How to solve this matrix using gauss-elimination by hand I feel like i am having a brain fart. I have been given this $Ax=b$ system:
$A= \begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix}$
$b= \begin{pmatrix} 0.254 \\ 0.217 \end{pmatrix}$
I know the answer is $x_1 = 1$ and $x_2 = -1$ but for some reason when i try to solve this using Gaussian elimination, the entire bottom row goes to $0$ giving me $1$ equation with $2$ variables.
Can someone get me out of this please? 
 A: Your cerebral flatulence may arise from the fact that
$$913\cdot563-780\cdot659=-1$$
so that the determinant of your matrix is $(-1)\times10^{-6}$, which might be causing roundoff errors.  I would recommend scaling everything up to 3-digit integers and seeing if you have better luck.  (Scaling things up won't cure the ill-conditioned nature of the matrix, it'll hopefully just make things a little more clear.  The problem is small enough that doing it in exact arithmetic, with fractions rather than decimals, is eminently practical.)
A: Your matrix is ill conditioned.  That means that although the matrix is regular but very small changes can make it singular.  I guess this problem was given in some class on numerical linear algebra.
If you multiply the first row of $A$  with $\frac{0.780}{0.913}$ you get $\begin{pmatrix} 0.780 & 0.5630011 \\ 0.780 & 0.563 \end{pmatrix}$. So you see that rows of $A$ are "almost" linear dependent. That means the matrix has "almost" rank one.  Maybe that gives you an intuitive idea of its ill conditioned nature.
A: Alright I worked it out step by step, so bear with me (I'll use A followed by b notation):
\begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix} \begin{pmatrix} 0.254\\ 0.217 \end{pmatrix}
We multiply the top row by the bottom and vice versa:
\begin{pmatrix} 0.71214 & 0.51402 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} 0.19812\\ 0.198121 \end{pmatrix}
Subtract the bottom from top:
\begin{pmatrix} 0 & 0.000001 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} -0.000001\\ 0.198121 \end{pmatrix}
Divide the top row by 0.000001 to reduce it to:
\begin{pmatrix} 0 & 1 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} -1\\ 0.198121 \end{pmatrix}
Multiply the top row by 0.514019 and subtract it from the bottom to get:
\begin{pmatrix} 0 & 1 \\ 0.71214 & 0\end{pmatrix} \begin{pmatrix} -1\\ 0.71214\end{pmatrix}
Diving the bottom row by 0.71214 we end with:
\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix} -1\\ 1\end{pmatrix}
from which we can readily see x1 = 1 and x2 = -1
