Table of Contents
- 0 1 0
- 0 0 1
- Product c*Ta will yield, 1 0 0 1 0 2
- 0 1 0 5 2 1
- 0 0 1 5 7 11
- Buy Linear Algebra paper online
- 5 2 1 0 1 0
- 5 7 11 0 0 1
- 5 2 1
- 5 7 11
- Ta is the transpose of A which is obtained as
- 0 1 0 5 2 1
- 0 0 1 5 7 11
- 5 2 1 1 2 1
- 5 7 11 1 1 2
- Therefore c[Ta]b will be; 12 9 16
- 16 11 15
- 16 16 25
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a)Find c[Ta]c
The matrix A is = | [ 1 5 5] [ 0 2 7] [ 2 1 11] |
. We find the matrix Ta=AT which is the transposed matrix of A |
Dimensions of matrix A are 3 x 3 thus the size of the matrix Ta will also be 3 x 3.
We find the elements of the matrix Ta to be:
c1 1 = a1 1 = 1; |
c2 1 = a1 2 = 5; |
c3 1 = a1 3 = 5; |
c1 2 = a2 1 = 0; |
c2 2 = a2 2 = 2; |
c3 2 = a2 3 = 7; |
c1 3 = a3 1 = 2; |
c2 3 = a3 2 = 1; |
c3 3 = a3 3 = 11; |
So, Ta = | [ 1 0 2] [ 5 2 1] [ 5 7 11] |
This is the transpose, Ta.
Therefore c(Ta)c where c= 1 0 0
0 1 0
0 0 1
We first get the product of c and Ta after which we will then multiply by matrix c once more.
Product c*Ta will yield, 1 0 0 1 0 2
0 1 0 5 2 1
0 0 1 5 7 11
The first matrix is a 3*3 matrix and the second one is also a 3*3 thus the answer should also be a 3*3 matrix. Therefore multiplying the first row of the first row of the first matrix by the first column of the second matrix and so on, we get the product is
(1*1)+(0*5)+(0*5) (1*0)+(0*2)+(0*7) (1*2)+(0*1)+(0*11)
(0*1)+(1*5)+(0*5) (0*0)+(1*2)+(0*7 (0*2)+(1*1)+(0*11)
(0*1)+(0*5)+(1*5) (0*0)+(0*2)+(1*7) (0*2)+(0*1)+(1*11)
This will be equal to, 1 0 2 multiplying this by c= 1 0 0 we get the final answer as
5 2 1 0 1 0
5 7 11 0 0 1
(1*1)+(0*5)+(0*5) (1*0)+(0*2)+(0*7) (1*2)+(0*1)+(0*11)
(0*1)+(1*5)+(0*5) (0*0)+(1*2)+(0*7 (0*2)+(1*1)+(0*11)
(0*1)+(0*5)+(1*5) (0*0)+(0*2)+(1*7) (0*2)+(0*1)+(1*11)
Thereforec(Ta)c will be, 1 0 2
5 2 1
5 7 11
b) Find c[Ta]b
Ta is the transpose of A which is obtained as
The matrix A is = | [ 1 5 5] [ 0 2 7] [ 2 1 11] |
. We find the matrix C=AT which is the transposed matrix A |
Dimensions of matrix A are 3 x 3 thus the size of the matrix Ta is also 3 x 3.
We find the elements of the matrix Ta to be:
c1 1 = a1 1 = 1; |
c2 1 = a1 2 = 5; |
c3 1 = a1 3 = 5; |
c1 2 = a2 1 = 0; |
c2 2 = a2 2 = 2; |
c3 2 = a2 3 = 7; |
c1 3 = a3 1 = 2; |
c2 3 = a3 2 = 1; |
c3 3 = a3 3 = 11; |
So, Ta = | [ 1 0 2] [ 5 2 1] [ 5 7 11] |
This is the transpose, Ta.
The product c[Ta]b will be obtained by:
Multiplying c by Ta first we get, we know that c= 1 0 0 and Ta= 1 0 2
0 1 0 5 2 1
0 0 1 5 7 11
This product will be:
The first matrix is a 3*3 matrix and the second one is also a 3*3 thus the answer should also be a 3*3 matrix. Therefore multiplying the first row of the first matrix by the first column of the second matrix and so on, we get the product will be
(1*1)+(0*5)+(0*5) (1*0)+(0*2)+(0*7) (1*2)+(0*1)+(0*11)
(0*1)+(1*5)+(0*5) (0*0)+(1*2)+(0*7 (0*2)+(1*1)+(0*11)
(0*1)+(0*5)+(1*5) (0*0)+(0*2)+(1*7) (0*2)+(0*1)+(1*11)
= 1 0 2 then multiplying this by matrix b= 2 1 1
5 2 1 1 2 1
5 7 11 1 1 2
The product yields:
(2*1)+(1*5)+(1*5) (2*0)+(1*2)+(1*7) (2*2)+(1*1)+(1*11)
(1*1)+(2*5)+(1*5) (1*0)+(2*2)+(1*7) (1*2)+(2*1)+(1*11)
(1*1)+(1*5)+(2*5) (1*0)+(1*2)+(2*7) (1*2)+(1*1)+(2*11)
Therefore c[Ta]b will be; 12 9 16
16 11 15
16 16 25
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