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 $u$?
Best Answer
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.