excellent pdf for intermediate students. This elementary linear algebra pdf file provides many simple problems and answers in linear algebra. Larson pdf for elementary linear algebra is useful for learning basic linear algebra problems.

Example 1:
Solve the linear equation
x + 10 = 25
Solution:
x + 10 = 25
Subtract 10 on both sides,
X + 10 – 10 = 25 – 10
X = 15
The answer is 15.
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Explain radius and diameter of a circle

Introduction:

                A circle is a 2D image. Diameter of a circle circumference article deals with the definition of the diameter and the circumference and the model problem related to it. Circumference is defined as the distance around the circle. Both the diameter and the circumference are measured in the units (like cm, m) etc. The dimensions of the circle are radius and diameter. Diameter of the circle is the distance between two points on a circle which should passes through the center of a circle. Radius is same as the half of the diameter and it is a distance between a center of a circle and any point on the circle using formula for circumference of a circle.

Example Diagram for Radius and Diameter of a Circle:

Radius = diameter/2

Diameter = 2* radius.

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Square Root Formula of 180

Introduction:

           The square root is the radical form of the method that are used in the calculation mathematical problems. The symbol for the square root is meant by root √. This could be in the form to describe their nature of working with the square root formula. There are numerous methods are available in the rooting, they are square (second) root `sqrt(x)` , cube (Third) root `root(3)(x)` up to nth root `root(n)(x)` . Here we are going to see about the formula method to solve the square root. There are number of methods available to solve square root. Here we are using newton's method to solve the square root formula of 180 and the problem solved.

                                                        

Procedure for Square Root by Newton's Method:

• Form the equation from the given function and differentiate with respect to x.
• Assume the value for the first initial value for x as x₀.
• And substitute the x₀ value in the formula to find the value for x₁.
• Repeat the above step up to the required result for the given function we get.

Formula for calculating the square root using the newton's method:
      Let the equation for the given function is f(x). Find the first derivative for the equation.
                          `sum_(n=0)^N x_(n+1) ` = `sum_(n=0)^N ( x_n - f(x_n)/(f'(x_n))) `
            Assume an value for the initial value for x₀

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Fundamental Theorem of Algebra

Introduction:

         The Fundamental Theorem of Algebra is a for equation solving. It means that every polynomial equation over the field of complex numbers of degree higher than 1 has a complex solution. Polynomial equations are in the form

                                               P(x) = a_nxⁿ + a_n-1 x ^n-1 + ... + a₁x + a₀ = 0,
where an is assumed non-zero in which case n is called the degree of the polynomial P and of the equation above.

Fundamental Theorem of Algebra:

       A number a is a solution to the equation P(x) = 0 if substituting a for x makes it identity: P(a) = 0. The coefficients are assumed to belong to a exact set of numbers where we also look for a solution. The polynomial form is very general but frequently studying P(x) = Q(x) is more suitable.
      To see how it works let's start with the counting numbers (N=numbers 1,2,3,...), and the simplest equation             x + a = b. For example, x + 5 = 12 has a solution x = 12 - 5 = 7. Introduction of negative numbers eases the problem:
Any equation x + a = b where a, b belongs to N has a solution x belongs to Z, where Z is the set of integers numbers like plus, minus whole numbers and zero.
Once we accepted the negatives, we have a stronger result:
Any equation x + a = b where a, b belongs to Z has a solution x belongs to Z.
Even if the coefficients are permitted to be negative, the equation still has a solution in Z. Now let's consider other equations over Z: 11x + 11= 0, x= -11 /11 = -1. 
   Any equation ax + b = 0 where a,b belongs to z has a solution x belongs to Q.
Moreover, we again get a stronger result,
Any equation ax + b = 0 where a,b belongs to Q has a solution x belongs to Q.
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Trigonometric Complex Numbers with example

Introduction:
                   In mathematics, the trigonometric functions are also called as circular functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Trigonometric functions are the most important in the study of triangles and modeling periodic phenomena, among many other applications. Now let us discuss about the trigonometric complex numbers.

Problems on Trigonometric Complex Numbers:

Complex trigonometric functions
 Through Euler's formula we know that
 e^ix = Cos x + i Sin x   and hence, e^-ix = Cos x - i Sin x
 This gives us the two identities:
 Cos x = ½(e^ix + e^-ix)   and Sin x = ½(e^ix - e-^ix)
             Then complex trigonometric functions can be defined by analogy as: Cos z = ½(eiz + e-iz)  and Sin z = ½(eiz - e-iz);  while other complex trigonometric functions
1. Find Trigonometry complex   ( 3 + 5i) - (8 +3i)
Answer:          
               =(3 + 5i) - (8+ 3i)
               =(3 - 8) + (5i - 3i)
               =-5 + 2i

2. Solve the trigonometry complex triangle function of the sides a=8, b=5 and c=4 of the trigonometry area triangle
Solution:
             Trigonometry function of s= (`1/2` ) (a+b+c) =8.5
                Function area =√ [s(s-a) (s-b) (s-c)] = 9.045
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Trigonometric Functions of an Angle

Introduction of Trigonometric Functions of an Angle:
             The word trigonometry is derived from Greek meaning study of a triangle, a right triangle in particular.
Trigonometric function means a relation of the sides of a right triangle. The angle that is made by the hypotenuse with the base of a right triangle determines the ratio of different sides and hence it is important to study the trigonometric function of any angle.
With the rotation of the hypotenuse, the angle varies and hence the trigonometric function is also called as circular function.

Trigonometric Function of an Angle – Basic Concepts

            The basic trigonometric functions of any angle are Sine functions, Cosine functions and Tangent functions. The are abbreviated as sin, cos and tan respectively.

            The Sine function gives the ratio of the vertical component of the hypotenuse(rise) and the hypotenuse. The cosine function gives the ratio of the horizontal component of the hypotenuse(run) and the hypotenuse. The Tangent function gives the ratio of the vertical component of the hypotenuse (rise) and the horizontal component of the hypotenuse(run).
The reciprocal functions of Sine, Cosine and Tangent are called Cosecant, Secant and Cotangent functions respectively.