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

For more Mathematics Related Contents click here

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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MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS

Hello every one ,Welcome once again, Today we are going to discuss What is the  matrix, elements of a matrix, order of a matrix, different types of matrix , transpose of matrix, ad joint of matrix, determinant  and  how to find the determinant of a matrix .

What is  Matrix

Matrix definition

A Matrix is a set of elements ( Numbers ) arranged in a particular numbers of Rows and columns in a rectangular table. Matrices inside parentheses ( ) or brackets [ ] is the matrix notation.

Here we have an examples of Matrices . 

The elements which are written in horizontal lines are called Row and elements which are written in Vertical lines are called Column. In the matrix A above ↑ the elements 5, 3, -2 are written in Row wise whereas the elements 5,4,3 are written columns wise, Similarly the elements 4,-1,7 are written in Row wise whereas the elements 3,-1,4 are written columns wise .

Order Of a Matrix

If any matrix have "m" number of rows and "n" number of columns , then "m×n" will be the order of that matrix. It is also called matrix dimensions . For  matrices  given below ,

[      5      6    -4    2    ]   This  matrix has 1×4 order,

    3

[  8 ]  Matrix has 3×1 order ,

   -2 

  [  -1 ] Matrix has 1×1 order.

And the matrices A,B,C and D  above have 3×3 ,2 ×2 , 3×4 and 3×2  respectively, as Matrix A has 3 rows and 3 columns, matrix B has 2 rows and 2 columns, Matrix C has 3 Rows and 4 Columns Similarly Matrix D has 3 rows and 2 columns.

Elements of a matrix

The Elements are entries or numbers used in the matrix . These are denoted by aij , where i is the row's number and j is the column's number in which the element is lying.

Consider the matrix A above which has elements  5 ,3,-2 in the 1st rows 4, -1,7 in the 2nd row and   3,4, -1 in the 3rd rows.

Types of matrix

Equality of Matrices

Two matrices are said to be equal ,if they have same order and same  elements at corresponding positions.

                                                                                                                                                              

If we compare matrices A,B,C and D given above then the matrices A and B may be equal to each other provide x = 6 and y = 3, Similarly Matrices C and D may be equal to each other provided x = 8 ,y = 7 and z = 3 .

Note that Matrices A and C can not be equal to each other, because A and C have not same order. Similarly B and D can not be equal to each other as B and D have different order.

Square Matrix

Any Matrix which have equal numbers of rows and columns,Then   That Matrix is called Square matrix .  A,B,C,D,E and F Matrices given below and above  are the examples of Square Matrix.

Zero , Scalar And Diagonal Matrix

Zero Matrix

If all the elements of a matrix are equal to zero then the matrix  is called Null Matrix ,It is also called Void Or Null Matrix. Matrices A  ,B and C given below are the example of Zero Matrices, As all these matrices have all the elements  zeros. so these matrices are Zero Matrices.

Diagonal Matrix

A Diagonal matrix is a square matrix in which all the elements are zero except diagonal elements . Matrices D ,E and F given below are examples of   Diagonal matrix of  2 × 2 Diagonal matrix and 3×3 orders . 

Scalar Matrix

A Diagonal matrix  matrix in which all the elements are zero and diagonal elements are equal to each others. Matrices A,B,C,D,E and F are examples of Scalar Matrices .Given Matrices D ,E and F given below are examples of   Diagonal matrix of  2 × 2 Diagonal matrix ,3×3  and 3×3 orders respectively . 

Identity Matrix

The Identity Matrix, symbolised as I, is a square matrix. where all the elements are 0 except the diagonal elements  , and all the diagonal  elements equal to one. Identity matrix is also called Unit Matrix.

Above two  examples are  3 × 3 and 2×2 Identity matrices respectively . 1st matrix is an example of  3 × 3 matrix and 2nd is an example of 2×2 Identity matrix. Because both the matrices have diagonals elements one and remaining elements  zeros. Identity Matrices and Unit Matrices are also the examples of Scalar Matrices

Row matrix

Any matrix which has only one row is called Row Matrix.  The  Row matrix may have any numbers of  columns .

[      5      6    -4    2    ] ,     [  2    5   ] ,    [-4    6    -6    0 ]

,[      7      4     3    -3    ]

Column Matrix

Any matrix which has only one column is called Column Matrix. The  Column matrix may have any numbers of Rows.

  

    3

[  8 ] ,      [  -1 ] 

    -2

Transpose of matrix

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.

To find the transpose of any matrix , Shift all the elements of all the Rows into respective Columns. Given below matrices have their transpose written on the right side of them.

It  can be seen that in all the examples given above all the elements of particular row have been changed to corresponding column. So A' is the transpose of  Matrix A ,similarly B' is the transpose of  Matrix B .

Determinant of a matrix

A determinant is a square array of numbers which represents a certain sum of products. we can find out a fixed value of determinant, consider an example of a 3 × 3 determinant , it has 3 rows and 3 columns).

Minor of an Element

The minor of an element aij is the determinant obtained by deleting the ith row and the jth column and is denoted by Mij .

Co factor of an Element

The co-factor of an element aij is the determinant obtained by deleting the ith row and the jth column and is multiplied by(-1)^(i+j)and is denoted by Cij. and Cij= (-1)^(i+j)×Mij

We can find the value of a 2 × 2 determinant   as follows  ,1st we multiply the  top left × bottom right first then subtract from it the product of top right element and left bottom. or 

(1st element in 1st row) × [Its Co factor] - (2nd element in 2nd row)× [Its Co factor].

 

Determinant value of above matrix is 4×5-(-3)×5= 20+15=35

How to find the determinant of a 3x3 matrix  

To find the determinant value 3×3 determinant .

1st element in 1st row ×[Its Co factor]-2nd element in 2nd row[Its Co factor]+3rd element in 3rd row [Its Co factor ].

Let us calculate the determinant value of 3×3 matrix  given above 

=2[(-4×-7) - (2×5)] - (-1) [3×(-7) - (2×5)]+ 5[(3×5) - (-4)×5]

=2[28-10]+1[-21-10]+5[15+20]

=36 - 31+75

=80

Ad joint of a Matrix

The transpose of a co factor matrix of any matrix is called ad joint of the Matrix.To find the ad joint Matrix ,1st find the co factors of  all the elements of given Matrix A.
Co factors of 1st row of  Matrix A are 7 and 5
Co factors of 2nd row of Matrix  A are -2 and -3

,then form the Matrix of these  co factors and name it Co factor  matrix, and after that take the transpose of the co factor matrix so formed.Then we have  transpose of the Ad joint of Matrix.

Conclusion

Thanks for giving your valuable time to the post "What is matrix, element of matrix, dimension of matrix, different types of matrix, transpose of matrix, ad joint of matrix, what is a determinant ,  determinant of 3x3 matrix ,  determinant of a 2x2 matrix " of this blog .If you found this post helpful to you , then share it with yours friends and family members . Also follow me on my blog for notifications of next posts.We shall meet again in next interesting and educating post , till then Good Bye. Take care ....

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SHORTEST METHOD OF MULTIPLICATION FOR TWO NUMBERS

We are going to learn shortest method of multiplication for two numbers which are very close to 100 ,how to multiply two numbers quickly, which can save our time in competitive exams .

Let us understand this multiplication shortcut with the help of this video.

Let us choose two numbers   98 and  96 nearer to 100 
98*96 = (98-4)(2*4) = (94)(08)=9408 (Only in mind )
Step 1 

Consider both the numbers as Num 1 = 98 and Num2 = 96.

Subtract Num 1 from 100 and write its Result 1 as one place .

Also Subtract Num 2 from 100 and write its as Result 2 second place .

Here Result 1 = 100 - 98 = 2
Result 2 = 100 - 96 = 4
Now multiply the results so obtained and mark it as Stepresult1.Stepresult1 = 4*2 = 8 = 08

Step 2 

Subtract the result of  Result 1 (blue Answer) from Num 2 i. e.(96 ).

Here we have
Stepresult2 = 96-2= 94

Final Answer = (Ist two digits are Stepresult2)(2nd two digits are Stepresult1)

= 9408
The result so obtained i. e. 9408 is the answer of product of two numbers

Let us consider  one more example   to fast way to multiply two digit numbers.

Another Example

92 * 91 = ?

Step 1

Last two digits of the answer = (100-92)(100-91)
= 8* 9 = 72

Step 2

1st two digits of the answer = (Greater number as Red highlighted in last step - Smaller number in last step Blue highlighted = 92 - 9 = 83

Final Answer = (Answer of Step 1)(Answer of Step 2) = 8372

 Example   89 * 92 = (92-11)(11*8)=8188

Let us discuss step by step the process of  multiplying these two numbers quickly

Step 1

Split Both the numbers in such a way that one the term must be 100 and other be 100-1st number. i. e. 89 will be written as 100-11 and 92 will be written as 100-8

Step 2

Now select either of the given number (suppose 92 ) and subtract the number ( other than 100 ) which is in other factor in previous step i. e. 11.

Step 3

The difference of 92-11 will the 1st two digits of our answer and product of other two digits are the 3rd and 4th digits of the answer.

One more Example

88×93 = (Answer of Step 1)(Answer of Step 2)
               =   (93-12)  [(100-88) * (100-93)]
            = (93-12)[12*7]
             = (81)(84)
               =8184
94×84 =   (As 94 is 6 less than 100)(As 84 is 16 less than 100)
             = (subtract 16 from 94)(Multiply 6 with 16)
            = (94-16 gives 1st two digits of product) (96 will be last two      digits of product)
Final answer = 7896
You can also learn it as 
93 * 87 = (Greater number - How much second number closer to 100 ) ( (100-93)(100-87))

93 * 87 = (93-13)(7*13)

93*87 = 8091

Conclusion

This method was how to multiply two numbers easily and to multiply two numbers using a shortcut . Thanks for spending your precious time to read this post ,If you liked this post . Please share it with your friends and also follow me on my blog to encourage me to do better than best  for multiplication,  multiplication tricks and   
vedic maths tricks .See your in next post, till then Bye.....

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HOW TO MULTIPLY A NUMBER BY 11 USING VEDIC MATHS TRICKS

Let us discuss the how to multiply two numbers quickly, how to multiply two numbers easily,how to multiply two numbers fastly , fast way to multiply two digit numbers, fastest way to multiply two digit numbers any number however large with 11 in fraction of seconds. With the help of this method , we can multiply any number in just 2 - 3 seconds.

1 Place zero at right side of the multiplicand.

2 Keep on adding each digit of the multiplicand from extreme right to its neighbour till end , if any stage sum is obtained greater than 10 then carry of 1 will be added to next step.

Let us understand it better with the help of video

• Examples 

52324 × 11= ?

1 Place zero at right end of the multiplicand like this 523240

2 Now add 0 to its neighbour 4 as 0 + 4 = 4

3 Now add 4 to its neighbour 2 as 4 + 2 = 6

4 Now add 2 to its neighbour 3 as 2 + 3 = 5

5 Now add 3 to its neighbour 2 as 3 + 2 = 5

6 Now add 2 to its neighbour 5 as 2 + 5 = 7

7 Place left most digit as it is = 5

8 Write the digits so obtained  from top to bottom as right to left

So Answer will be 5,75,564

This video demonstrate the process of multiplication very easily

• Examples 

4543423 × 11= ?

1 Place zero at right end of the multiplicand like this 45434230


2 Now add 0 to its left neighbour 3 as 0 + 3 = 3


3 Now add 3 to its left neighbour 2 as 3 + 2 = 5


4 Now add 2 to its left neighbour 4 as 2 + 4 = 6


5 Now add 4 to its left neighbour 3 as 4 + 3 = 7


6 Now add 3 to its left neighbour 4 as 3 + 4 = 7


6 Now add 4 to its left neighbour 5 as 4+ 5 = 9


6 Now add 5 to its left neighbour 4 as 5 + 4 = 9


7 Place left most digit as it is = 4

8 Write the digits so obtained  from top to bottom as right to left

So Answer will be 4,99,77653

3598678 × 11= ?

1 Place zero at right end of the multiplicand like this 3598678

2 Now add 0 to its neighbour 8 as 0+8= 8

3 Now add 8 to its neighbour 7 as 8+7=15 write 5 and carry 1 to next step

4 Now add 7 to its neighbour 6 as 7+6=13+1 (carry)=14 write 4 and carry 1 to next step

5 Now add 6 to its neighbour 8 as 6+8 = 14+1(carry) = 15 write 5 and carry 1 to next step

6 Now add 8 to its neighbour 9 as 8+9=17+1(carry) =18 write 8 and carry 1 to next step

7 Now add 9 to its neighbour 5 as 9+5=14+1(carry) = 15 write 5 and carry 1 to next step

8 Now add 5 to its neighbour 3 as 5+3=8+1(carry) = 9

9 Place left most digit as it is = 3

11 Write the digits so obtained from top to bottom as right to left


So Answer will be 3,95,85458

8923586 × 11 = ?

1 Place zero at right end of the multiplicand like this 89235860

2 Now add 0 to its neighbour 6 as 0+6 = 6

3 Now add 6 to its neighbour 8 as 6+8=14

write 4 and carry over 1 to next step

4 Now add 8 to its neighbour 5 as 8+5=13+1(carry) = 14 

write 4 and carry over 1 to next step

5 Now add 5 to its neighbour 3 as 5+3=8+(1)carry= 9

6 Now add 3 to its neighbour 2 as 3 + 2 = 5

7 Now add 2 to its neighbour 9 as 2+9=11 

write 1 and carry over 1 to next step

8 Now add 9 to its neighbour 8 as 9+8=17+(1)carry = 18 

write 8 and carry over 1 to next step

9 Now add 1(carry ) to its neighbour 8 as 1+8 = 9

Write all  the digits so obtained  from top to bottom as right to left.

So answer will be 98,159,446

35681237 ×11 = ?

1 Place right most digit 7 of multiplicand as right most digit of answer.

2 Keep on adding right sided digit to its left sided digit in pairwise.

3 If the sum at any time is found to be more than 10, then take "1" as carry over to next step every time.

4 Repeat the process till last digit. So After 1st step we shall have 7

After 2nd step we shall have 7+3=10 =0 (right sided digit of 10 ) and 1 as carry to next step.

After 3rd step we shall have 3+2=5+1=6 and no number as carry to next step.

After 4th step we shall have 2+1=3 and no number as carry to next step.


Similarly we get 1+8=9,


and 8 + 6 = 14 = 4 as (right sided digit of 14 ) and 1 as carry to next step.


5 + 6 = 11 + 1 = 12 = 2 (right sided digit of 12), and


3+8 = 8 + 1 = 9;

And the last digit = 3

Now write all the highlighted digits from bottom to top .

So Answer will be    392493067

These are some of the examples demonstrated in the video given below

Application of this Method

If we have to multiply 666854×55
then rewrite given product as 666854×(11×5)
Now multiply 666854×11 as follows
Step 1

Place right most digit 4 as result and keep on adding the digits to its left one by one which gives 7335394, and
Step 2

Now place 0 as right most digit of this result i. e.73353940 ,


Step 3 

Now divide with 2 we get 36676970 and this is the Final answer.
Example

Let us multiply 35987604 × 55

Rewrite 35987604 × (11×5)

Multiply 35987604 × 11 = 395863644

Now place "0" at extreme right of this number it become 3958636440 , 

Now divide this number with 2 to get the Answer 1979318220.
Now Fast multiplication with one more Example

69852364639×55

Step 1 

 1st multiply the given number with 11 by placing and adding digits from left to right 9,12,9,10,10,9,5,7,13,17,15,6 (if total greater than 10 ,carry 1 to next number) like this 9,2,0,1,1,0,6,7,3,8,6,7.

Step 2

 Write these numbers from left to right, place zero at end and divide by 2 to get the answer like this  7683760110290 → 384,188,005,5145.

Conclusion

This shortcuts method was to multiply a number with 11 . Thanks for spending your precious time to read this post ,If you liked this post . Please share it with your friends and also follow me on my blog to encourage me to do better than best. See your in next post, till then Bye.....

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