Types of number

There are various types of number as follow

1. Natural Number

Nature number are counting number and these are denoted by  $\mathbb{N}$ ,

i.e.,             $\mathbb{N}=\left&space;\{&space;1,2,3,.....&space;\right&space;\}.$

* All natural number are positive.

* Zero is not a natural number, therefore 1 is the smallest natural number.

2. Whole number

All natural number and zero form the set of whole number and these are denoted by W,

i.e,                $W=\left&space;\{&space;0,1,2,3,....&space;\right&space;\}.$

* Zero is the smallest whole number

* Whole number are also called as non-negative integers.

3. Integers

Whole number and negative number form the set of integers and there are denoted by $\mathbb{I}$,

i.e.,         $\mathbb{I}=\left&space;\{&space;......,-4,-3,-2,-1,0,1,2,3,4,......&space;\right&space;\}.$

Integers are of following two types

1. Positive integers:- Natural numbers are called as positive integers and these are denoted by  $\mathbb{I}^{+}$,

i.e.,        $\mathbb{I}^{+}=\left&space;\{&space;1,2,3,4,.......&space;\right&space;\}.$

2. Negative integers:- Negative of natural number are called as negative integers and these are denoted by  $\mathbb{I}^{-}$

i.e.,             $\mathbb{I}^{-}=\left&space;\{&space;-1,-2,-3,-4,.....&space;\right&space;\}.$

* 0 is neither +ve nor -ve integers.

4. Even number

A counting number which is divisible by 2, is called an even number

For example, 2,4,6,8,10,12,....etc.

* The unit's place of every even number will be 0,2,4,6, or 8.

5. Odd number

A counting number which is not divisible by 2, is called an odd number

For example, 1,3,5,7,9,11,13,15,17,....etc.

* The unit's place of every odd number will be 1,3,5,7 or 9.

6. Prime number

A counting number is called a prime number when it is exactly divisible by only 1 and itself.

For example 2,3,5,7,11,13,.... etc.

* 2 is the only even number which is prime.

* A prime number is always greater then 1.

* 1 is not a prime number therefore the lowest odd prime number is 3.

* Every prime number greater than 3 can be represented by 6n+1, where n is integer.

7. Composite number

Composite number are non-prime natural number. They must have atleast one factor apart from 1 and itself.

For example 4,6,8,9,..etc.

* composite number can be both odd and even.

* 1 is neither a prime number nor a composite number.

8. Coprime

Two natural number are said to be  coprime, if their common divisor is 1

For example (7,9), (15,16), etc.

* coprime number may or may not be prime.

* Every pair of consecutive number is coprime.

9. Rational number

A number that can be expressed in the form of p/q is called a rational number, where p,q are integers and  $q\neq&space;0$.

For example   $\frac{3}{5},\frac{7}{9},\frac{8}{9},\frac{13}{15},\:&space;\:&space;etc.$

10. Irrational number

The number that cannot be expressed i  form of p/q are called irrational number, where p,q are integers and $q\neq&space;0$.

For example   $\sqrt{2},\sqrt{3},\sqrt{7},\sqrt{11}\:&space;etc.$

*  $\pi$ is an irrational number as 22/7 is not the actual value of  $\pi$ but it is its nearest value.

* Non-periodic infinite decimal fractions are called irrational numbers.

11. Real numbers

Real number  include both rational and irrational number. They are denoted by $\mathbb{R}$.

For example      $\frac{7}{9},\sqrt{2},\sqrt{5},\pi&space;,\frac{8}{9},\:&space;\:&space;etc.$

Rank of a matrix

In linear algebra, the rank of a matrix is the dimension of its row space or column space. It is important to note that the row space and column space of a matrix have equal dimensions.

Definition of rank of a matrix

A nimber r is said to be a rank of a non-zero matrix A if

1. There exist at least one minor of order r of A which does not vanish and

2. Every minor of higher order than r is zero.

The rank of a matrix A is denoted by  $\rho&space;(A)$.
We have $\rho&space;(A)=r$

Another definition of rank of a matrix

The rank of a non-zero matrix is largest order of any non-vanishing minor of the matrix.

Remarks

From the above definition of rank of a matrix, we observe that

1. The rank of zero matrix is zero i.e.,  $\rho&space;(O)=0$ where O is a zero matrix,

2. The rank of a non-singular matrix of order n is n,

3.  $\rho&space;(A)\leq&space;r$, if every minor of order r+1 vanishes,

4.  $\rho&space;(A)\geq&space;r$, if there is a minor of order r which does not vanish.

☆Finding rank of matrix

1. Minor method

2. Normal form

3. Echelon form of matrix

1. Example on minor method

Q1. Determine the rank of matrix
$\begin{bmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&&space;5&space;&&space;2\\&space;12&-4&&space;5&space;\end{bmatrix}$
Ans

$A=\begin{bmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&&space;5&space;&&space;2\\&space;12&-4&&space;5&space;\end{bmatrix}$
$|A|=\begin{vmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&space;&&space;5&space;&&space;2\\&space;12&&space;-4&space;&5&space;\end{vmatrix}$
$=4\begin{vmatrix}&space;5&space;&&space;2\\&space;-4&space;&&space;5&space;\end{vmatrix}-2\begin{vmatrix}&space;8&space;&&space;2\\&space;12&&space;5&space;\end{vmatrix}+3\begin{vmatrix}&space;8&space;&&space;5\\&space;12&&space;-4&space;\end{vmatrix}$
$=4(25+8)-2(40-24)+3(-32-60)$
$=4(33)-2(16)+3(-92)$
$=132-32-276\neq&space;0$
$\therefore&space;\,&space;\:&space;\:&space;\:&space;\;&space;\;&space;\;&space;|A|\neq&space;0$
A is non singular
$\rho&space;(A)=3$

Q2. Determine the rank of matrix
$\begin{bmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&12&space;\\&space;2&space;&&space;4&space;&&space;8&space;\end{bmatrix}$
Ans.
$Let\:&space;\:&space;A=\begin{bmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&12&space;\\&space;2&space;&&space;4&space;&&space;8&space;\end{bmatrix}$
$|A|=\begin{vmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&&space;12\\&space;2&&space;4&space;&&space;8&space;\end{vmatrix}$
$=2\begin{vmatrix}&space;6&space;&&space;12\\&space;4&space;&&space;8&space;\end{vmatrix}-3\begin{vmatrix}&space;3&space;&&space;12\\&space;2&space;&&space;8&space;\end{vmatrix}+3\begin{vmatrix}&space;3&space;&&space;6\\&space;2&space;&&space;4&space;\end{vmatrix}$
$=2(48-48)-3(24-24)+3(12-12)$
$=0-0+0=0$
$|A|=0$ i.e, only minor of order 3 of A vanishes.
Now we consider any minor of order 2.

Consider   $\begin{vmatrix}&space;2&space;&&space;3\\&space;3&space;&&space;6&space;\end{vmatrix}=12-9=3\neq&space;0$
There is a minor of order 2 of A which does not vanish

$\rho&space;(A)=2$

2. Normal form of a matrix

$\begin{bmatrix}&space;I_{r}&space;&O&space;\\&space;O&space;&&space;O&space;\end{bmatrix},\:&space;\begin{bmatrix}&space;I_{r}&space;&&space;O&space;\end{bmatrix},\:&space;\begin{bmatrix}&space;I_{r}&space;\\&space;O&space;\end{bmatrix}$ are called the normal formes of matrix

Example

Q3. Prove that the matrix  $\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;2&space;&&space;3&space;&&space;0\\&space;0&space;&&space;1&space;&&space;2&space;\end{bmatrix}$ is equivalent to  $I_{3&space;}$.
Ans
$Let\:&space;\:&space;\:&space;\:&space;A=\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;2&space;&&space;3&space;&&space;0\\&space;0&space;&&space;1&space;&&space;2&space;\end{bmatrix}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;0&space;&&space;-1&space;&&space;-6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;R_{2}\rightarrow&space;R_{2}-2R_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;-1&space;&&space;-6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;C_{2}\rightarrow&space;C_{2}-2C_{1},\:&space;C_{3}\rightarrow&space;C_{3}-3C_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;R_{2}\rightarrow-&space;R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;6\\&space;0&&space;0&space;&&space;-4&space;\end{bmatrix},&space;by\:&space;\:&space;R_{3}\rightarrow&space;R_{3}-R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;0\\&space;0&&space;0&space;&&space;-4&space;\end{bmatrix},&space;by\:&space;\:&space;C_{3}\rightarrow&space;C_{3}-6C_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;0\\&space;0&&space;0&space;&&space;1&space;\end{bmatrix},&space;by\:&space;\:&space;R_{3}\rightarrow-\frac{1}{4}&space;R_{3}$
$A\sim&space;I_{3}$
Given matrix is equivalent to  $I_{3&space;}$.

3. Echelon form

A matrix  $A=\left&space;[&space;a_{ij}&space;\right&space;]$ is said to be in echelon form if

1. The zero rows of A occur below all the non-zero rows of A

2. The number of rows before the first non-zero element in a row is less than the number of such zero in the next row.

Example

Q4. Reduce to row echelon form the matrix
$A=\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;2&space;&&space;-4&space;&&space;3&space;&5&space;\\&space;-1&space;&&space;2&space;&&space;6&space;&&space;-7&space;\end{bmatrix}$
Ans.
$A=\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;2&space;&&space;-4&space;&&space;3&space;&5&space;\\&space;-1&space;&&space;2&space;&&space;6&space;&&space;-7&space;\end{bmatrix}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;5&space;&-3\\&space;0&space;&&space;0&space;&&space;5&space;&&space;-3&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{2}\rightarrow&space;R_{2}-2R_{1},\:&space;R_{3}\rightarrow&space;R_{3}+R_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;1&space;&-\frac{3}{5}\\&space;0&space;&&space;0&space;&&space;5&space;&&space;-3&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{2}\rightarrow&space;\frac{1}{5}R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;1&space;&-\frac{3}{5}\\&space;0&space;&&space;0&space;&&space;0&space;&&space;0&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{3}\rightarrow&space;R_{3}-5R_{2}$

Which is in row echelon form.
Since there are two non-zero rows in row echelon form.

$\rho&space;(A)=2$

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