How do you solve the system 3y + 2z = 4, 2x − y − 3z = 3, 2x + 2y − z = 7?

Nov 27, 2015

Answer:

$\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = t \left[\begin{matrix}- 7 \\ 4 \\ - 6\end{matrix}\right] + \left[\begin{matrix}\frac{9}{2} \\ 0 \\ 2\end{matrix}\right] , \forall t \in m a t h \boldsymbol{R}$

Explanation:

$\left[\begin{matrix}0 & 3 & 2 \\ 2 & - 1 & - 3 \\ 2 & 2 & - 1\end{matrix}\right] \cdot \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}4 \\ 3 \\ 7\end{matrix}\right]$

${L}_{3} ' : = {L}_{2} - {L}_{3} : \left[2 - 2 , - 1 - 2 , - 3 + 1\right] \cdot \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = 3 - 7$

${L}_{3} ' : \left[0 , - 3 , - 2\right] \cdot \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = - 4$

${L}_{3} ' ' = {L}_{1} + {L}_{4} : \left[0 , 3 - 3 , 2 - 2\right] \cdot \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = 4 - 4$

This is true for all $\left(x , y , z\right) \in m a t h {\boldsymbol{R}}^{3}$

${L}_{1} R i g h t a r r o w z \left(y\right) = \frac{4 - 3 y}{2}$

${L}_{2} R i g h t a r r o w 2 x - y - 3 \frac{4 - 3 y}{2} = 3$ and we want x(y).

$4 x - 2 y - 12 + 9 y = 6$

$4 x = 18 - 7 y$

$\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}\frac{18 - 7 y}{4} \\ y \\ \frac{4 - 3 y}{2}\end{matrix}\right] = y \left[\begin{matrix}- \frac{7}{4} \\ 1 \\ - \frac{3}{2}\end{matrix}\right] + \left[\begin{matrix}\frac{18}{4} \\ 0 \\ \frac{4}{2}\end{matrix}\right]$

$y = 4 t$