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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.

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

HOW TO UNDERSTAND SYMMETRIC AND SKEW SYMMETRIC MATRICES

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 .If you liked the post then share it with your friends and follow me on my blog to boost me to do more and more for you. We shall meet in next post ,till then BYE.....................

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.

How to find symmetric and skew symmetric matrices

How to find symmetric and skew symmetric matrices

If you want to know the answer of this problem , just click here

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.

For 2 × 2 Matrix ,

How to Construct 3 × 3 Symmetric Matrix

For 3 × 3 Matrix ,If

Skew Symmetric Matrix