# How do you solve x+3z=-2, 2x+2y+z=4, and 3x+y-2z=5 using matrices?

Jul 15, 2016

$\textcolor{b l u e}{x = \frac{8}{17} \text{ : "y=33/17" ; } z = - \frac{14}{17}}$

Answer approach sheet 1 of 2

#### Explanation:

These are a blighter to format!!!!

$\text{ "xcolor(white)(.)ycolor(white)(.)zcolor(white)(.)" | "ans}$
" "[(1,0,3" | "-2),(2,2,1" | "4),(3,1,-2" | "5)]
$\text{ } R o w 2 - 2 \left(R o w 1\right)$
$\text{ } R o w 3 - 3 \left(R o w 1\right)$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,3" | "-2),(0,2,-5" | "8),(0,1,-11" | "11)]
$\text{ } R o w 2 - 2 \left(R o w 3\right)$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,3" | "-2),(0,0,17" | "-14),(0,1,-11" | "11)]
$\text{ } R o w 2 \div 17$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,3" | "-2),(0,0,1" | "-14/17),(0,1,-11" | "11)]
$\text{ } R o w 1 - 3 \left(R o w 2\right)$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,0" | "8/17),(0,0,1" | "-14/17),(0,1,-11" | "11)]
$\text{ } R o w 3 + 11 \left(R o w 2\right)$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,0" | "8/17),(0,0,1" | "-14/17),(0,1,0" | "33/17)]
$\text{ Swap "Row3" & } R o w 2$
$\text{ } \downarrow$

$\textcolor{w h i t e}{.}$
" "[(1,0,0" | "8/17),(0,1,0" | "33/17),(0,0,1" | "-14/17)]

$\textcolor{b l u e}{x = \frac{8}{17} \text{ : "y=33/17" ; } z = - \frac{14}{17}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: Jul 16, 2016

Answer approach sheet 2 of 2
Using Linear Algebra
This is a guide only to get you started

#### Explanation:

Let $A = \left(\begin{matrix}1 & 0 & 3 \\ 2 & 2 & 1 \\ 3 & 1 & - 2\end{matrix}\right)$

Let $B = \left(\begin{matrix}- 2 \\ 4 \\ 5\end{matrix}\right)$

Then ${A}^{-} . B = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You will have to investigate how to obtain ${A}^{-}$ as I can not remember but I can give you the matrix. There is an algorithm. The hard way is to use Gauss Jordon elimination on the LHS of the matrix below and the inverse will be in the RHS

$\left(\begin{matrix}1 & 0 & 3 & \text{|" & 1 & 0 & 0 \\ 2 & 2 & 1 & "|" & 0 & 1 & 0 \\ 3 & 1 & 2 & "|} & 0 & 0 & 1\end{matrix}\right)$ 