<s>[INST] Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 & 1& 4 \end{matrix}\right)^T$ Consider the linear map $T:\mathbb{R}^2->\mathbb{R}^3$ with 

$$T\left(\begin{matrix} x \\ y \end{matrix}\right)=\left(\begin{matrix} x-2y \\ 2x+y \\ -5x+8y \end{matrix}\right)$$ Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}\right)$

My try:
$$\left(\begin{matrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ -5 & 8 & 4 \end{matrix}\right)_{R_2->R_2-2R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ -5 & 8 & 4 \end{matrix}\right)_{R_3->R_3+5R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ 0 & -2 & 19 \end{matrix}\right)_{R_2->\frac{R_2}{5}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{matrix}\right)_{R_3->R_3+2R_2}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 17 \end{matrix}\right)_{R_3->\frac{R_3}{17}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}\right)_{R_2->R_2+R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1-3R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1+2R_2}$$
$$=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$
Then how to find $u$?
 [/INST] You just what to eliminate in the first two columns.
You are fine through here.
$\begin{array}{cc|c} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{array}$
But then then you hit the problem.  The vector you have been asked to find is not in the column space of the matrix.  Your problem has no solution.
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