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Matrix of linear transformation

  • Thread starter andrey21
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  • #1
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Matrix of linear transformation (urgent)

Identify the matrix of the transformation for the following:

a) (x,y,z) = (2x-y+4z,x+y-z,x-z)

b) (x,y) = (x,2x)

c) (x,y,z) = (x-2y,3x-6y)

Here are my attempts
a)
2,-1, 4
1, 1,-1
1, 0,01

b)
1,0
2,0

c)

1,-2, 0
3,-6, 0


Are theses correct? Thanks in advance:)
 

Answers and Replies

  • #2
33,505
5,191


Identify the matrix of the transformation for the following:

a) (x,y,z) = (2x-y+4z,x+y-z,x-z)

b) (x,y) = (x,2x)

c) (x,y,z) = (x-2y,3x-6y)

Here are my attempts
a)
2,-1, 4
1, 1,-1
1, 0,01

b)
1,0
2,0

c)

1,-2, 0
3,-6, 0


Are theses correct? Thanks in advance:)
In the first one you have what looks like a typo. The last term in the third row should be -1. You probably hit the wrong key. The other two are fine.
 
  • #3
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Ye sorry that should be -1 :) Next the question asks to find the basis for the Kernal for each. How do I go about doing this?
 
  • #4
33,505
5,191


That's kernel. The kernel is the subspace consisting of all vectors that are solutions to the equation Ax = 0. Start with that.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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955


By the way, it is a very bad idea to write something like:
"(x,y,z) = (2x-y+4z,x+y-z,x-z)"

That says the two are equal which is NOT what you mean. What you mean is that the linear transformation maps (changes) (x, y, z) into (2x- y+ 4z, x+ y- z, x- z) which might be written
T(x,y,z) = (2x-y+4z,x+y-z,x-z)

or simply (x,y,z) => (2x-y+4z,x+y-z,x-z) where the "arrow", =>, indicates the change.

The kernel of the matrix [math]\begin{bmatrix}2 & -1 & 4 \\ 1 & 1 & -1 \\ 1 & 0 & -1\end{bmatrix}[/math] is the set of vectors, [math]\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/math] such that
[math]\begin{bmatrix}2 & -1 & 4 \\ 1 & 1 & -1 \\ 1 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}[/math]

which is the same as the equations 2x- y+ 4z= 0, x+ y- z= 0, x- z= 0, which you could have got from the original definition of the linear transformation. Solve those equations for x, y, and z. Notice that one "obvious" solution is x= y= z= 0, the "trivial" solution. If there are other, non-trivial solutions, then there will be an infinite number of them. It is easy to show that the kernel is a subspace of the vector space- if u and v are non-zero members of the kernel of a linear transformation, so is any linear combination of them.
 
  • #6
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Sorry halls of ivy but ur post goes a little strange in the middle:)
 
  • #7
466
0


Ok so I have worked out the follwowing:

a) 2x-y+4z =0
b) x+y-z=0
c) x-z=0

therefore

x = z

equation b) becomes:

y = 0

and a) becomes:

6x=0

so us the basis of kernal x(000)??
 

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