# How do you solve using gaussian elimination or gauss-jordan elimination, 3x+2y = -9, -10x + 5y = - 5?

Mar 16, 2017

See below

#### Explanation:

First create the augmented matrix:

$\left(\begin{matrix}3 & 2 \\ - 10 & 5\end{matrix}\right) \left(\begin{matrix}- 9 \\ - 5\end{matrix}\right)$

For Gaussian elimination , we eliminate by combining rows to get a lower triangle matrix.

$R 2 \to R 2 + \frac{10}{3} R 1$
$\left(\begin{matrix}3 & 2 \\ 0 & \frac{35}{3}\end{matrix}\right) \left(\begin{matrix}- 9 \\ - 35\end{matrix}\right)$

Then back substitute, so from the second row:

$\frac{35}{3} y = - 35 \implies y - 3$

From first row:

$3 x + 2 y = - 9 \implies x = \frac{- 9 - 2 \left(- 3\right)}{3} = - 1$

For Gauss-Jordan elimination , we row reduce to the identity matrix ie $A m a t h b f x = m a t h b f b \to I m a t h b f x = m a t h b f b '$. So the first step can be the same:

$R 2 \to R 2 + \frac{10}{3} R 1$
$\left(\begin{matrix}3 & 2 \\ 0 & \frac{35}{3}\end{matrix}\right) \left(\begin{matrix}- 9 \\ - 35\end{matrix}\right)$

Then we can do this:
$R 1 \to R 1 - R 2 \cdot \frac{6}{35}$
$\left(\begin{matrix}3 & 0 \\ 0 & \frac{35}{3}\end{matrix}\right) \left(\begin{matrix}- 3 \\ - 35\end{matrix}\right)$

Finally we divide each row by its pivot, ie:

$\left(\begin{matrix}\frac{3}{3} & 0 \\ 0 & \frac{\frac{35}{3}}{\frac{35}{3}}\end{matrix}\right) \left(\begin{matrix}- \frac{3}{3} \\ - \frac{35}{\frac{35}{3}}\end{matrix}\right)$

$\implies \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}- 1 \\ - 3\end{matrix}\right)$