# How do you solve using gaussian elimination or gauss-jordan elimination, 3x + y =1 , -7x - 2y = -1?

Feb 1, 2016

This can be rewritten into an augmented matrix form:

$\left[\begin{matrix}3 & 1 & | & 1 \\ - 7 & - 2 & | & - 1\end{matrix}\right]$

where each entry is a coefficient associated with each variable, and the last column contains the answers to each equality.

To get this into row echelon form, we need to get leading $1$'s in each row such that the matrix is shaped like an upside-down stepladder and entries below each $1$ are $0$. If any rows are all $0$'s, then it has to be at the bottom.

To get it into reduced-row echelon form, get all numbers above AND below each leading $1$ to be $0$.

I am using the notation where the rightmost indicated row is the one that is operated upon.

$\stackrel{3 {R}_{1} + {R}_{2} \text{ }}{\to} \left[\begin{matrix}3 & 1 & | & 1 \\ 2 & 1 & | & 2\end{matrix}\right]$

$\stackrel{- {R}_{2} + {R}_{1} \text{ }}{\to} \left[\begin{matrix}1 & 0 & | & - 1 \\ 2 & 1 & | & 2\end{matrix}\right]$

$\stackrel{- 2 {R}_{1} + {R}_{2} \text{ }}{\to} \left[\begin{matrix}\textcolor{b l u e}{1} & 0 & | & \textcolor{b l u e}{- 1} \\ 0 & \textcolor{b l u e}{1} & | & \textcolor{b l u e}{4}\end{matrix}\right]$

From here we can just bring it back to algebraic form to get:

$\textcolor{b l u e}{x = - 1}$
$\textcolor{b l u e}{y = 4}$

We can check to see that this is correct by plugging these back in:

$3 \left(- 1\right) + \left(4\right) = - 3 + 4 = 1$
$- 7 \left(- 1\right) - 2 \left(4\right) = 7 - 8 = - 1$