# How do you solve 4r + 5s - 3t = 9, 6r - 3s + 1t = 67, 2r + 0s + 1t = 22?

Mar 5, 2016

$r = 9 , s = - 3 , t = 4$

#### Explanation:

This system of variables is equal to
$\left[\begin{matrix}4 & 5 & - 3 \\ 6 & - 3 & 1 \\ 2 & 0 & 1\end{matrix}\right] \cdot \left[\begin{matrix}r \\ s \\ t\end{matrix}\right] = \left[\begin{matrix}9 \\ 67 \\ 22\end{matrix}\right]$
We can solve it in this way
$\Delta = \left[\begin{matrix}4 & 5 & - 3 \\ 6 & - 3 & 1 \\ 2 & 0 & 1\end{matrix}\right] = - 12 + 10 + 0 - \left(18 + 30 + 0\right) = - 2 - 48 = - 50$
$\Delta r = \left[\begin{matrix}9 & 5 & - 3 \\ 67 & - 3 & 1 \\ 22 & 0 & 1\end{matrix}\right] = - 27 + 110 + 0 - \left(198 + 335 + 0\right) = 83 - 533 = - 450$
$\Delta s = \left[\begin{matrix}4 & 9 & - 3 \\ 6 & 67 & 1 \\ 2 & 22 & 1\end{matrix}\right] = 268 + 18 - 396 - \left(- 402 + 88 + 54\right) = - 110 + 260 = 150$
$\Delta t = \left[\begin{matrix}4 & 5 & 9 \\ 6 & - 3 & 67 \\ 2 & 0 & 22\end{matrix}\right] = - 264 + 670 + 0 - \left(- 54 + 660 + 0\right) = 406 - 606 = - 200$

Finally

$r = \frac{\Delta r}{\Delta} = \frac{- 450}{-} 50 = 9$
$s = \frac{\Delta s}{\Delta} = \frac{150}{-} 50 = - 3$
$t = \frac{\Delta t}{\Delta} = \frac{- 200}{-} 50 = 4$