Could someone explain in the simplest terms, how to multiply matrices?

Question

email me at [e-mail] if you like. Thanks for your help.

Answer 1

Any matrix is a rectangular table of numbers and it used to describe linear equations. The horizontal lines in a matrix are called rows and the vertical lines are called columns. A matrix with m rows and n columns is called an m-by-n matrix (written m×n) and m and n are called its dimensions. The dimensions of a matrix are always given with the number of rows first, then the number of columns.

Multiplication of two matrices is well-defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix (m rows, p columns) .
now how to make the multiplication process. assume any two matrices A 3x2 , B 2x2 first we check if we can get AB . and we can see the number of columns of A is the same as the number of rows of B. then the multiolication is possible . now you will multiply each row of A by each row of B to get the elements of AB ... If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix (m rows, p columns)
so , AB is a 3x2 Matrix

A=[ a b , c d , e f ], row1A= a b , row2A= c d , row3A= e f
B=[g h , i j ] --- row1B= g h , row2B= i j


AB= [ c1 c2 , c3 c4 , c5 c6 ]

the mult. process ,

row1A x row1B will gives c1= (a x g) + ( b x h )
row1A x row2B will gives c2= (a x i) + ( b x j)

row2A x row1B will gives c3= (c x g) + ( d x h )
row2A x row2B will gives c4= (c x i) + ( d x j )

row3A x row1B will gives c5= (e x g) + ( f x h )
row3A x row2B will gives c6= (e x i) + ( f x j )

Matrix multiplication has the following properties:

1- (AB)C = A(BC) for all k-by-m matrices A, m-by-n matrices B and n-by-p matrices C ("associativity").

2- (A + B)C = AC + BC for all m-by-n matrices A and B and n-by-k matrices C ("right distributivity").

3- C(A + B) = CA + CB for all m-by-n matrices A and B and k-by-m matrices C ("left distributivity").

It is important to note that commutativity does not generally hold; that is, given matrices A and B and their product defined, then generally AB ≠ BA


thats all.. i hope that i could help you :) you can contact me if you have any questions

Answer 2

First thing first. Check the dimensions of the individual matrices you are going to multiply. Dimensions means row number followed by column number.

If the column number of the first matrix is equal to the row number of the second matrix, then you can multiply them. If they are not equal then you cannot even do the multiplication. For example, a 2x3 matrix can be multiplied with a 3x2 matrix to produce a (please note) 2x2 matrix. In the dimension designation, the first number is the row and the second is the column. This means you will have four elements in your resultant(product) matrix.

The order of multipliplying elements is very, very important. Refer to an Algebra book with great illustrations showing the multiplication with arrows.

In any case, consider the matrices with the following elements,

A=
a b c
d e f
B =
g h
j k
m n

multiply a and g plus b times j plus c times m. The result you get is the first term of product matrix. Do the same format of multiplying and adding and you will have four elements in the final (answer) matrix.

Answer 3

You multiply them from corresponding row to column. So in case you have [2 3] and [5], then you definately multiply 2 by skill of 5 and a couple of. and 3 by skill of 5 and a couple of. write [2] you have got 2 gadgets of a million*2 matrices.

Answer 4

Though it is simple it may not be easy to explain without drawing

with drawing it is easy.

sat A Is of size (M,N) and B be of size (N,K) and terms are a(i,j) and
b(i,j) i mean ith row jth coulm.

the product shall be of C of size(M,K) element c(i,j)

now for c(i,j) pick the row i of A and column j of B.

multiply 1st element of row i of A with 1st element of column j of B and 2nd with 2nd till the last and add them you get C(i,j)

Answer 5

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