Polynomial Matrices

A polynomial matrices or matrix polynomial is a  matrix whose elements are univariate or multivariate polynomials.A univariate polynomial matrix P of degree p is define like:sum_(n=0)^pA(n)x^(n)=A(0)+A(1)x+A(2)x^(2)+....+A(p)x^(p) where A(p) is non-zero and A(i) indicate a matrix of constant coefficients. Hence a polynomial matrix is the matrix-equivalent of a polynomial, by means of every one element of the matrix satisfying the classification of a polynomial of degree p.

Properties of polynomial matrices

A polynomial matrix in excess of a field with determinant equivalent to a non-zero constant is called unimodular, and have an inverse, which is also a polynomial matrix.
Note, that the simply scalar unimodular polynomials are polynomials of degree 0 - nonzero constants, for the reason that an inverse of an arbitrary polynomial of high degree is a rational function.
The roots of a polynomial matrix in excess of the complex numbers are the points in the complex plane wherever the matrix loses rank.

Characteristic polynomial of a product of two matrices

If A and B are two square n×n matrices then,attribute polynomials of AB and BA match:

PAB(t)=PBA(t).

If A is m×n-matrix and B is n×m matrices such that m
PAB(t)=tn-mPAB(t)

Polynomial in t and in the entry of A and B is a general polynomial identity. It consequently suffice to verify it on an open set ofparameter value in the complex numbers.

The tuples (A,B,t) wherever A is an invertible complex n by n matrix,

B is any complex n by n matrix,

and t is any complex number since an open set in complex space of dimension 2n2 + 1. When A is non-singular our result follow from the fact that AB and BA are similar:

BA=A-1(AB)A.

Example 1:

An example the 3x3 polynomial matrices

P=[[1,x^(2),x],[0,2x,2],[8x+2,x^2-1,0]]

=[[1,0,0],[0,0,2],[2,-1,0]]+[[0,0,1],[0,2,0],[3,0,0]]x+[[0,1,0],[0,0,0],[0,1,0]]xx^(2)

Example 2:

Find the eign value of given polynomial matrices

P=[[3,3],[0,6]]

The polynomial has the characteristic equation

0=det(P-λI)

=det[[3-lambda,3],[0,6-lambda]]

18-6lambda -3lambda + λ2

18-9lambda +18

λ2-3λ-6λ+18

λ(λ-3)-6(λ-3)

(λ-3)(λ-6)

λ=3,andλ=6

The eigenvalues of these matrices are 3,6

Example 3: Find the product of the given matrices M1=[[1,2],[3,4]] and M2=[[8,3],[2,7]]

The given polynomial is

M1=[[1,2],[3,4]]


M2=[[8,3],[2,7]]

The product of the given matrices M1 and M2  =M1xM2

M1xM2    =[[1,2],[3,4]]xx [[8,3],[2,7]]

The product of the given  matrices is=[[12,17],[32,37]]

Least Common Denominator

If the denominators are unlike, then we can find LCD (Least common Denominator) of the given denominators.

Least Common Denominator is the smallest positive(least) integer which is common in multiples of the denominators.

For example, given fractions are 1/3 and 1/6. Find LCD.

List the multiples of 3:   3, 6, 9, 12, 15, 18, 21,...

Multiples of 6:   6, 12, 18, 24,...

Here, 6 is the lowest common term for both the multiples of 3 and multiples of 6.

The answer is 6, and that is the Least Common Denominator.

There are five examples for least common denominator. From these examples for least common denominator, we can get clear view about least common denominator. Let us see the examples for least common denominator in the following section.

Examples on examples for least common denominator:

We are going to explain examples for least common denominator.

Example 1:

Find the least common denominator of the fractions; 1/5,1/3

Solution:

Here, the denominators are 5 and 3.

The common denominator of 5 and 3 is 15.

Multiples of 5: 5,10,15,20,…..

Multiples of 3: 3,6,9,12,15,18,….

Here, 15 is the lowest common term for both the multiples of 5 and multiples of 3.

The answer is 15, and that is the Least Common Denominator.

Example 2:

If the given fractions are 3/4,1/3, find the least common denominator.

Solution:

Here, the denominators are 4 and 3.

The common denominator of 4 and 3 is 12.

Multiples of 4: 4,8,12,16,20….

Multiples of 3: 3,6,9,12,15….

Here, 12 is the lowest common term for both the multiples of 4 and multiples of 3.

So, the Least Common Denominator is 12.

Example 3:

Find the least common denominator of ; 5/6, 2/15

Solution:

Here, the denominators are 6 and 15.

The common denominator of 6 and 15 is 30.

Multiples of 6:  6,12,18,24,30…

Multiples of 15: 15,30,45,60….

Here, 30 is the lowest common term for both the multiples of 6 and multiples of 15.

So, the Least common denominator is 30.

Example 4 on examples for least common denominator:

What is the least common denominator of the fractions;  5/12, 11/18

Solution:

Here, the denominators are 12 and 18.

The common denominator of 12 and 18 is 36.

Multiples of 12:  12,24,36,48,….

Multiples of 18: 18,36,54,….

Here, 36 is the lowest common term for both the multiples of 12 and multiples of 18.

So, the Least common denominator is 36.

Example 5:

1/5 + 1/6 + 1/15  What is the LCD?

Solution:

First we list the multiples of each denominator.

Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...

Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...

Multiples of 15 are 30, 45, 60, 75, 90,....

Here, 30 is the lowest common term for both the multiples of 5 and multiples of 6 and multiples of 15 .

So, the Least common denominator is 30.

Therefore, Examples for least common denominator are explained.

Learning Parallel Lines

Learning parallel lines is very easy. As the name it self indicates that there will be some lines that are parallel to each other. Think logically if two lines are parallel will they ever meet at some point. The answer is no. Because the two parallel lines maintain equal distance between them so there is no chance that they meet at some point. So we can conclude that two lines are parallel if and only if they are coplanar and never meet each other that is they maintain same distance apart.

Transversals are very important while learning parallel lines.

Tranversal can be defined as a line that intersects two or more  lines, at different points. Simply a line that crosses two or more lines  is called transversal. With the help of transversals we can say whether the two given lines are parallel or not. Figure below shows how a transversal looks.



How can we know the lines are parallel or not?

Answer is with the help of some angles that are formed when transversal passes through pair of coplanar lines. These angles are very important while learning parallel lines.The angles are

Corresponding Angles
Alternate Interior Angles
Alternate Exterior Angles
Co-Interior Angles


Corresponding angles:

In our figure the corresponding angles are " a,e " , " d, h" , "b , f" , "c,g".

Alternate Interior angles:

The name it self indicates that alternate means on the other side and interior means inner angles. The Alternate Interior angles are "d,f" and "c,e".

Alternate exterior angles:


The name it self indicates that alternate means on the other side and exterior means outer angles. Alternate exterior angles in the given figure are "a,g" and "b,h".

Co-interior angles:

In the given figure Co-interior interior angles are "d,e" and "c,f".

Now for two lines to be parallel corresponding angles, alternate Interior angles, alternate exterior angles must be equal and sum of Co-interior  angles must be equal to 180. Even if one condition is satisfied it is enough as automatically all the other will satisfy.

This is all about learning parallel lines. Hope you enjoyed it......

Definition of Probability Distribution

Probability distribution function referred as p(x) which is a function that satisfies below properties:

Probability that x can obtain particular value is p(x).

p[X,x]=p(x)=px

p(x) positive for all real x.
Sum of p(x) over all feasible values of x is 1, that is

`sum_(pj)pj=1`

Where, j- all possible values that x can have and pj is probability at xj.

One significance of properties 2,3 is that 0 <= p(x) <= 1.

Steps used for how to determine probability distribution:

Using following steps used to how determine probability distribution:

Step 1 for how to determine probability distribution:

Plot the data for a image illustration of the data type.

Step 2 for how to determine probability distribution:

First steps is to determining what data distribution one has - and thus the equation type to use to model the data - is to rule out what it cannot be.

The data sets cannot be a discrete uniform distribution if there are any crest in the data set.
The data is not Poisson or binomial if the data has more than one crest.
If it contains a single arc, no secondary crests, and contains a slow slope on each side, it may be Poisson or a gamma distribution. But it is not discrete uniform distribution.
If the data is regularly distributed, and it is without a slant in the direction of one side, it is secure to rule out a gamma or Weibull distribution.
If the function has an even distribution or a crest in the center of the graphed outcomes, it is not a geometric distribution or an exponential distribution.
If the incidence of a factor differ with an environmental variable, it probably is not a Poisson distribution.


Step 3 for how to determine probability distribution:

After probability distribution type has been tapering downward, do an R squared examination of each probable type of probability distribution. The one with the maximum R squared value is most possible correct.

Step 4 for how to determine probability distribution:

Remove one outlier data point. Now recalculate R squared. If the same probability distribution form comes up as the neighboring match, then there is high confidence that this is the correct probability distribution to use for group of data.

From above steps we can understand how to use probability distribution.

Polynomial Exponent

Polynomial of a single term is called as Monomial. If the Polynomial has the two terms Sum or Difference then it is called a Bi-nomial, the Sums or Differences of a three-term Polynomial is called a Trinomial. The Sums and/or Differences of polynomial of four or more terms are simply called polynomial.

Two or more terms of a Polynomial exponent that has the equivalent variable and precisely the same whole number polynomial exponent are called Like Terms. For suitable example the terms; 6X3 and -7X3, are Like Terms, since the variable X is the similar variable in both terms, and the Whole number exponent, 3, is absolutely the similar polynomial exponent in both terms. The terms;4X3 and 4X4 are not Like Terms although the variable X is the same in both terms but the Whole number polynomial exponent are completely different from other.

Operation of polynomial exponent:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

(2x2+3x3)+(x2+7x3)

=2x2+x2+3x3+7x3

The variable and exponent must be same then we add the polynomial exponent,

=3x2+10x3

So the result is  =3x2+10x3

Subtraction of polynomial exponent

Example2:

(3x2+3x3)-(x2+7x3)

=3x2-x2+3x3-7x3

The variable and exponent must be same then we subtract the polynomial exponent,

=2x2-4x3

So the result is =2x2-4x3

Multiplication of polynomial exponent:

Two or more polynomial, so multiply their exponent terms. For reasonable example to multiply the following two Terms, 2X2 and 7X2, Then the following terms then first multiply the coefficients (2) (7) and we multiply (X2) (X2) which can be expressed as the following term 14X4, that is, when we multiply variables that are the same.


Different variables in polynomial exponent:

The variables are different and the exponent also different, so we can write the variables and exponents adjoining to each other.

For Example, (7X2)(3Y2)is equal to 21X2Y2.

So the result is 21X2Y2

Similar to all the different variable, polynomial exponent.

To multiply the two polynomial exponent, for example (X-2Y)(X-2Y), then the result is x2-2xy+4y2 . Here we take the first term X of the first Binomial and multiply each term in the second Binomial, then we take the second term in the first Binomial and multiply each term in the second Binomial, and then we add all the terms. Similar to all the polynomial categories, for example trinomial, binomial exponents.

Inverse Trigonometry Definition

The three trigonometry functions are arcsin(x), arccos(x), arctan(x). The inverse trigonometry function is the inverse functions of the trigonometry, written as cos^-1 x , cot^-1 x , csc^-1 x , sec^-1 x , sin^-1 x , and tan^-1 x .

Sine:

H (x) = sin(x)   where   x is in [-pi/2 , pi/2]

Cosine:

G (x) = cos(x) where   x is in [0 , pi ]

Tangent:

F (x) = tan(x)   where   x is in (-pi/2 , pi/2 )

Understanding Inverse Trig Function is always challenging for me but thanks to all math help websites to help me out.

Cosecant Definition

Csc theta = (hypotenuse)/(opposite)

Secant Definition

Sec theta = (hypotenuse)/(adjacent)

Cotangent Definition

Cot theta = (adjacent)/(opposite)

Examples of Inverse trigonometry functions:

Example 1:

Find the angle of x in the below diagram. Give the answer in four decimal points.

Solution:

sin x =2.3/8.15

x = sin -1(2.3/8.15 )

= 16.3921˚

Example 2:

Solve sin(cos-1 x )

Solution:
Let z = sin ( cos-1 x ) and y = cos-1 x so that z = sin  y. y = cos-1 x may also be marked as


cos y = x with pi / 2 <= y <=- pi / 2

Also

sin2y + cos2y = 1

Substitute cos y by x and solve for cos y to obtain

sin y = + or - sqrt (1 - x^2)

But pi / 2  <= y <= - pi / 2 so that cos y is positive

z = sin y = sin(cos-1 x) = sqrt (1 - x^2)

Example 3:

Evaluate the following sin-1( cos ((7 pi) / 4 ))

Solution:
sin-1( cos ( y ) ) = y only for  pi / 2 <=  y <= -pi / 2 . So we initial transform the given expression noting that cos ((7pi) / 4 ) = cos (-pi / 4 ) as follows

sin-1( cos ((7 pi) / 4 )) = sin-1( cos ( pi / 4 ))  - pi / 4  was selected since it satisfies the condition  pi / 2 <=  y <= - pi / 2 . Hence

sin-1( cos ((7 pi) / 4 )) =pi/4

Example 4:

With the given value find inverse of tan.

Solution:

Since tangent, again, is the trig function associated with opposite and adjacent sides, we apply the inverse of tangent to calculate the measure of this angle.

Tan 16 = (opposite)/(adjacent)

Tan 16 = ((bar(AC))/(bar(CB)))

Tan-1(7/24 ) = 16 o

Associative Property Sum

In mathematics, if x, y, z be any three variables involving the addition operation and then satisfy the following condition,

x + (y + z) = (x + y) + z

This kind of property is called associative property sum or addition. In associative property sum, the sum of algebraic expression provides the same result in changing the order of brackets in this expression. Such as

x + y + z = (x + y) + z = x + (y + z)



Examples for associative property sum:

Example 1 on associative property sum:

Prove the associative property sum of the given expression.

(a) 4 + 6 + 7

Solution:

(a) Let x, y, and z be 4, 6 and 7 respectively.

Therefore, the given expression satisfy the following condition,

x + y + z = x + (y + z) = (x + y) + z

Here,

Prove the following expression

4 + (6 + 7) = (4 + 6) + 7

L.H.S = 4 + (6 + 7)        (first perform the addition operation within the bracket from left to right,

because brackets are higher priority of the order of operations)

= 4 + (13)      (second perform the addition operation outside of the expression from left to right)

= 17                    ------ (1)

R.H.S = (4 + 6) + 7

= (10) + 7

= 17             --------- (2)

From equation (1) and (2) we get,

L.H.S = R.H.S

i.e., 4 + (6 + 7) = (4 + 6) + 7

Hence, the given expressions satisfy the associative property with sum of numbers.

Example 2 on associative property sum:

(a) 23 + 67 + 34, prove the associative property of sum or addition.

Solution:

(a) Let m, n, and l be 23, 67 and 34 respectively.

Therefore, the given expression satisfy the following condition,

m + n + l = m + (n + l) = (m + n) + l

Here,

Prove the following expression

23 + 67 + 34 = 23 + (67 + 34) = (23 + 67) + 34

23 + 67 + 34 = 124       ---------------- (1)

We consider the remaining expression, such as

23 + (67 + 34) = (23 + 67) + 34

L.H.S = 23 + (67 + 34)

= 23 + (101)

= 124                    ------ (2)

R.H.S = (23 + 67) + 34

= (90) + 34

= 124                  --------- (2)

from equation (1) , (2) and (3) we get,

L.H.S = R.H.S = 124

i.e.,  4 + (6 + 7) = (4 + 6) + 7 = 4 + 6 + 7

Hence, the given expression of sum is satisfied the associative property.