HOW TO UNDERSTAND BINARY OPERATIONS IN RELATIONS AND FUNCTIONS

Hello Friends Welcome 

Today  we shall discuss  Binary operation and how to  understand binary operation with the help of some Examples. Because with the help of Binary operations we can crack many quizzes like .

3                 4                      15

6                 7                      49 
9                 ?                      99
If someone ask us  to solve the problem given below,
            

Binary Operations

Then how will you solve this problem or such types of problems?  so with the of help Binary operations we can solve such problems,

 Commutative Property

If a person leaves for his office at 9 am daily ,which is 5 KM from his home , and comes back  home at 6 pm , then its distance from home to office and back office to home is same 5 KM , then this Property is called commutative Property .

For example 
(1) If  Shayam slaps Ram twice , and in return Ram also slaps Shyam then it is called commutative.
(2) if we add 4  to 5 we shall get 9 and in other case if we add 5  to 4 then  we also  get 9. implies 4 + 5 = 5 + 4 = 9

In Mathematics it written as a ∗ b = b∗ a  for all  values of a ,b  .
but if we subtract 4 from 5 we shall get  1 but if we subtract 5 from 4 we shall get -1, then this relation is not called commutative ,because answer are not same in both the cases.

Associative Property

If we have three numbers 4, 5, and 6 , and we have to add these three numbers in two  ways that 1st we add 4 and 5 and then 6 will be added to the result obtained in last step. and in second ways we shall 5 and 6 then resultant will be added to 4 . And in both the cases result comes out same then this  will be called Associated property. 

( 4 + 5 ) + 6 = 4 + ( 5 + 6 )  

           9 + 6 = 4 + 11 =15

But if we subtract these numbers in place of sum ,then these numbers do not satisfy the property of Associativity .

( 4 - 5 ) -  6  ≠  4 -  ( 5 -  6 )  

      (-1) -  6 ≠   4 -  (-1) 

          -1 - 6   ≠  4 + 1

              -7  ≠ 5
This concept can be better understand with the help of this video

  

Binary Operations

A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a, b) by a ∗ b

Question

Let ∗ be binary operation on the set Q of rational numbers defined below

(1)  a ∗ b = a – b
(2)  a ∗ b = ab + 1
(3)  a ∗ b = a + ab
(4)  a * b = | a - b |


Determine whether ∗ is binary, commutative or associative.

Solution

(1) a ∗ b = a – b

 Commutative Property

 a ∗ b = b∗ a 

Now 

 a ∗ b = a – b   -----------------(1)

b*a = b - a = - (a-b) -----------(2)

Therefore   a ∗ b ≠ b∗ a 

Therefore * is not a commutative operation under Q

Associative Property

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c = (a – b)*c         Replaced a*b with a - b as given

Now assume "a-b" as 1st number(blue) and "c" as 2nd number(red)  Again using eq (1)

(a ∗ b )*c =(a – b) * c    

                 = (a - b ) - c

                 = a - b - c ------------(3)

 From (1) and (2)

   a * (b * c)  = a*(b - c)

Now assume "a" as 1st number(blue) and "b - c" as 2nd number(red) 

a * (b * c)  = (a) - (b - c )

                = a - b + c ------------(4)

From (1) and (2)

(a ∗ b) *c  ≠ a ∗ (b * c) 

Therefore * is not a associative  operation under Q

(2) a ∗ b = ab + 1   (Product of two numbers and plus one)

 For commutative Property

 a ∗ b = b∗ a 

Now 

 a ∗ b = ab + 1   ------------------(5)

b ∗ a = ba + 1 = ab +1 --------(6)

Therefore   a ∗ b = b∗a 

Therefore * is a commutative operation under Q

For Associative Property

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c =(ab + 1) * c         Replaced a*b with ab+1 as given

Now assume " ab+1 " as 1st number(blue) and "c" as 2nd number(red)

(a ∗ b )*c =(ab + 1)*c    (Product of two numbers and plus one)

                 = (ab + 1)(c) + 1

                 =  (ab + 1)c +1

                 = abc + c + 1   -------(7)

   a * (b*c)  = a*(bc + 1)   (Product of two numbers and plus one)

Now assume "a" as 1st number(blue) and "bc+1" as 2nd number(red) 

a * (b*c)  = (a)(bc + 1) + 1

                =abc + a + 1  ------------------------(8)

From (7) and (8)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

(3)  a ∗ b =a + ab  (Product of two numbers and plus 1st num ) 

so if we put all the values of aand  b according to table then we can solve problem discussed in begining of the post.

 Binary Operations

For commutative Property

 a ∗ b = b ∗ a 

Now 

 a ∗ b = a + ab   -----------------(9)

b ∗ a = b + ba = b + ab -------(10)

Therefore   a ∗ b ≠ b ∗ a 

Therefore * is a  not commutative operation under Q

For Associativeness

(a ∗ b) *c = a∗ (b*c) 

(a ∗ b )*c =(a+ab)*c         Replaced a*b with a+ab as given

Now assume "a + ab" as 1st number(blue) and "c" as 2nd number(red)

(a ∗ b )*c =(a + ab) * c   (Product of two numbers and plus 1st number , here 1st num is a + ab and 2nd num is c)

                 = (a + ab ) + ( a + ab )c

                 =  a + ab + ( a + ab )c

                 = a + ab + ac + abc           ------------------------(11)

   a * (b*c)  = a*( b + bc)   (Product of two numbers and plus 1st number )

Now assume "a" as 1st number(blue) and "b + bc" as 2nd number(red) 

a * (b * c)  = (a) + a( b + bc)

                =a + a( b + bc)     

                =a + ab + abc    ------------------------(12)

From (11) and (12)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

(4) 

 For commutative Property

 a ∗ b = b ∗ a 

Now 

 a ∗ b = |a - b | ---------------------(13)

b ∗ a = |b - a |= |a - b | ---------------(14)

Therefore   a ∗ b = b ∗ a 

Therefore * is a  commutative operation under Q

For Associativeness

( a ∗ b ) * c = a ∗ (b * c) 

(a ∗ b )*c  = (| a - b | ) * c         Replaced a*b with | a - b | as given

Now assume "| a - b |" as 1st number and "c" as 2nd number

(a ∗ b )*c =(| a - b | ) * c   ( modulus of difference of two numbers , here 1st num is | a - b | and 2nd num is c)

                 =  | (| a - b |) - c |

                 = | | a - b |  - c |           ------------------------(15)

   a * (b*c)  = a* ( |b-c |)   (Modulus of difference of  two numbers )

Now assume "a" as 1st number(blue) and "|b-c |" as 2nd number(red) 

a * (b * c)  = | (a) - |b-c | |

                 =  

                 = |  a - |b-c | | -------- (16)

From (15) and (16)

(a ∗ b) *c ≠ a∗ (b*c) 

Therefore * is not a associative  operation under Q

Conclusion

Thanks for devoting your precious time to read this post. I hope this post on How to understand     Binary Operations , commutative , Associative has helped you more , If you find this post little bit of your concern then, then follow me on my blog and read my other posts . We shall meet  in next post, till then Bye.

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HOW TO FIND THE INVERSE OF 2×2 AND 3×3 MATRIX USING SHORTCUT METHOD

Hello and Welcome to this post ,Today we are going to discuss the shortest and easiest methods of finding the Inverse of 2×2 matrix and 3×3 Matrix. Usually when we have to find the Inverse  of any  Matrix  then we follow the following steps .

1 Check whether the determinant value of the given Matrix is Non Zero.

 2  Find out   the co-factors of all the elements of the Matrix.

 3 Put these co-factors in co-factor Matrix.

 4 Find the Ad joint of this matrix by taking the  Transpose of a Matrix  of the co-factor matrix.

  5 Now  Multiply  Ad Joint of Matrix   with the reciprocal of               Determinant value of  the given Matrix.

This Method is very confusing, Long  and time Consuming.  So Let us have a New,  Easy and Shortcut Method .

Method For 2×2 Matrix

If we have to find the Inverse of  2×2 Matrix then Follows these steps.

 1 Interchange the position of the elements which are  a11  and a22 .

2  Change the Magnitude of the elements  which are in position a12 and  a21   .

3  Divide  every elements of the given Matrix with its Determinant value.

Example

 To find the Inverse of this matrix just interchange the position of elements a₁₁ and a₂₂  i.e   Interchange the positions of elements  5  and -3 and in second step change the magnitude of the elements which are  in positions a12 and  a21   i.e. change the sign of 9 and 4.

Now divide each elements with determinants value of the matrix which is  (5)(-3) - (9)(4) = -15 -36 = -51

So The Inverse of the given Matrix A  will  be 

Then after interchanging the positions of 8 and 2 change the magnitude of  7 and -6 and divide every elements with its determinant value (8)( 2) - (7)*(-6) = 16+42 = 58

The Inverse of  B is

 After interchanging the position of -3 and -6 and changing the magnitude of  -4 and -5 and at last dividing every elements with its determinant value (-3)×(-6) - (-4)×(-5) = 18 - 20 =  -2

The Inverse of  C is   

This video Explains all about Inverse of 2×2 and 3×3 Matrix

READ  MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS

Method for 3×3  Matrix        

Ist of all  Write the given Matrix in five columns by adding the 4^th column as repetition of 1^st column and 5^th column as repetition of 2^nd column, then

C₁    C2     C3      C4     C5
5      -1       4       5      -1
2       3       5       2        3
5      -2      6        5      -2

Now Expanding this Matrix to 5×5 Matrix by adding 4th Row as repetition of 1st Rows and adding 5 Row as repetition of 2nd column as what we received in last step.

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3

 Now to find the Inverse  of the given Matrix ,we have to find the cofactor of every elements

1 Find the co-factor of 1st element of Row 1 i. e. 5, determinant value of the Matrix (RED below ) obtained by eliminating the 1st Row and 1st Column which will be (3×6)-{(5)×(-2)} = 28,write these co-factor value in 1st column of 1st Row. (we are evaluating co-factors row wise and writing Column wise)

R₁      5      -1      4        5          -1
R₂    2       3      5        2           3
R₃    5      -2      6       5          -2
R₄    5      -1      4       5          -1
R5     2       3      5       2           3

2 Now Find the co-factor of 2nd element of 1st Row i. e. -1,which is equal to determinant value of the Matrix  (RED below) obtained by eliminating the 1st Row and 2nd Column which will be 5*5-(2)*(6) =13,write this co-factor value in  2nd Row of 1st column .(we are evaluating co-factors row wise and writing Column wise) 

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3 

3 Now Find the co-factor of 3rd element of 1st Row  i.e. 4, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 1st Row and 3rd Column which will be 2*(-2)-(3)*(5) = -19,write this co-factor value in  3rd Row of 1st column.(we are evaluating co factors row wise and writing Column wise)

R₁           5            -1            4             5         -1

R₂           2             3             5            2           3

R₃           5            -2             6            5          -2

R₄           5            -1             4            5          -1

R5           2             3             5            2           3

4  Now Find the co-factor of 1st element of 2nd Row  i. e. 2, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 2nd  Row and 1st Column which will be -2*(4)-(6)*(-1) = -2,write this co-factor value in  2nd Column of 1st Row .(we are evaluating co factors row wise and writing Column wise) .

R₁         5         -1           4             5           -1

R₂         2          3            5             2            3

R₃         5         -2            6            5           -2

R₄         5         -1            4            5           -1

R5         2          3            5            2             3

5 Now Find the co-factor of 2nd element of 2nd Row i.e 3, which is equal to determinant value of the matrix (RED below ) obtained by eliminating the 2nd Row and 2nd column which will be 6*(5)-(5)*(4) = 10,write this co-factor value in 2nd Row of 2nd column

  R₁        5           -1          4          5          -1

  R₂        2            3           5         2           3

  R₃        5           -2          6          5          -2

  R₄        5           -1          4          5          -1

  R5        2            3          5          2           3

6 Find the co-factor of 3rd element of 2nd Row i. e.5, which is equal to determinant value of the Matrix (RED ) obtained by eliminating the 2nd Row and 2nd Column which will be 5× (-1)-(-2) × (5) = 5,write this co-factor value this 2nd Column of 3rd Row, write this co-factor value in 2nd Column of 1st Row . (we are evaluating co factors row wise and writing Column wise ) .

 R₁         5          -1           4           5          -1

 R₂         2           3           5           2            3

 R₃         5          -2           6           5          -2

 R₄         5          -1           4           5          -1

 R5         2           3           5            2           3

Similarly for 1st , 2nd, 3rd element the co-factor values will be as follows

      For  A₃1 i.e  5

R₁       5            -1             4            5          -1

R₂       2             3              5           2           3

R₃       5            -2             6            5          -2

R₄       5            -1             4            5          -1

R5       2             3              5           2           3

For A₃₂ i.e   -2
R₁          5         -1            4            5          -1 
R2          2          3            5            2           3

R₃          5         -2            6            5          -2

R₄          5          -1           4            5          -1

R5          2           3           5            2           3

  for  A₃₃  i.e. 6

R₁          5           -1           4           5          -1

R₂          2            3           5           2           3

R₃          5           -2           6           5         -2

R₄          5           -1           4           5          -1

R5          2            3           5           2           3 

so we have  -17 ,-17 and 17 as co-factors of 3rd Row, write these co factors in 3rd column .

(we are evaluating co factors row wise and writing Column wise)

Ad joint   A   =   

                                        
⎾ 28          -2          -17 ⏋
⎹⎸ 13          10         -17 ⎹
⎿ -19         5            17  ⏌

Now divide with the determinant value of given 3×3 Matrix , which will be 5(28)-1(-13) + 4(-19) = 140 + 13 -7 6 = 77.

Now divide each element of Ad joint Matrix obtained in previous step with determinant value 77,

Then   A⁻¹  =   

Conclusion

This  post was regarding short cut methods of finding Inverse of  2×2 and 3×3 Matrices , If you liked this post ,Please  share your precious views on this topic and share this post with your friends to benefit them. we shall Meet in the next post ,till then BYE . 

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