# How do you solve x + 2y + 2z = 0 and 3x - 2y + 3z = -4 and 2x - y + z = -1 using matrices?

Feb 17, 2016

Use Cramer's Rule (see below)
to get $x = \frac{2}{3} , y = 1 , z = - \frac{4}{3}$

#### Explanation:

Given:
$\left(\begin{matrix}x & y & z \\ \textcolor{red}{1} & \textcolor{b l u e}{2} & \textcolor{g r e e n}{2} \\ \textcolor{red}{3} & \textcolor{b l u e}{- 2} & \textcolor{g r e e n}{3} \\ \textcolor{red}{2} & \textcolor{b l u e}{- 1} & \textcolor{g r e e n}{1}\end{matrix}\right) = \left(\begin{matrix}c \\ \textcolor{b r o w n}{0} \\ \textcolor{b r o w n}{- 4} \\ \textcolor{b r o w n}{- 1}\end{matrix}\right)$

The determinant:
$| \left(\textcolor{red}{1} , \textcolor{b l u e}{2} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) |$

$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{1} \cdot | \left(\textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) | - \textcolor{b l u e}{2} \cdot | \left(\textcolor{red}{3} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{g r e e n}{1}\right) | + \textcolor{g r e e n}{2} \cdot | \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1}\right) |$

$\textcolor{w h i t e}{\text{XXX}} = \left(- 2 + 3\right) - 2 \left(3 - 6\right) + 2 \left(- 3 + 4\right)$

$\textcolor{w h i t e}{\text{XXX}} = 9$

Other determinants can be evaluated in a similar method.

Cramer's Rule tells us that for the given values:
$x = \frac{| \left(\textcolor{b r o w n}{0} , \textcolor{b l u e}{2} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{b r o w n}{- 4} , \textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{b r o w n}{- 1} , \textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) |}{| \left(\textcolor{red}{1} , \textcolor{b l u e}{2} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) |} = \frac{6}{9} = \frac{2}{3}$

$y = \frac{| \left(\textcolor{red}{1} , \textcolor{b r o w n}{0} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{red}{3} , \textcolor{b r o w n}{- 4} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{b r o w n}{- 1} , \textcolor{g r e e n}{1}\right) |}{| \left(\textcolor{red}{1} , \textcolor{b l u e}{2} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) |} = \frac{9}{9} = 1$

$z = \frac{| \left(\textcolor{red}{1} , \textcolor{b l u e}{2} , \textcolor{b r o w n}{0}\right) , \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2} , \textcolor{b r o w n}{- 4}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1} , \textcolor{b r o w n}{- 1}\right) |}{| \left(\textcolor{red}{1} , \textcolor{b l u e}{2} , \textcolor{g r e e n}{2}\right) , \left(\textcolor{red}{3} , \textcolor{b l u e}{- 2} , \textcolor{g r e e n}{3}\right) , \left(\textcolor{red}{2} , \textcolor{b l u e}{- 1} , \textcolor{g r e e n}{1}\right) |} = - \frac{12}{9} = - \frac{4}{3}$