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I need help with solving this hw problem for my matrix algebra class.

Consider the transformation $T: \Bbb R^3 \to \Bbb R^2$ defined by:

$$ T(x) = T \pmatrix{x_1\\x_2\\ x_3} = (2x_1 + x_3) \pmatrix{1\\2} + (x_2 - 3x_3)\pmatrix{-1\\1} $$

1a) determine the matrix of the above transformation

1b) determine the reduced row echelon form of the matrix found in 1a

1c) based on your answer to part 1b, is the transformation T onto?

1d) based on your answer to part 1b, is the transformation T one-to-one?

1e) based on your answer to part 1b, determine the set of vectors x in R^3 for which T(x) = 0. Write your answer in parametric vector form

Would sincerely appreciate any help.

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  • $\begingroup$ What parts did you try? $\endgroup$ Oct 5 '16 at 13:38
  • $\begingroup$ im not sure how to find the matrix of the transformation $\endgroup$
    – Demuze28
    Oct 5 '16 at 13:39
  • $\begingroup$ Look up how to find a matrix for a linear transformation. The images of the basis vectors are the rows ... etc. $\endgroup$ Oct 5 '16 at 13:40
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Hint: For part a), it helps to rewrite $$ T(x) = \pmatrix{2x_1 - x_2 + 4x_3\\ 4x_1 + x_2 - x_3} $$ Now, for which matrix $A$ do we have $T(x) = Ax$?

If you go through the rest of the problem correctly, you should find that the transformation is onto, but not one-to-one. Solving part e) is the same thing as finding the nullspace of the matrix.

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  • $\begingroup$ is it not one-to-one because there are 2 solutions? $\endgroup$
    – Demuze28
    Oct 5 '16 at 14:09
  • $\begingroup$ It is not one-to-one because there are at least 2 solutions to the equation $T(x) = \vec 0$. $\endgroup$ Oct 5 '16 at 14:43
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For part (a), notice that

$$T(x) = (2x_1 + x_3)\begin{bmatrix} 1\\2\end{bmatrix} + (x_2 -3x_3)\begin{bmatrix} -1\\1\end{bmatrix} = \begin{bmatrix} 2x_1 + x_3\\4x_1 +2x_3\end{bmatrix} +\begin{bmatrix} -x_2 +3x_3\\x_2 -3x_3\end{bmatrix} = \begin{bmatrix} 2x_1 -x_2 + 4x_3\\4x_1 + x_2 -x_3\end{bmatrix} = \begin{bmatrix} 2 & -1 & 4\\4 & 1 & -1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}$$

Which implies that the matrix defining $T(x)$ is $$\begin{bmatrix} 2 & -1 & 4\\4 & 1 & -1 \end{bmatrix}$$

For 1e) I'm getting the reduced-form matrix as :

$$\begin{bmatrix} 2 & -1 & 4\\ 4 & 1 &-1 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & \frac{1}{2}\\ 0 & 1 &-3 \end{bmatrix} $$

Which gives the system of equations

$$ x_1 + \frac{1}{2}x_3 = 0 $$ $$ x_2 -3x_3 = 0 $$

Which implies that $$ x_1 = -\frac{1}{2}x_3 $$ $$ x_2 =3x_3 $$

Lastly, if we define $x_3 = t$, to solution in parametric form is then

$$ x_1 = -\frac{1}{2}t $$ $$ x_2 =3t $$ $$ x_3 =t $$

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  • $\begingroup$ Could you help me with 1e? I got [1 0 .5] as the rref [0 1 -3] $\endgroup$
    – Demuze28
    Oct 5 '16 at 14:13
  • $\begingroup$ Hopefully that's correct. I'd double check the reduced-form matrix part, as I did it rather quickly. The rest will show how to get it into parametric form, though. $\endgroup$ Oct 5 '16 at 14:25
  • $\begingroup$ the rref is wrong should be R1 = [1 0 .5] and R2 = [0 1 -3], also do i have to set that aug matrix to 0? in order to find the set of vectors x in R^3 for which T(x) = 0. $\endgroup$
    – Demuze28
    Oct 5 '16 at 14:39
  • $\begingroup$ So $T(x) = A[x_1, x_3, x_3]^T = 0$, where $A$ is the rref given above. If you do that out you'll end up with the last three equations I wrote. You can write this in vector notation as $x = [-\frac{1}{2}, 3, 1]^T t$ $\endgroup$ Oct 5 '16 at 14:46
  • $\begingroup$ r1 should be [1 0 .5] and is that parametric vector form? $\endgroup$
    – Demuze28
    Oct 5 '16 at 14:49

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