How do you solve 2r + 3s - 4t = 20, 4r - s + 5t = 13, 3r + 2s + 4 = 15?

Dec 6, 2015

$\left(5 , 2 , - 1\right)$

Explanation:

You meant $\left[\begin{matrix}2 & 3 & - 4 \\ 4 & - 1 & 5 \\ 3 & 2 & 4\end{matrix}\right] \cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}20 \\ 13 \\ 15\end{matrix}\right]$

${L}_{3} : = \frac{{L}_{3} + {L}_{1}}{5}$

${L}_{2} : = 4 \cdot {L}_{2} + 5 \cdot {L}_{1}$

$\left[\begin{matrix}2 & 3 & - 4 \\ 4 \cdot 4 + 5 \cdot 2 & 4 \cdot \left(- 1\right) + 5 \cdot 3 & 4 \cdot 5 + 5 \cdot \left(- 4\right) \\ \frac{3 + 2}{5} & \frac{2 + 3}{5} & \frac{4 - 4}{5}\end{matrix}\right]$

$\cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}20 \\ 4 \cdot 13 + 5 \cdot 20 \\ \frac{15 + 20}{5}\end{matrix}\right]$

$\left[\begin{matrix}2 & 3 & - 4 \\ 26 & 11 & 0 \\ 1 & 1 & 0\end{matrix}\right] \cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}20 \\ 152 \\ 7\end{matrix}\right]$

${L}_{2} : = {L}_{2} - 11 \cdot {L}_{3}$

$\left[\begin{matrix}2 & 3 & - 4 \\ 26 - 11 & 11 - 11 & 0 \\ 1 & 1 & 0\end{matrix}\right] \cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}20 \\ 152 - 11 \cdot 7 \\ 7\end{matrix}\right]$

$\left[\begin{matrix}2 & 3 & - 4 \\ 15 & 0 & 0 \\ 1 & 1 & 0\end{matrix}\right] \cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}20 \\ 75 \\ 7\end{matrix}\right]$

$15 r = 75 R i g h t a r r o w r = 5$

$r + s = 7 R i g h t a r r o w s = 7 - 5 = 2$

$2 r + 3 s - 4 t = 20$

$10 + 6 - 4 t = 20 R i g h t a r r o w t = - 1$