Category: Maths Notes PPT

Grouping of data is made easy by the use of frequency polygons:

When we are working on a particular thing, it is necessary to make sure that we record all the data that we have. This is going to be useful for the following reasons:

 

1. The data is going to be preserved and others who need it can use it at a later stage.

2. It is going to act as a check for the ones working on it to compare and contrast all the possible entries. 

 

Hence, recording all the data is very much important as such. There are many types in which one can choose to represent the data. Here are a few of the forms: 

 

1. Pictorial Representation: It is a general belief that people will remember things in the form of pictures very easily. Therefore, people have decided to represent the data in the form of pictures. This is one of the best methods but the major disadvantage is that pictures are going to consume a lot of time and the entire idea is going to take a lot of effort. 

 

2. Sheets: This representation of data wherever possible is one the best methods as the data can be retrieved easily. There is one major drawback with this form. All the data is in a clustered form and it is very difficult for the people to see to it that they are going to segregate the data. 

 

3. Graphical Representation: This is found to be one of the best forms of representation of data. This is used widely by many people all over the world. This is one the most efficient methods of representation as such.

 

Frequency Polygons

Frequency polygons are one type of graphical representation of data. There are many ways in which the data can be graphically represented and frequency polygons are the best and the most efficient of all.

 

The data that is present on a sheet of paper in the form of tables is sorted out and made in such a way that it can be plotted on a graph paper. This is going to be very much useful for the people who are reading the data. The graph so plotted is also represented in the form of bars so that it is clearer for the people to understand the statistics as such.

All this helps to understand the frequency distribution of the data in a much better way. 

 

What are the advantages of Frequency Polygons?

1. The frequency polygons not only help to make sure that the data is sorted out and represented, but they are also going to make it easier for the people to compare and contrast all the results. 

2. These are much easier to understand and they give a clear picture of the distribution of data. It is necessary that people should be able to read these graphs to understand them as such. 

3. This method is not even time taking like a few other processes. If the techniques that are used to draw and represent the data are known, this becomes easier as such.

 

Terms associated with Frequency Polygons:

1. Class Interval: It is necessary that the people should select a proper class interval. Class interval is nothing but a specific range in which the data is going to fall. This should make the data appear simple and easy. 

 

2. Midpoint: This is the center point of the bar is to be drawn for the data and this should be identified for the symmetry of the graph as such. 

 

3. Class Mark: It is nothing but the midpoint. It is the mean of the upper limit and the lower limit of the class interval.

  

4. Upper Limit: The upper limit is the ending boundary condition of the class interval. 

 

5. Lower Limit: The lower limit is the starting boundary condition of the class interval.

 

 

How to Draw a Frequency Polygon?

1. Select a suitable class interval for the entire data that is available. Usually, the class interval is plotted on the X-axis or the horizontal line and the frequencies that are corresponding to the class intervals are plotted on the Y-axis or the vertical line. 

2. Once the classmark or the midpoint is chosen for all the class intervals, it becomes very easy to plot the graph. The classmark is represented as follows:

Class mark = (U.L + L.L)/2

Where U.L is the upper limit and L.L is the lower limit. 

3. It is the choice of the person to either mark the class marks or the class intervals on the horizontal axis. The frequencies are also marked on the vertical axis. 

4. There are a few important features to note about the frequency polygons. The height from the horizontal line is always the frequency. It is also important to note that, when the class intervals are being plotted on the X-axis, the midpoint should be considered for the sake of plotting against the frequencies. It should not be plotted against the upper or lower limits. If the class marks are considered directly, then there is nothing to worry about. 

5. The obtained points are joined using the lines and this forms the frequency polygon. 

6. Considering the height as the frequency and the width as the class intervals, the bars are drawn on the chart to represent the data as such.

 

Construction of Frequency Polygon with Histogram 

1. First, get the frequency distribution from the given data.

2. Draw a histogram and join the midpoints of the tops of adjacent rectangles of the histogram with the help of line segments.

3. Then, get the midpoints of 2 assumed class intervals of zero frequency, one next to the first bar on its left and another next to the last bar on its right. These class intervals are referred to as imagined class intervals.

4. Now complete the polygon by joining the midpoints of the first and the last class intervals to the midpoint of the imagined class intervals adjacent to them.

 

Precautions to be Taken While Drawing the Curves:

1. The points should be joined in such a way that they are straight. Freehand curves should be avoided.  

2. There are some people who skip out on drawing rectangles. It is necessary that they should make it a point to draw them because these are going to provide a wider picture of the data that is being represented on the graph. 

These frequency distributions have been a great way to group and organize the data. It would be great if the people make sure that they know how to read all these graphs for that matter.

 

Use of Frequency Polygons

The frequency polygons are used in the following cases and situations: (i) When the data is large, extensive, and continuous in form, and (ii) for comparing the two different sets of data having the same nature.

 

Important points

The following things are  needed to be kept in mind while drawing a frequency polygon: 

• While drawing frequency polygons, joining points should be in a straight manner

• Be sure to avoid free-hand curves, and skip drawing rectangles

• Frequency polygons provide a clear picture of the data that is being represented on the graph.

• A frequency polygon is one of the easiest ways for grouping the data. It is a visual way of presenting quantitative data and its frequency.

• It is used to compare data sets or represent a cumulative frequency distribution. The formula for calculating Classmark for each interval is – 

In mathematical movement (G.P.), the grouping is mathematical and is a consequence of the amount of G.P. A mathematical series is the amount of the multitude of terms of mathematical arrangement. Prior to going to figure out how to observe the amount of a given Geometric Progression, first know what a GP is exhaustively.

What is Geometric Progression?

In the event that in a succession of terms, each succeeding term is produced by duplicating each first term with a consistent worth, then, at that point, the grouping is known as a mathematical movement. (GP), though the consistent worth is known as the normal proportion. For instance, 2, 4, 8, 16, 32, 64 … is a GP, where the normal proportion is 2.

Amount of Nth terms of G.P.

A mathematical series is an amount of a limitless number of terms with the end goal that the proportion between progressive terms is consistent. In this segment, we will figure out how to track down the amount of mathematical series

Determination of Sum of GP

Since, we know, in a G.P., the normal proportion between the progressive terms is steady, so we will consider a mathematical series of limited terms to infer the equation to track down the amount of Geometric Progression

Let’s assume that there are 10 terms in a sequence. If each term is a multiple of the next term then this sequence is said to be in Geometric Progression or Geometric Sum of G.P. Here, the number which is used to constantly multiply is known as the common ratio. In Arithmetic Progression, when the nth term is subtracted from (n – 1)th term, there will be a constant difference between any two numbers. A.P can be written as:

x, x + b, x + 2b, x + 3b, x + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x + (n – 1) b

Consider the sequence: 

3, 6, 12, 24, 48, . . . . . . . . . . . . . . . . . . . . . and the list goes on..

Observe this:

6/3 = 2

12 / 6 = 2

24 / 12 = 2

48 / 24 = 2

In the same manner: 

Consider the series:  1, ⅓, ⅙, ⅟₁₂,  ⅟₂₄, ⅟₄₈, . . . . . . . . . . . . . . . . . . . . . . . .

⅓ / 1 = ⅟₂

¼ / ½ = ½

⅛ / ¼ = ½

⅟₁₆ / ⅟₈ = ½ 

⅟₂₄ / ⅟₁₆ = ½

⅟₄₈ / ⅟₂₄ = ½

In the examples shown above, you can see that the one thing that is constant is the ratio between any two sequential numbers. These sequences are called Geometric Progression. 

A sequence x₁, x₂, x₃, x₄, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xₙ is a Geometric Progression, then aₖ₊₁ / aₖ = r  {k is always greater than 1}

Where,

R = a constant ration between two numbers. 

The General Term of Geometric Progression

The nth term of Arithmetic Progression was found out to be:

xₙ  = x + (n – 1) b

In the case of Geometric Progression, let’s assume that x is the first number and “r” is the common ratio between all the numbers.

So, the second term would be: x₂ = x * r 

The third term would be: x₃ = x₂ * r = x * r * r = xr²

In the same manner, the nth term would be xn  = x * rⁿ⁻¹

The general term = xₙ  = x * rⁿ⁻¹

Common Term

Consider the following sequence: x, xr, xr² , xr³, . . . . . . . . . . . . . 

Here, 

The First term is = x

The Second term is = xr

The Third term is = xr₂

Similarly nth term, xₙ = x * rⁿ⁻¹

Hence, the common ratio (r) = (Any given term) / (the next preceding term)

= xₙ  / x * rⁿ⁻¹

= (rⁿ⁻¹) /(rⁿ⁻²)

= r

Thus, the general term of a G.P is given by  rⁿ⁻¹ and the general form of a G.P is x + xr + xr² + . . . . . . . . . . . . . 

Finite and Infinite 

The finite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹

The infinite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹ can be called the finite geometric progression series.

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . can be called as the infinite geometric progression series.

Sum of GP of n terms

Let assume, 

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹

Let’s assume,

The sum of gp terms =  Sₙ

First Term = x

Ration = r

Now, 

Sₙ = x, xr, xr², xr³, . . . . . . . . . . . . . xrⁿ⁻¹- – – – – – – – – – – – — – – – – – – – – – – (1)

Here, r = 1 or r ≠ 1

If r = 1, then Sₙ = x, x, x, x . . . . . . . . . . . . . . . = nx

If r ≠ 1, then rSₙ = x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹ + xrⁿ – – – – – – – – – – – – – (2)

Now subtract equation (1) from equation (2), it gives you:

rSₙ- Sₙ = ( xr + xr² + xr³ +. . . . . . . + xrⁿ⁻¹+ xrⁿ) – (x + xr + xr² + xr³+ .. . . . . .+ xrⁿ⁻² + xrⁿ⁻¹)

(r – 1)Sₙ = xrⁿ
– x = x(rⁿ- 1)

Sₙ= x(rⁿ – 1) / (r – 1)

Sₙ= x(1 – rⁿ)/ ( 1 – r)

Sum of n terms in GP

Sₙ= x(1 – rⁿ)/ ( 1 – r) here,r≠1here,r≠1

Solved Problems

Example 1: Find the value of n, if the nth term of Geometric Progression is 2, 4, 8, 16, . . . . . . . . . . . . . . . . , 128.

Solution:

Step 1: Find the ratio

r = 4/2 = 2

Step 2: The first term: x = 2.

xⁿ = xrⁿ⁻¹

128 = 2 x 2ⁿ⁻¹

2ⁿ⁻¹ = 128 / 2

2ⁿ⁻ = 64

2ⁿ⁻¹ = (2)6

n – 1 = 6

n = 6 + 1

n = 7

Hence, the 7th  term here is 128.

Example 2: If the Geometric Progression is 5, 10, 20, 40  . . . . . . . , find the sum of 8 terms.

Solution: 

Given, 

x = 5

n = 8

r = 10/5 = 2

Sum of n terms in gp,

Sₙ = x( r[^{n}]- 1) / (r – 1)

S₈ = 5(2⁸- 1) / (2 – 1)

S₈ = 3 (6561) / 2

S₈ = 3 (256) / 2

S₈ = 768 / 2

S₈ = 384

What do you mean by Greater than and Less than Symbols?

• The greater than and less than symbols are mathematical signs which are used to denote an inequality between any two values.

• They are used to compare values.

• The greater than and the less symbols reduce the time complexity and it makes understanding easier.

What is Greater than Sign?

• “>” is greater than sign, it means that the value on the left side is greater than the value on the right side.

• The symbol consists of two strokes of equal length connecting in an acute angle at the right.

• For example, 9>1 which means 9 is greater than 1.

What is Less than Sign?

• “<” is the less than sign, it means that the value on the left side is less than the value on the right side.

• The symbol consists of two strokes of equal length connecting in an acute angle at the left.

• For example, 15<16 which means 15 is less than 16.

What is Equal to?

• Equal to is denoted by the symbol “=”.

• It is used to show the equality between two values.

• While writing equations, we also use the equal to symbol.

Comparison of Numbers

For example,

12 is greater than 9, 4 is less than 10

Instead of writing in words, we will use ‘greater than’ sign or the ‘more than’ symbol and ‘less than’ sign,

12> 9 and 4<10

Tricks to Remember when to use more than Symbol and the Less than Symbol

Method 1

Understand the symbol from the left side to the right side,

On the left hand, it has two points and on the right side it has one point, which can be written as,

Denotes greater than 

This denotes the greater than sign.

Understand the symbol from the left side to the right side,

On the left hand, it has one point and on the right side it has two points, which can be written as:                           

Denotes Less than

Method 2

• The crocodile or the alligator method is a very famous method.

• We assume the < to be a crocodile and the number on both sides to be the fishes. 

• The crocodile always wants to eat a larger number of fishes, so the alligator mouth always opens towards the greater number.

• We assume that the values on both sides represent the number of fish.

• For example, 9>2

Here, the alligator mouth opens towards the value 9 which means that 9 is greater than 2.

Method 3

• The letter ‘L’ looks like the less than symbol “<” 

• The trick to remembering how the less than sign looks is very simple. As “less than” starts with l, the symbol < looks more like the letter L.

Table Showing the Symbols along with their Name

Here are a few Examples of the use of Greater than and Less than Symbols

Questions to be Solved on Greater than and Less than

Question 1) Which sign do we use with these numbers?

Solution) We need to fill in the blanks with greater t
han or less than symbols,

Since 2 is less than 8, we will use the less than symbol (<)

2 < 8

Since 15 is greater than 9, we will use the greater than symbol (>)

15 > 9

Question 2) Rani has 17 apples and Liza has 29 apples. Find out who has a greater number of apples.

Solution) Let’s list down the given information,

Number of apples Rani has = 17

Number of apples Liza has = 29

Since 29 is greater than 17, 29 > 17

Therefore, Liza has more apples than Rani.

Question 3) Rahul’s mother bought 5 roses and Aditya’s mother bought 2 roses. Find out who’s mother bought a greater number of roses.

Solution) Let’s list down the given information,

Number of roses Rahul’s mother has = 5

Number of roses Aditya’s mother has = 2

Since, 5 is greater than 2, 5>2

Therefore, Rahul’s mother has a greater number of roses than Aditya’s mother.

Question 4) Nirmal got 40 marks in her Maths exam whereas her friend Ritu got 24 marks in the same exam. Who scored fewer marks?

Solution) Let’s list down the given information,

Marks obtained by Nirmal = 40

Marks obtained by Ritu = 24

Since, 24 is less than 40, 24<40

Therefore, Ritu scored less than Nirmal.

Conclusion

This is how we can easily understand the process of using the ‘greater than’ and ‘less than’ sign in mathematics. Concentrate on the methods of remembering the symbols and use them accordingly to solve the problems. 

Important Questions Class 9 Maths Chapter 12 Heron’s Formula is a very important resource for students preparing for IX Board Examination. Here we have provided Solutions for Important questions of Class 9 Maths Chapter 12 heron’s formula. Students who are preparing for their final exams for the academic sessions can practice these questions to get better marks. All the problems are based on the NCERT book and CBSE syllabus, presented by our Math experts. In this article, we will solve heron’s formula extra questions with answers.

Heron’s Formula Extra Questions

Here we will solve class 9th heron’s formula extra questions with answers

Q1: Find the Area of a Triangle whose two sides are 18 cm and 10 cm respectively and the perimeter is 42cm.

Solution:

Let us consider the third side of the triangle to be “c”.

Now, the three sides of the triangle are a = 18 cm, b = 10 cm, and “c” cm 

It is given that the perimeter of the triangle = 42cm

So, 

18 + 10 + c = 42

c = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle (s) = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle A = [sqrt{s(s-a)(s-b)(s-c)}]

= [sqrt{21(21 – 18) (21 – 10) (21 – 14)}]

=  [sqrt{21 × 3 × 11 × 7}]

= [sqrt{21 × 21 × 11}]

= [21sqrt{11}]cm2

Q2: Sides of a Triangle are in the ratio of 14 : 20: 25 and its perimeter is 590cm. Find its area.

Solution:

The ratio of the sides of the triangle is given as 14: 20: 25

Let us consider the common ratio between the sides of the triangle be “a”

∴ The sides are 14a, 20a and 25a

It is also given that the perimeter of the triangle = 590 cm

12a + 17a + 25a = 590 

=> 59a = 590

So, a = 10

Now, the sides of the triangle are 140 cm, 200 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 590/2 = 295 cm

Using Heron’s formula for Area of the triangle

= [sqrt{s(s-a)(s-b)(s-c)}]

= [sqrt{295(295-140)(295-200)(295-250)}]

= [sqrt{295 × 155 × 95 × 45}]

= [sqrt{195,474,375}]

= 13981.21cm2

Q3: A field is in the Shape of a Trapezium whose parallel sides are 22 m and 10 m. The non-parallel sides are given as 13 m and 14 m. Find the area of the field.

Solution:

Draw a line segment BE line AD. Then, draw a perpendicular on the line segment CD from point B

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = DE = 10 m

AD = BE = 13 m

EC = DC– ED 

= 22 – 10 = 12 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+ 14 + 12)/2 

= 39/2 m

=19.5m

By using Heron’s formula,

Area of ΔBEC =

= [sqrt{s(s-a)(s-b)(s-c)}]

= [sqrt{19.5(19.5-13)(19.5-14)(19.5-15)}]

= [sqrt{19.5 × 6.5 × 5.5 × 4.5}]

= [sqrt{3137.06 }]

= 56m2

We also know that the area of ΔBEC = (½) × EC × BF

56 cm2 = (½) × 12 × BF

BF = 56 x 2 /12cm 

     = 9.3 cm

So, the total area of ABED will be BF × DE i.e. 9.3 × 10 = 93 m2

∴ Area of the field = 93 + 56 = 149 m2

Q4: Find the Area of the Triangular field of sides 55 m, 60 m, and 65 m. Find the cost of laying the grass in the triangular field at the rate of Rs 8 per m2.

Solution:

Given that 

Sides of the triangular field are 50 m, 60 m and 65 m.

Cost of laying grass in a triangular field = Rs 8 per m2

Let a = 55, b = 60, c = 65

Semi- Perimeter s = (a + b + c)/2

⇒ s = (55 + 60 + 65)/2

= 180/2

= 90.

By using Heron’s formula,

Area of ΔBEC =

= [sqrt{s(s-a)(s-b)(s-c)}]

= [sqrt{90(90-55)(90-60)(90-65)}]

= [sqrt{90×35×30×25}]

=[sqrt{2362500}]

= 1537m2

Cost of laying grass = Area of triangle × Cost of laying grass per m2

= 1537×8

= Rs.12296

Q5: The Perimeter of an Isosceles triangle is 42 cm. The ratio of the equal side to its base is 3: 4. Find the area of the triangle.

Solution:

Given that,

The perimeter of the isosceles triangle = 42 cm

It is also given that,

Ratio of equal side to base = 3 : 4

Let the equal side = 3x

So, base = 4x

Perimeter of the triangle = 42

⇒ 3x + 3x + 4x = 42

⇒ 10x = 42

⇒ x = 4.2

Equal side = 3x = 3×4.2 = 12.6

Base = 4x = 4×4.2 = 16.8

The sides of the triangle = 12.6cm, 12.6cm and 16.8cm.

Let a = 12.6, b = 12.6, c = 16.8

s = (a + b + c)/2

⇒ s = (12.6 + 12.6 + 16.8)/2

= 42/2

= 21.

By using Heron’s formula,

Area of ΔBEC =

= [sqrt{s(s-a)(s-b)(s-c)}]

= [sqrt{21(21-12.6)(21 – 12.6)(21 – 16.8)}]

=[sqrt{21 × 8.4 × 8.4 × 4.2}]

= [sqrt{6223.39}]

= 78.88cm2

Extra Questions of Heron’s Formula 9th class

1. A parallelogram has sides 30 m and 40 m and one of its diagonals is 20m long. Find the area of the parallelogram.

2. The sides of a quadrilat
   eral ABCD are 4cm, 8cm, 12cm, and 16cm respectively. Find its area.

3. A rhombus-shaped sheet with perimeter 50cm and one diagonal 10 cm, is painted on both sides at the rate of Rs. 7 per m2. Find the cost of painting.

4. From a point in the interior of an equilateral triangle, perpendicular are drawn on the three sides. The lengths of the perpendicular are 12cm, 14cm, and 8cm. Find the area of the triangle.

The square root of a number x is the value which when multiplied by itself gives the original number x as its product. The square root of a number x is denoted by [sqrt{x}]. 

Mathematically, if y is the square root of  a number x, i.e., [sqrt{x}] = y, then  multiplying y with itself gives x as its product, i.e., y × y = [y^2] = x.

For example, the square root of 4 = [sqrt{4}] = 2, the square root of 25 = [sqrt{25}] = 5, etc.

In this article, you will learn different methods to find the square root of a given number.

There are two methods which are generally used for finding the square root of a number. These two methods are: 

(1) Prime Factorisation Method

(2) Long-Division Method

METHOD 1

Finding Square Root of a Perfect square number by Prime Factorisation Method

When a given number is a perfect square, the following steps are used to find its square root:

Step 1: Resolve the given number into its prime factors.

Step 2: Make pairs of similar prime factors.

Step 3: Separate out one factor from every pair.

Step 4: Multiply these separate out factors, their product will give the required square root.

For Example: Find the square root of 256 by Prime factorisation method.

We will follow the above steps to find the square root of 256:

Step 1: Resolve the given number 256 into its prime factors.

So, 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 

Step 2: Make pairs of similar prime factors.

256 = [underline{2 × 2}] ×  [underline{2 × 2}] ×  [underline{2 × 2}] ×  [underline{2 × 2}]

Step 3: Separate out one factor from every pair.

Step 4: Multiplying these separate out factors, we get their product as; 2 × 2 × 2 × 2 = 16

Therefore, the square root of 256 = [sqrt{256}] = 2 × 2 × 2 × 2 = 16.

METHOD 2

Finding Square Root of a Perfect square number by Long Division Method

To learn the long division method for finding the square root of a number, let us consider an example, Find the square root of 529 by long division method.

Step 1: Place a bar over every pair of digits starting from the digit at the unit’s place. If the number of digits in the given number is odd, then the left-most single digit too will have a bar. Thus we have, [overline{5}] [overline{29}].

Step 2: Find the largest number whose square is less than or equal to the number under the left-most bar (22 < 5 < 32 ). Take this number as the divisor and the number under the extreme left bar as the dividend (here 5) for division and get the remainder.

Step 3: Write down the number under the next bar, which is 29 in this case, to the right of the remainder. So the new dividend is 129. 

Step 4: Double the quotient and write it with a blank on its right.

Step 5: Choose a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product obtained is less than or equal to the dividend. In this case 42 × 2 = 84. As 43 × 3 = 129. So we choose the new digit as 3. Perform the division with new divisor 43 and dividend 129 to get the remainder.

Step 6: Since the remainder obtained is 0 and no digits are left in the given number, therefore, the square root of 529 = [sqrt{529}]  = 23.

How to Solve the Square Root Equations?

Let us consider an example to learn how to solve the square root equations. Consider a square root equation below and find the value of x:

[sqrt{x-2}] = 5.

To solve the above equation, we will first remove the square root from LHS of the equation by squaring both LHS and RHS.

  [(sqrt{x-2})^2] = (5)2

⇒ x – 2 = 25

⇒ x = 25 + 2

⇒ x  = 27

Therefore, the required value of x for the given square root equation is 27.

The Identity Matrix is ​​a multiplicative unit of the Matrix. Before learning what the Identity Matrix is, let’s recall the importance of units in Mathematics. Identity is a Mathematical quantity, and when manipulated in a particular quantity, the same quantity remains. Consider the following Example. 

The additive Identity element is 0. This is because adding any number to 0 gives the same number as the total. For Example, 3 + 0 = 3, 0 + (1) = 1. Multiplying any number by 1 yield the same number as the product, so the multiplication Identity is 1. For Example, 3 × 1 = 3, 1 × (1) = 1. Similarly, of course, if you add the zero Matrix to any 2×2 Matrix, you’ll see that you get the same Matrix, or zero. The Matrix becomes an additive Identity Matrix called. But what is the multiplicative identity in terms of Matrices? Let’s learn this in detail here.  

• The Identity Matrix is ​​a Square Matrix in which each element of the main diagonal is 1 and each other element is 0. Also known as the Identity Matrix. Represents an Identity Matrix of degree n × n (or n) as I. 

• This is sometimes referred to simply as I. 

Identity Matrix Properties 

1. Let us discuss the properties of the Identity Matrix.

Identity Matrix is always in the form of a Square Matrix.

The Identity Matrix is called a Square Matrix because it has the same number of rows and columns. For any given whole number n, the Identity Matrix is given by n x n.

2. Multiplying a given Matrix with the Identity Matrix would result in the Matrix itself.

Since the multiplication is not always defined, the size of the Matrix matters when you work on the Matrix multiplication.

For Example, for the given m x n Matrix C, you get

[ C = begin{bmatrix} 1 &  2 & 3 &  4 \ 5 & 6 & 7 & 8end{bmatrix}]

The above Matrix is a 2 x 4 Matrix since it contains 2 rows and 4 columns.

3. When multiplying two inverse Matrices, you would get an Identity Matrix.

If you multiply two Matrices that are inverses of each other you would get an Identity Matrix.

The properties of the Identity Matrix based on the definition are: 

• All Identity Matrices are Square Matrices. All identities are diagonal because only the elements of the main diagonal are non-zero. 

• All identities are scalar Matrices because all elements of the main diagonal are the same and all other elements are zero. 

• Multiply the Identity Matrix by another Matrix to get the same Matrix. According to the study, the Determinant of the Identity Matrix is ​​1. 

• The Identity Matrix is ​​symmetric with respect to IT = I. The inverse of the Identity Matrix is ​​itself I * I1 = I1 * I = I. In = I, integer ‘n’. 

• That is if we count deeply the Square of the Identity Matrix is ​​equal to itself, the cube of the Identity Matrix is ​​equal to itself, and so on. 

• Multiply the Matrix by its reciprocal to get the Identity Matrix (see this in the next section). Find the inverse Matrix using the Identity Matrix 

 The inverse of Matrix A (denoted as A1) is Matrix B (and vice versa) only if AB = BA = I. Where A, B, and I are Square Matrices of the same degree. Given A and B, you can easily check that they are the opposite of each other by simply checking AB = BA = I. But given Matrix A, how can we find the inverse B? You can find the inverse of the Matrix by following the steps below. 

• Step 1: Use an Identity Matrix of the same degree to describe an augmented Matrix adjacent to the specified Matrix and separate these two Matrices with a line.  

• Step 2: Apply a row operation aimed at converting the left Matrix (A) to an Identity Matrix.  

• Step 3: The Matrix on the right and left is the inverse Matrix itself. The procedure for finding the inverse Matrix using the Identity Matrix is ​​shown. 

See the Identity Matrix Example section below for an Example of finding the inverse Matrix using these steps. 

Identity Matrix Application 

• Identity Matrices are used in linear algebra for several purposes. The usage of the Identity Matrix is ​​as follows: The 

• Identity Matrix is ​​used to check if two given Matrices are opposite to each other. The Identity Matrix is ​​also used to figure out the inverse of the Matrix. The Identity Matrix is ​​used to find eigenvalues ​​and eigenvectors. When solving simultaneous equations using basic row operations, the Identity Matrix is ​​used. Important information about the Identity Matrix 

• There are some important points to keep in mind regarding the Identity Matrix. 

• If you see an Identity Matrix with no operations specified, it should be the Identity Matrix by default in terms of multiplication.  To write an Identity Matrix of a particular order, first, write an empty Matrix of the specified order, write 1 in place of the main diagonal element, and finally write 0 in place of all other elements.  If AB = BA = I, then A and B are opposite to each other.  To find the inverse of a Matrix, write it next to the Identity Matrix of the same order on the right. Apply row operations to the entire expanded Matrix to create the Matrix on the left as an Identity Matrix. Then the Matrix on the right is the inverse of the specified Matrix.

The Identity Matrix is known as the Matrix that is in the form of the n × n Square Matrix in which the diagonal contains the ones and all the other elements are zeros. It is also referred to as a unit Matrix or an elementary Matrix. It is denoted as In or just I, wherein n is the size of the
Square Matrix. To explain the Identity Matrix definition part by part, let us start by reminding you that the Square Matrix refers to the Matrix that contains the same amount of rows and columns. The order of the Matrix comes from its dimensions, and the main diagonal is the array of the elements that are inside the Matrix that form the inclined line starting from the top left corner and extending to the bottom right corner. Given the characteristics of the Identity Matrix, you can also conclude that these types of Matrices are also called diagonal Matrices. In this article, we will learn about what is an Identity Matrix, the Determinant of Identity Matrix, Identity Matrix properties, the Identity Matrix in c, and learn about the Identity Matrix Example.

Identity Matrix Definition

An Identity Matrix refers to a type of the Square Matrix in which its diagonal entries are equal to 1 and the off-diagonal entries are equal to 0.

Identity Matrices play a vital role in linear algebra. In particular, their role in the Matrix multiplication is similar to the role that is played by the number 1 when it comes to the multiplication of the real numbers:

1. The real number remains unchanged if it is multiplied by 1

2. The Matrix remains unchanged if it is multiplied by an Identity Matrix

Identity Matrix Example

Some Examples of Identity Matrices are as follows:

The 2 x 2 Identity Matrix is given by

[ I = begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix}]

The Identity Matrix of order 3 is represented in the following manner:

[ I = begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix}]

Solved Examples

Example 1

Write an Identity Matrix of the order 4

Solution:

The Identity Matrix of the order 4 x 4 is given as

[ I = begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 end{bmatrix}]

Example 2

Determine if the given Matrix is an Identity Matrix or not.

[ C = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 end{bmatrix}]

Solution:

No, the given Matrix is not an Identity Matrix since it is not a Square Matrix. The number of rows is not equal to the number of columns. The given Matrix is of the order 2 x 3.