# How do you solve ((2, 2, 0), (4, -3, 2), (0, -3, 5))x=((-10), (2), (-9))?

Mar 1, 2016

x_1 = -17/29; x_2=-128/29; x_3 = -129/29

#### Explanation:

Firs and foremost correct the question the way it is written is meaningless. What the question should read is:
$\left(\begin{matrix}2 & 2 & 0 \\ 4 & - 3 & 2 \\ 0 & - 3 & 5\end{matrix}\right) \left(\begin{matrix}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{matrix}\right) = \left(\begin{matrix}- 10 \\ 2 \\ - 9\end{matrix}\right)$
x is a vector $\vec{x} = \left(\begin{matrix}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{matrix}\right)$ OK?
Now Either use a determinant or process of elimination to:
$\left(\begin{matrix}2 & 2 & 0 & | & - 10 \\ 4 & - 3 & 2 & | & 2 \\ 0 & - 3 & 5 & | & - 9\end{matrix}\right)$
$\textcolor{b l u e}{{R}_{2} \implies {R}_{2} - 2 \cdot {R}_{1}}$
$\left(\begin{matrix}2 & 2 & 0 & | & - 10 \\ \textcolor{b l u e}{0} & \textcolor{b l u e}{- 7} & \textcolor{b l u e}{2} & | & \textcolor{b l u e}{22} \\ 0 & - 3 & 5 & | & - 9\end{matrix}\right)$
$\textcolor{red}{{R}_{3} \implies {R}_{3} - \frac{3}{7} \cdot {R}_{2}}$
$\left(\begin{matrix}2 & 2 & 0 & | & - 10 \\ 0 & - 7 & 2 & | & 22 \\ 0 & 0 & \textcolor{red}{\frac{29}{7}} & | & \textcolor{red}{- \frac{129}{7}}\end{matrix}\right)$
you're almost there:
color(red)(29/7 x_3 = -129/7; x_3 = -129/29
Now work your way back to determine ${x}_{2}$
color(blue) (-7x_2 -2*129/29 = 22; x_2=-128/29
and last equation say
2x_1 - 2*128/29 = -10; x_1 = -17/29