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

symmetrizable 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.....................

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HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES BY MATRIX METHOD

Today we are going to learn, how to solve system of linear equations of 2 and 3 variables using matrix method. So using matrix method to solve system of linear equation , we must know some topics such as co-factor of element, Transpose of matrix , Ad joint of a Matrix, Multiplication of two Matrices ,Determinant value of a Matrix ,Inverse of a matrix etc. Let us understand the process of finding the solution of system of linear equations with the help of some examples.

To Solve the system  of Linear Equations using 2×2 Matrix Method

x - 5y = 4

2x + 5y = −2

Writing this system of equation in Matrix form

AX = B 

where X = A-1 B------------------------(1)


And A-1 = (1/Det A ) ( Ad joint A)



Where

We need the inverse of   A , which we write as   A^-1 

To find the inverse 1st find out Co-factor Matrix of A

Ad joint  A  =   (     5    5​ -2   1   ​)

Co factor   A  =   (     5    −2​ 5   1   ​)

As we know the Ad joint Matrix of any matrix can be found by taking the transpose of the Co Factor matrix.

Now let us find the determinant of  A

∣A∣ = 5 − (-10) = 15 , which is non Zero,


Therefore  A^-1  Exists     ,So

Now putting the value of inverse of Matrix A in equation (1)

Now putting the elements of Matrix X , and 

By the equality of two Matrices ,their elements in respective positions are equal to each others,

Hence x= 2/3  and y = -2/3

How to check the correctness of a solution

put x=  2/3  and y = -2/3  in the given  equations  

x - 5y = 4 implies 2/3 - 5(-2/3) = 2/3+10/3 = 4

and 2x + 5y  = -2 implies 2*(2/3) + 5(-2/3) = 4/3-10/3 = -6/3= -2

therefore both equations are satisfied by these values of 'x' and 'y'

To Solve the system of Linear Equations using 3×3 Matrix Method

 x  - y + z   = 4
2x + y - 3z  = 0
 x + y + z    = 2

Converting this system of Linear equations into Matrix Form

AX    =  B 

X  =    A^-1  B ----------------------------------(1)
Where

Now we have to find the   A^-1 . , and it will exist if its determinants value |A| must be non Zero. Let us find |A|

    




 |A|= 1{1×1-(-3)×1}-(-1){2×1-(3)×1}+1(2×1-1×1)

|A|= 1 {1+3}+1 {2+3}+1 {2-1}

|A|= 4+5+1 = 10

Since |A| is non Zero , Therefore inverse of A exists .

We have to find the Co factors of all the elements of matrix A . Let us find out the co factors of all the elements of matrix row wise.

Co-Factors of 1st Row

     

            co-factor of A11  =   4
            co-factor of A12  =  -5
            co-factor of A13  =   1

Co-Factors of 2nd Row

            co-factor of A21  =   2 

            co-factor of A22  =   0
            co-factor of A23  =  -2

Co-Factors of 3rd Row

              co-factor of A31  =  2
              co-factor of A32  =  5
              co-factor of A33  =  3

Now co Factor matrix of A can be written as follows




Now to find the Ad joint of this Matrix, Take the transpose of this Matrix, 

As we know the Inverse of a Matrix  is the scalar  multiplication of Ad Joint Matrix of Matrix A and reciprocal of Determinant value of the Matrix A.

Putting the values of Inverse of A and Matrix B in equation (1), So after Multiplication of these two matrices ,we get

$\backslash displaystyle=\{\backslash left(\backslash begin\{matrix\}\{22\}\backslash \backslash -\{16\}\backslash \backslash -\{16\}\backslash end\{matrix\}\backslash right)\}$

Now putting the values of elements  of   Matrix X, and equating the elements in their respective positions .

Now using the property of equality of two Matrices , All the elements in their respective position are equal to each other , we get

 x = 2  , y = -1, z = 1  is the solution of the system of linear Equations.

Conclusion

In this post I  discussed the method of solving the System of linear Equations with the help of Matrix Method , If you liked the post please share it with your friends , And in case of any improvement please make use of Comment Box . To keep on supporting me follow my Blog . Thanks for your valuable time to read this post .We shall meet you in next post ,till then BYE.

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HOW TO MULTIPLY TWO MATRICES || PRODUCT OF TWO MATRICES

In this post we are  going to discuss  multiplication of two matrices , i.e. when two matrices can be multiplied with each other or when two matrices are NOT eligible for multiplication, what are the necessary conditions for the multiplication of two matrices to be multiplied.

Conditions for Matrix Multiplication

Before multiplication of two matrices we have to check whether multiplication is possible or not , If it is possible then matrices will be multiplied to each other. Necessary condition for multiplication of two matrices is if " The number of columns in the first matrix is the same as the number of rows in the second matrix ". We must know  different types of matrices ,Rows and Column.

Note : The commutation may or may not be possible for multiplication of matrices, That is  in some case AB = BA but In general AB is not equal to BA.

Example 1

1. Multiplication of 2 × 3 matrix with 3 × 4 matrix is possible as number of columns in 1st Matrix is equal to numbers of rows in 2nd Matrix and Resultant matrix will be of 2 × 4 order .

2. Multiplication of 5 × 1 matrix with 1 × 2 matrix is also possible as it gives 5× 2 matrix as resultant Matrix.

3. Multiplication of 4 ×3 matrix with  2 × 3 matrix is NOT possible. Because red colour numbers 3 and 2 do not match .

How to Multiply Matrices 

If condition for matrix multiplication is satisfied, then below is the method of multiplication for two matrices .

Let's take  two matrices A and B  of  2 × 3 and 3×2  orders respectively. The resultant matrix will be  2 × 2 matrix. we start across the 1st row of the first matrix multiplying down the 1st column of the second matrix, element by element. Then we add the resulting products. Our product will be written in position a11 (top left) of the answer matrix which will be equal to   "au+bw+cy".

We shall repeat similar process for the 1st row of the first matrix and the 2nd column of the second matrix. The result will be written  in position a12 which will comes out "av+bx+cz".

Now for the 2nd row of the first matrix and the 1st column of the second matrix. The result will be placed in position a21  as   "du+ew+fy" .Finally, Multiply the 2nd row of the first matrix and the 2nd column of the second matrix to get the product as "dv+ex+fz" The result is placed in position a22. So the resultant matrix will be written as :--

Let us take one  example to Multiply 2×3 and 3×2 Matrices ,The order of resultant matrix will be 2 × 2.

A11 element will be 4×(-3)+1×5+4×6 =  -12+5+24 =17

A12 element will be   4×1+1×6+4×4  =  4+6+16 = 26

A21 element will be  2×(-3)-5×5+7×6 = -6-25+42 = 11

A22 element will be  2×1+(-5)×6+7×4 = 2-30+28 = 0

Since the resultant Matrix  2× 2   as follows

A11  element will be 8×(-2)+9×4      = -16 +36 =20

A12   element will be 8×3+9×0            = 24+ 0 = 24

A21  element will be 5×(-2)+(-1)×4   = -10-4 = -14

A22  element will be 5×3+(-1)×0         = 15+0 = 15
so the resultant matrix will be written as 

To find AB

Since Numbers of columns (3) in 1st matrix (Here A Matrix ) is not equal to numbers of Rows (1) in 2nd Matrix (Here Matrix B ). Therefore matrix multiplication is NOT possible in this case.

To Find BA

Since Numbers of columns (3) in 1st matrix (Here B Matrix ) is equal to numbers of Rows (3) in 2nd Matrix (Here A Matrix ). And resultant Matrix shall be 1×3 order . Therefore matrix multiplication is possible in this case and can be calculated in the following manner :-

BA = ( 4*2+3*3+8*3      4*1+3*(-5)+8*2        4*(-7)+3*0+8*(-1) )

BA = ( 8+9+24    4-15+16    -28+0-8 )

BA = ( 41    5     -36 )

Therefore AB is not equal to BA
This video can better demonstrate the multiplication of two matrices .

Important Note

And what about the matrix multiplication of AB and BA of the matrices given below whose order are 4×3 and 3×3 respectively . 

Then Matrix Multiplication BA is not possible as the numbers of columns(3) in matrix  B is   not  equal to the numbers of rows (4) in  matrix A .

But If we want to Multiply the matrix A with Matrix B as AB then it is possible as the numbers of columns(3) in A matrix is equal to the numbers of rows(3) in B matrix, so it can be calculated as explained earlier. 

It means the order of the matrices play very important role to decide whether matrix multiplication is possible or not. And if possible it is again the order of both the matrices to decide what will be the order of the resultant matrix.

Therefore in  multiplication of matrices of 4×3 and 3×3 .The order of resulting matrix will be 4×3 ,The outer most numbers ( Marked red ). 

Very Important 

If we have two matrices A and B of order 1×3  and 3×1 respectively ,  and if we have find their product AB  and BA , then  

In  1st case  where we have to find AB , let us check its multiplication would be possible or not , as number of columns (3) in 1st matrix is equal to numbers of rows (3) in 2nd Matrix , so AB would be possible and order of resulting matrix would be 1×1 and it can be found using matrix multiplication .

And if we have find their product BA , then would it be possible or not ? let us check .....

In 2nd case where we have to find BA , let us check its multiplication would be possible or not , as number of columns (1)  in 1st matrix is equal to numbers of rows (1) in 2nd Matrix , so AB would be possible and order of resulting matrix would be 3×3 and it can be found using matrix multiplication .

Read this one also  HOW TO UNDERSTAND  BINARY OPERATIONS  ,RELATIONS AND FUNCTIONS   ||  COMMUTATIVE || ASSOCIATIVE

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Conclusion 

This post was about multiplication of two Matrices, Eligibility conditions for product of matrices, Thanks for giving your valuable time to this post . If you liked this post,Please share your valuable opinions about this post. See you in next post ,till then Bye.

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$$\displaystyle{A}={\left(\begin{matrix}-{2}&{1}&{7}\\{3}&-{1}&{0}\\{0}&{2}&-{1}\end{matrix}\right)}    
$$\displaystyle{A}={\left(\begin{matrix}-{2}&{1}&{7}\\{3}&-{1}&{0}\\{0}&{2}&-{1}\end{matrix}\right)}   

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HOW TO FIND THE TRANSPOSE OF MATRIX WITH AN EASY METHOD

In this post we are going to learn "what is transpose of matrix, calculate  transpose ,define transpose, meaning of transpose " .  As it is clear from its name  transpose means trans + pose i. e. transfer + position ( transfer of position ) ,  the transpose   of any matrix is obtained by transfer of Rows into Columns And vice versa. i.e transforming 1st row to 1st column and transforming 2nd row to 2nd column and so on for 3rd ,4th and 5th rows and columns. This is very easy and interesting topic in matrices and determinants.

 To find the transpose of the matrix 

1^st of all  shift all the elements which are in 1st row to 1^st column as

5

5

2
,then shift the elements which are in      2nd row to 2^nd column as  

-1

-3

 7

similarly  shift all the elements which are in 3rd row to 3^rd column as

4

2

8

And the matrix so obtained is the transpose matrix. We can  check that same colour row have been  transformed to same colour Column
        

Now we shall take one  example to find the transpose matrix

Here Given matrix have 3 rows and 4 columns .It means we shall have 4 rows and 3  columns  in transpose matrix.

1st shift all the elements which are in 1st row to 1^st column as

5
8               
6
-4

shift all the elements which are in 2nd row to 2nd column as

3
5
8
3

shift all the elements which are in 3rd row to 3rd column as 

-3
8
-7
6

so shifting the corresponding Rows into Corresponding columns. We can  check that same colour rows are transformed to same colour columns.

Now we shall take one more  example to find the transpose of matrix .Here Given matrix have 3 rows and 4 columns,It means  we shall have 4 rows and  3 columns in transpose of that matrix.

1st shift all the elements which are in 1st row to 1^st column as

 7

-1

-2

 5

shift all the elements which are in 2nd row to 2nd column as

4

4

3

3

shift all the elements which are in 3rd row to 3rd column as 

  8

 9

 6

-1

so shifting the corresponding Rows into Corresponding columns. We can that check same colour rows are transformed to same colour column.

Let us take an example  Where  A =

Step 1 Then  on transforming 1st Column  to 1st Row  ,we have 

4             -3         9          as 1st  Row

Step 2 Then  on transforming  2nd Column  to 2nd Row  ,we have 

5             2          -2          as  2nd Row

||ly    on transforming   3rd  Column  to 3rd Row  ,we have 

7          3          8             as 3rd Row

After  taking Transpose  A' will be  

Let us take more example to find out the transpose of matrices. These are two examples , both of which are of  3×3 orders. Hence the transpose of these matrices will be  again 3×3. 


1st consider matrix G , after transforming its 1st row into column , the 1st column of the transpose matrix of  G '  will be  -2,  -5,  4.
After transforming its 2nd row into column , the 2nd column of the transpose matrix of  G '  will be   -5,  7 ,  3. And after transforming its 3rd row into column , the 3rd  column of the transpose matrix of  G '  will be  4,  3,  8. If we write the transpose of  matrix G then we can see that there is no difference between the matrix G and the transpose of the matrix G.

Now consider matrix H , after transforming its 1st row into column , the 1st column of the transpose matrix of  G '  will be  2,  3,  4.      
After transforming its 2nd row into column , the 2nd column of the transpose matrix of  H '  will be   3, 5 ,  6. And after transforming its 3rd row into column , the 3rd  column of the transpose matrix of  H '  will be  4,  6,  7. Again in the case of matrix H , Matrix H and its Transpose matrix   H '  are same.

Here is more interesting Example of Transpose of this Matrix,This matrix have 3 Rows and 3 columns,after taking Transpose this matrix still have 3 Rows and 3 columns,

Now Take Transpose of this Matrix 

what ?

Surprise to see that the Transpose of some of the Matrices are the Matrices Itself, i. e. if A¹= A

Such Matrices are called Symmetric  Matrices.

Again if we take the transpose of the matrix given below and take -1 common from the matrix so obtained i .e  this matrix will be equal to transpose of the negative of the transpose of the given matrix

then   A¹ =  - A ,  Such Matrices are called  Skew Matrices.

Final words

This post was regarding what is transpose of matrix, calculate  transpose , define transpose,meaning of transpose , If you learn  something from this post then share it  with your friends  and also follow me on my blog ,We  shall meet  again in next post , till then  Good Bye ..................................

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