How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 3y = -2, -6x + y = -14?

Jun 11, 2017

$\textcolor{b r o w n}{\left(\begin{matrix}1 & 0 & | & + 2 \\ 0 & 1 & | & - 2\end{matrix}\right)}$

$x = 2 \mathmr{and} y = - 2$

Explanation:

Using the form:

$\left(x , y , | , \text{answer}\right)$

$\textcolor{b r o w n}{\left(\begin{matrix}2 & 3 & | & - 2 \\ - 6 & + 1 & | & - 14\end{matrix}\right)}$
$\text{ } \left(- 3\right) \times R o {w}_{1}$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}- 6 & - 9 & | & + 6 \\ - 6 & + 1 & | & - 14\end{matrix}\right)}$
$\text{ } R o {w}_{1} - R o {w}_{2}$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}0 & - 10 & | & + 20 \\ - 6 & + 1 & | & - 14\end{matrix}\right)}$
$\text{ } R o {w}_{1} \div \left(- 10\right)$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}0 & + 1 & | & - 2 \\ - 6 & + 1 & | & - 14\end{matrix}\right)}$
$\text{ } R o {w}_{2} - R o {w}_{1}$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}0 & + 1 & | & - 2 \\ - 6 & 0 & | & - 12\end{matrix}\right)}$
$\text{ } R o {w}_{2} \div \left(- 6\right)$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}0 & + 1 & | & - 2 \\ + 1 & 0 & | & + 2\end{matrix}\right)}$

Write in row echelon form

$\textcolor{b r o w n}{\left(\begin{matrix}1 & 0 & | & + 2 \\ 0 & 1 & | & - 2\end{matrix}\right)}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check that $x = 2 \mathmr{and} y = - 2$

$2 x + 3 y = - 2 \text{ "->" } L H S \to 2 \left(2\right) + 3 \left(- 2\right) = 4 - 6 = - 2$

$- 6 x + y = - 14 \text{ "->" } L H S \to - 6 \left(2\right) + \left(- 2\right) = - 14$