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 Linear Algebra essay

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## Linear Algebra. Custom Linear Algebra Essay Writing Service || Linear Algebra Essay samples, help

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

### Linear Algebra. Custom Linear Algebra Essay Writing Service || Linear Algebra Essay samples, help

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