# How do you solve 3x+3y=15 and 2x+3y=-5 using matrices?

Feb 25, 2016

Either use the Gauss-Jordan method or Cramer's Rule:
$\textcolor{w h i t e}{\text{XXX}} \left(x , y\right) = \left(20 , - 15\right)$

#### Explanation:

In either case we will need to rewrite the given equations into matrix form:

$\left(\begin{matrix}3 & 3 & \textcolor{b l u e}{15} \\ 2 & 3 & \textcolor{b l u e}{- 5}\end{matrix}\right)$

Gauss-Jordan
Divide row 1 (by $3$ so that the leading term is 1
$\textcolor{w h i t e}{\text{XXX}} \left(\begin{matrix}1 & 1 & 5 \\ 2 & 3 & - 5\end{matrix}\right)$
Subtract $2 \times$ this new row 1 from row 2 to reduce its leading term to 0
$\textcolor{w h i t e}{\text{XXX}} \left(\begin{matrix}1 & 1 & 5 \\ 0 & 1 & - 15\end{matrix}\right)$

Subtract this new row 2 from row 1:
$\textcolor{w h i t e}{\text{XXX}} \left(\begin{matrix}1 & 0 & 20 \\ 0 & 1 & - 15\end{matrix}\right)$
The first column is $x$, the second is $y$ and the third is the value they are equal to.

Cramer's Rule
$D = \left(\begin{matrix}3 & 3 \\ 2 & 3\end{matrix}\right) \textcolor{w h i t e}{\text{XX")D_x=((color(blue)(15),3),(color(blue)(-5),3))color(white)("XX}} {D}_{y} = \left(\begin{matrix}3 & \textcolor{b l u e}{15} \\ 2 & \textcolor{b l u e}{- 5}\end{matrix}\right)$

The determinants:
$\textcolor{w h i t e}{\text{XXX}} \left\mid D \right\mid = 3 \times 3 - 2 \times 3 = 3$

$\textcolor{w h i t e}{\text{XXX}} \left\mid {D}_{x} \right\mid = 15 \times 3 - \left(- 5\right) \times 3 = 60$

$\textcolor{w h i t e}{\text{XXX}} \left\mid {D}_{y} \right\mid = 3 \times \left(- 5\right) - 2 \times 15 = - 45$

$x = \frac{\left\mid {D}_{x} \right\mid}{\left\mid D \right\mid} = \frac{60}{3} = 20$

$y = \frac{\left\mid D y \right\mid}{\left\mid D \right\mid} = \frac{- 45}{3} = - 15$