HOW TO FIND SYMMETRIC AND SKEW SYMMETRIC MATRICES

CAN YOU SOLVE THIS  PUZZLE ?

At the end of this post, we shall be able to solve or crack this type of quiz, puzzle, or brain teaser,basis for skew symmetric matrices ,symmetric and skew symmetric matrix in hindi,symmetric skew-symmetric and orthogonal matrices,The symmetric and skew symmetric Matrices, How to know whether any given matrix is symmetric or skew symmetric and How to construct 2 × 2 and 3 × 3 Matrix which are Symmetric Matrix And Skew Symmetric Matrix. Before we proceed we must know what is  Transpose Of a Matrix .

Symmetric Matrix

Any square matrix is said to Symmetric Matrix if the transpose of that Matrix is equal to the matrix itself. That is if we transform all the Rows of the Matrix into respective column, even then we get same Matrix . Let us discuss this with the help of Some Examples.

 

How to Construct 2 × 2 Symmetric Matrix

1 Complete the 1^st  Row of the matrix with the elements of your choice.

2 Copy all  Elements which are in 1^st Row to  1^st Column.

3 Now place last element in   2^nd Column  of   2^nd Row with your choice.

The Matrix so obtained is Symmetric Matrix.

For 2 × 2 Matrix ,

If

As  Transpose of these Matrices  are equal the Matrices itself, Therefore these Matrices are Symmetric Matrix .

How to Construct 3 × 3 Symmetric Matrix

1 Complete the 1^st  Row of the matrix with the elements of your choice.

2   Copy all  Elements which are in 1^st Row to  1^st Column.

3   Complete the 2^nd   Row of the matrix with the elements of your choice.

4  Copy all  Elements which are in 2^nd  Row to  2^nd  Column.

5  Put the  last  element  in the   3rd Column of 3^rd Row of  your choice.

The Matrices so obtained are Symmetric Matrices.

For 3 × 3 Matrix ,If

As  Transpose of these Matrices are the Matrices itself, Therefore these Matrices are Symmetric Matrices.

 Read   Shortcut to find 2×2 and 3×3 Inverse Matrices     in a easy way.

Skew Symmetric Matrix

Any square matrix is said to Skew Symmetric Matrix if the transpose of that Matrix is equal to the negative of the matrix. That is if we transform all the Rows of the Matrix into respective columns, even then we get same matrix with change in magnitude. Let us discuss this with the help of Some Examples .

Watch this video to better understand this concept of symmetric and skew symmetric  Matrix

How to Construct 2 × 2 Skew Symmetric Matrix

1  Put all the elements equal to Zero in diagonal positions.

2 Complete the 1^st  Row of the matrix with the elements of your choice.

3 Copy all  Elements which are in 1^st Row to  1^st Column with change in magnitude of each element.

The Matrix so obtained is Skew  Symmetric Matrix

For 2 × 2 Matrix,If 

As  Transpose of each of the  Matrix written above  is negative of the Matrix itself, Therefore these Matrices are Skew Symmetric Matrices.

How to Construct 3 × 3 Skew Symmetric Matrix

1   Complete all Diagonal elements of the Matrix with Zero.

2   Complete the remaining elements the 1^st  Row with your Choice

3   Copy all  Elements which are in 1^st Row to  1^st Column.

4   Complete the remaining elements of the 2^nd   Row of the matrix with the elements of your choice.

5  Copy all  Elements which are in 2^nd  Row to  2^nd  Column with change in magnitude of every element.

6 As the elements in the 3^rd column of 3^rd Row is already taken as Zero.

For 3 × 3 Matrix

If

The Matrices so obtained are Skew Symmetric Matrices, as the negative of the  Transpose of each Matrices are equal to the  Matrices Constructed.

Conclusion

This post is about Symmetric Matrix And Skew Symmetric Matrix . How to Identify and construct 2 × 2 and 3 × 3 Matrices which are Symmetric Matrix And Skew Symmetric Matrix .basis for skew symmetric matrices, symmetric and skew symmetric matrix in hindi ,skew diagonal matrix,if a is skew symmetric then a^2 is symmetric ,symmetric skew-symmetric and orthogonal matrices

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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 Properperty .

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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