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Info given:

There is a linear transformation $T$: $\mathbb{R}^5 \rightarrow \mathbb{R}^4$ given by $$ T\left(\begin{array}{r} \mathbf{x} \end{array}\right) = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2_x2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right] \quad \text{for} \quad \mathbf{x} = \left[\begin{array}{r} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{array}\right] \in \mathbb{R}^5 $$

First I've found the matrix $A$ that fulfulls $T(x) = Ax$ for all $x \in \mathbb{R}^5$.

After that, I am told that we should let $y = ($$y_1$ $y_2$ $y_3$ $y_4$)$^T$ $\in \mathbb{R}^4$ be an arbitrary (but unknown) vector.

Question/task:

My task is then to find a vector $x \in \mathbb{R}^5$ (expressed by the unknowns $y_1, y_2, y_3, y_4$) that fulfills $T(x) = y$

My try:

I have tried myself and end up with the following: $$x = \frac{y}{A}$$

$$= \frac{\left(\begin{matrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{matrix}\right)}{\left(\begin{matrix} 2 & -4 & -1 & -3 & 2 \\ -1 & 2 & 1 & 0 & 1 \\ 1 & -2 & -1 & -3 & -1 \\ -1 & 4 & -1 & 0 & 1 \end{matrix}\right)}$$

$$=\left(\begin{matrix} \frac{y_1}{2-4-1-3+2} \\ \frac{y_2}{-1+2+1+1} \\ \frac{y_3}{1-2-1-3-1} \\ \frac{y_4}{-1+4-1+1} \end{matrix}\right)$$

I'm not sure if this is correct or if the approach is correct. Any help would be so appreciated!

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  • $\begingroup$ Dividing by a matrix? That doesn't make sense. $\endgroup$ – mattos May 15 '19 at 10:56
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With all due respect, but $x = \frac{y}{A}$ is non-sense !

To find $x$, you have to solve the linear system $Ax=y$.

https://en.wikipedia.org/wiki/System_of_linear_equations

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  • $\begingroup$ So glad I asked here first! Thanks, Fred $\endgroup$ – Lubbi May 15 '19 at 10:59
  • $\begingroup$ Fred do you mind show me your approach then? I am lost at this. $\endgroup$ – Lubbi May 15 '19 at 11:01
  • $\begingroup$ Can this answer be corret? $$\left(\begin{matrix} 2*y-11*x_5 \\ y-4*x_5 \\ y-4*x_5 \\ \frac{-2*y}{3} \\ x_5 \end{matrix}\right)$$ $\endgroup$ – jubibanna May 15 '19 at 14:07

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