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Let $T$, A linear transformation such that:

$$T\left[ {\begin{array}{*{20}{c}} x_1 \\ x_2 \\ x_3 \end{array}} \right] = \left[ \begin{array}{*{20}{c}} 2x_1 - x_2 + 5x_3 \\ - 4x_1 + 2x_2 - 10x_3 \end{array} \right]$$

What is the pre-image for $T^{ - 1}\left[ \begin{array}{*{20}{c}} 4 \\ - 8 \end{array} \right]$?

After row reduction we have:

$$\left[ \begin{array}{*{20}{c}} 1 & - \frac{1}{2} & \frac{5}{2} & 2 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Obviously, there are infinitely many solutions, but how to represent it?

Thanks in advance.

EDIT:


The offered book's solution is:

$$T^{ - 1} \left[\begin{array}{*{20}{c}} 4 \\ - 8 \end{array} \right] = \left\{ \left[ \begin{array}{*{20}{c}} 2 \\ 0 \\ 0 \end{array} \right] + x_2 \left[ \begin{array}{*{20}{c}} {1/2} \\ 1 \\ 0 \\ \end{array} \right] + {x_3}\left[ \begin{array}{*{20}{c}} -5/2 \\ 0 \\ 1 \end{array} \right] \mid x_{2,} x_3 \in F \right\}$$

Can you help me interpret it?

EDIT2:


Never mind, Got it. it's the same representation as offered here.

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    $\begingroup$ You have a map from R^3 to R^2. This shouldn't have 4 columns. $\endgroup$ – Batman Feb 10 '14 at 15:17
  • $\begingroup$ it's in the form of $Ax = b$, where the fourth row is $b$. I think it's fine. $\endgroup$ – AndrePoole Feb 10 '14 at 15:19
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    $\begingroup$ {x_{2,}}{x_3} instead of x_{2,} x_3 and a zillion other things like that. All those tons of extra curly braces can make editing more difficult. I deleted them. Maybe I missed some. $\endgroup$ – Michael Hardy Feb 10 '14 at 16:54
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Introduce free parameters s and t. $$ (x_1,x_2, x_3) = (2+s/2-5t/2,s,t) $$

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  • $\begingroup$ Firstly, Thanks. Can you have a look at my edit please? $\endgroup$ – AndrePoole Feb 10 '14 at 15:28

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