# How do you solve the system 4x+8y=7, 3x-3y=0 using matrix equation?

Apr 11, 2017

$x = \frac{7}{12}$ and $y = \frac{7}{12}$

#### Explanation:

In matrix notation, the system of linear equations

$4 x + 8 y = 7$ and $3 x - 3 y = 0$ can be written as

$\left(\begin{matrix}4 & 8 \\ 3 & - 3\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}7 \\ 0\end{matrix}\right)$

i.e. $A \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}7 \\ 0\end{matrix}\right)$, then

$\left(\begin{matrix}x \\ y\end{matrix}\right) = {A}^{- 1} \left(\begin{matrix}7 \\ 0\end{matrix}\right)$,

where ${A}^{- 1}$ is inverse of $A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ and for a 2X2 matrix is defined as ${A}^{- 1} = \frac{1}{a d - b c} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

Hence inverse of $\left(\begin{matrix}4 & 8 \\ 3 & - 3\end{matrix}\right)$ is $\frac{1}{- 12 - 24} \left(\begin{matrix}- 3 & - 8 \\ - 3 & 4\end{matrix}\right)$

i.e. $\frac{1}{- 36} \left(\begin{matrix}- 3 & - 8 \\ - 3 & 4\end{matrix}\right) = \left(\begin{matrix}\frac{1}{12} & \frac{2}{9} \\ \frac{1}{12} & - \frac{1}{9}\end{matrix}\right)$

and $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{1}{12} & \frac{2}{9} \\ \frac{1}{12} & - \frac{1}{9}\end{matrix}\right) \left(\begin{matrix}7 \\ 0\end{matrix}\right) = \left(\frac{7}{12} , \frac{7}{12}\right)$

i.e. $x = \frac{7}{12}$ and $y = \frac{7}{12}$