# How do you solve the system 3x-5y=-24, 5x+4y=-3 using matrix equation?

Feb 14, 2017

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 3 \\ 3\end{matrix}\right)$

#### Explanation:

$A m a t h b f x = m a t h b f b$

$\implies \left(\begin{matrix}3 & - 5 \\ 5 & 4\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 24 \\ - 3\end{matrix}\right)$

Here:
${A}^{- 1} = {\left(\begin{matrix}3 & - 5 \\ 5 & 4\end{matrix}\right)}^{- 1} = \frac{1}{3 \cdot 4 - 5 \cdot \left(- 5\right)} \left(\begin{matrix}4 & 5 \\ - 5 & 3\end{matrix}\right)$

$= \frac{1}{37} \left(\begin{matrix}4 & 5 \\ - 5 & 3\end{matrix}\right)$

So we can say that:

$\frac{1}{37} \left(\begin{matrix}4 & 5 \\ - 5 & 3\end{matrix}\right) \left(\begin{matrix}3 & - 5 \\ 5 & 4\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{37} \left(\begin{matrix}4 & 5 \\ - 5 & 3\end{matrix}\right) \left(\begin{matrix}- 24 \\ - 3\end{matrix}\right)$

$\implies \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{37} \left(\begin{matrix}4 & 5 \\ - 5 & 3\end{matrix}\right) \left(\begin{matrix}- 24 \\ - 3\end{matrix}\right)$

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{37} \left(\begin{matrix}- 111 \\ 111\end{matrix}\right) = \left(\begin{matrix}- 3 \\ 3\end{matrix}\right)$