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 $u$?
 A: According to your working (which is correct), $u$ i.e. the third clmn of the matrix, is not a linear combination of the first two columns.
EDIT:
We are trying to find $u=(x,y)$ such that 
\begin{equation}\begin{bmatrix}1&-2\\ 2 & 1 \\-5 & 8 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}\end{equation}
The augmented matrix for the above matrix equation (linear system where we are trying to solve for $x,y$ is as given in the question). Using results regarding solvability of the matrix equation above (for example, or using brute force to try and determine $x,y$), we find that there is no solution (Rank of the augmented matrix is $3>2$ (as found by you), where $2$ is the rank of the matrix on the left of the above matrix equation. 
On the other hand, if we wanted to find $u$ such that 
\begin{equation}\begin{bmatrix}1&-2\\ 2 & 1 \\-5 & 8 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 5\\ -2\end{bmatrix}\end{equation}
we would find that 
\begin{equation}\begin{bmatrix}1&-2 &0\\ 2 & 1 &5\\-5 & 8 &-2\end{bmatrix}\sim\begin{bmatrix}1&0 &2\\ 0 & 1 &1\\0 & 0 &0\end{bmatrix},\end{equation} so $u=(2,1)$ (the last column of the reduced form of the augmented matrix (this also needs some thinking). This time the rank of the augmented matrix is $2$ (did not 'increase' as we added a new column, which tells us that the 'added' columns is a linear combination of the columns on its left in the (augmented) matrix.
A: 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.
A: What you need to solve is the system of linear equations:
$$\left\{\begin{align} x-2y&=3, \\ 2x+y&=1, \\ -5x+8y&=4. \end{align}\right.$$
The point of the Gauss elimination method (I recommend you check out this Wikipedia page!) is that we "encode" a system of equations by its augmented matrix, and then we apply the so-called elementary row operations to simplify it in a sense. Each row of the augmented matrix represents the respective equation. And each operation on the matrix corresponds to what we can do to equations in the system.
You correctly set up the augmented matrix as
$$\begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ -5 & 8 & 4 \end{bmatrix},$$
but why? (Disclaimer: traditionally we would separate the last column of an augmented matrix with a bar or a dashed line, but I don't remember how to do it in LaTeX.) Look at the coefficients in the first equation and at the numbers in the first row:
$$x-2y=3 \quad \text{or} \quad \color{green}{1}x+\color{blue}{(-2)}y=\color{magenta}{3} \quad \text{is represented by} \quad \begin{matrix} \color{green}{1} & \color{blue}{-2} & \color{magenta}{3} \end{matrix}.$$
Similarly for all the other rows and equations. Then, as I said, each step of the Gauss elimination method is equivalent to doing something to the equations of the original system. Each new matrix you obtain corresponds in the same way to a new system of equations, but all these systems are equivalent to the original one — in the sense that the solution set is preserved. For example, the second matrix that you obtained describes the follows system of equations:
$$\begin{bmatrix} 1 & -2 & 3 \\ 0 & -5 & 5 \\ -5 & 8 & 4 \end{bmatrix} \quad \text{represents} \quad \left\{\begin{align} x-2y&=3, \\ -5y&=5, \\ -5x+8y&=4. \end{align}\right.$$
And this new system is equivalent to the original one.
Now look at the last row of the last matrix and think what is the equation represented by that row:
$$\begin{matrix} \color{green}{0} & \color{blue}{0} & \color{magenta}{1} \end{matrix} \quad \text{represents} \quad \color{green}{0}x+\color{blue}{0}y=\color{magenta}{1}.$$
What values of $x$ and $y$ make this equation true?
