Matrix Operations-part-2.
I have added more data as compared to the video content, especially, a simple way to multiply two matrices, by p to get a11,a12, and a13, we place the first row of the first matrix, then we put the three columns of the second matrix but in rows, this enables us to perform the multiplication process easily. please refer to the next slide image.
In a discussion of the properties of addition through a solved problem, the data are quoted from Prof. Kuldeep Singh’s handbook.
Powers of a matrix.
The next property of the matrix operations-part-2 is the powers of the matrix. For instance, if we have matrix A with dimension (2×2), it consists of 4 elements as follows(1,2,0,-1).
A raise to the power of 2 or A^2, can be done by multiplying matrix A by itself.
The matrix A dimension is(2×2), then the final matrix will be with dimension (2×2) as well.
Let us perform the multiplication, for the first element its value= (1*1+2*0). The first element at the first row with the second column=1.For the element of the 2nd row, with the first column, its value=(0*1+-1*0)=0.
For the first row, the second column element value can be found by multiplying (1*2+2*-1)=2+(-2)=0.
For the last element value of the second row, the second column can be estimated as (0*2+(-1*-1)=+1. The A^2 can be arranged as follows.
Another way is to write the first row of matrix A, above that line we can write the first column of B, which in this case is A itself as a row, and also another row which is the second column of B.ToGet a11, multiply(11+20)=1. While for the a12 multiply(12+2-1)=0.
To get the value of a21, write the second row of A as a row, above that row, write the first column of B as a row, and also the second column of B as a row as well. The values of a21 and a22 are shown.
Multiplication of 3×3 matrices.
For the next operation of multiplying two matrices AxB, each matrix has a dimension of (3×3) that is a part of the Matrix operations-part-2
We will write between brackets(3×3). While matrix B has a dimension of (3×4). For the final matrix or (AB). First, check that Figure 3, which is the number of columns in matrix A matches Figure 3 of matrix B, which is the number of rows.
The final matrix (AB) dimension can be obtained by taking the leftmost number in matrix A and multiplying it by the rightmost number in matrix B.
Matrix A is at the left side of matrix B, multiplication is to be performed. If we change the arrangement to let matrix B be on the left side of matrix A.
For the given example, the product can not be done, since (3×4)*(3*3) cannot be multiplied.
Even if these matrices have (3*3)*(3*x3) the product BA will not be the same as AB. We will proceed to do the multiplication of (AB), and we will find that the first element = 3*2+ 5*5+(-1*9)=22, as we can see.
To determine the place of the required element, assume that this element is the intersection of the tail of the arrow for the first arrow with the tail of the first column arrow. For the first row, the second element of(AB).
The element is located at the intersection between the horizontal arrow and the vertical arrow. The value=3*-2+5*0+(-1*-4))=-2. For the element at the first row, the third column, again.
The intersection between the horizontal arrow with the vertical arrow. The value=33+ 5 7-(1*1)=43.
While for the first row, the fourth column value. it will be=3*1+5*8-1*1=42. Now proceed to the second row, when intersecting with the first column, will give the element of the second row/ first column, the value = 4*2+0*5+2*9=26.
The same process will be carried on for the other elements. The final AB value is shown. The final dimension is(3×4) or three rows with four columns. We cannot perform the operation of multiplication of B by A, since the number of rows in matrix B is not the same as the number of columns in matrix A. To check that AB doesn’t equal BA.the next slide image illustrates the process of getting the Matrix BA.
This is the next post- Solved problem for matrix operations.
For a useful external link, math is fun for the matrix part.
