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D www.webassign.net/web/student/Assignment Responses submit?depals132329 My Notes o Ask Your Teacher -5 points WillLinAlgANTB 6 R012. Determine the kernel and range of the transformation defined by the matrix (Enter your answers as a comma separated list. Enter each vector in the form 2 4 (x1, x2, Use r for any arbitrary scalar ker(T) range (T show that dim ker dim range dim domain Th. dim ker dim ra dim domain Ask Your Teacher 5 points WilLunAlgalt8 6 R 013 Find bases for the kernel and range of the transformation fined by the matrix 0 1 5 Enter your answers as a comma separated list. Enter each vector in de the form x

Solution

4

8

2

4

We will reduce A to its RREf as under:

Multiply the 1st row by ¼

Add -2 times the 1st row to the 2nd row

Then the RREF of A is

1

2

0

0

Since the 2^nd column of A is a multiple of its 1^st column, Range (T) = Col(A) = span{(4,2)^T}. Also, Ker(T) = Null(A), is the set of solutions to theequation AX = 0. Let X = (x,y)^T. Then this equation is equivalent to x+2y = 0 so that x = -2y and X = (-2y,y)^T = y(-2,1)^T. Hence Ker(T) = span{(-2,1)^T}.

2. Let A =

1

2

5

0

1

-5

2

5

5

We will reduce A to its RREf as under:

Add -2 times the 1st row to the 3rd row

Add -1 times the 2nd row to the 3rd row

Add -2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

15

0

1

-5

0

0

0

Apparently, the 3^rd column of A is a linear combination of its first two columns.

Hence a basis for Range (T) = Col(A) is { (1,0,2)^T, (2,1,5)^T}.

Also, Ker(T) = Null(A), is the set of solutions to theequation AX = 0. Let X = (x,y,z)^T. Then this equation is equivalent to x +15z=0 and y-5z=0 so that x =-15z and y=5z and hence, X = (-15z,5z,z)^T = z(-15,5,1)^T.Thus, a basis for Ker (T) is { (-15,5,1)^T}.

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