i got this problem ( very trivial I guess)
$39x -51y =15$
$-52x + 68 = -20$
I've done the Gauss reduction and got this,
Matrix:
\begin{pmatrix} 1 & \frac{-17}{13} & \frac{5}{13} \\ 0 & 0 & 0 \\ \end{pmatrix}
The answer is this: the point where the line goes through is $(3,2)$ and directional vector is $(17,13)$ and i don't know how i get this.
The Gauss-reduction is right. The problem is that i don't know which method to use to find the answer. Can i get some help from here? Thanks
Since the matrix has rank one (txo rows; one of them consisting of zeros), the solution of the system is a line, defined by the elements of the first row: $$1\cdot x + \left(-\frac{17}{13}\right)\cdot y = \frac{5}{13}$$ That is: $$13x-17y = 5 \tag{L} $$
The directional vector of a line given by $ax+by=c$ is the vector $\vec{v}=(-b,a)$, or any multiple of it, $s\vec{v}$ with $s\in\mathbb{R}\setminus\{0\}$.
Regarding the point, take any $x\in\mathbb{R}$, substitute it in (L) and solve the remaining to obtain its corresponding $y$.