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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=A^T 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:

c_{1 1} = a_{1 1} = 1;

c_{2 1} = a_{1 2} = 5;

c_{3 1} = a_{1 3} = 5;

c_{1 2} = a_{2 1} = 0;

c_{2 2} = a_{2 2} = 2;

c_{3 2} = a_{2 3} = 7;

c_{1 3} = a_{3 1} = 2;

c_{2 3} = a_{3 2} = 1;

c_{3 3} = a_{3 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

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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=A^T 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:

c_{1 1} = a_{1 1} = 1;

c_{2 1} = a_{1 2} = 5;

c_{3 1} = a_{1 3} = 5;

c_{1 2} = a_{2 1} = 0;

c_{2 2} = a_{2 2} = 2;

c_{3 2} = a_{2 3} = 7;

c_{1 3} = a_{3 1} = 2;

c_{2 3} = a_{3 2} = 1;

c_{3 3} = a_{3 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

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(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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Free Custom «Linear Algebra» Essay Paper

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 Free Custom «Linear Algebra» Essay Paper 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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