# How do you solve 2x+2y = 2 and 4x+3y=7 using matrices?

Jan 28, 2017

The answer is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$

#### Explanation:

The equations are

$2 x + 2 y = 2$

$4 x + 3 y = 7$

In matrix form , we have

$\left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}2 \\ 7\end{matrix}\right)$

Let $A = \left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right)$

We must find the inverse matrix ${A}^{-} 1$

The determinant of matrix $A$ is

$\det A = | \left(2 , 2\right) , \left(4 , 3\right) | = 6 - 8 = - 2$

As $\det A \ne 0$, the matrix is invertible

${A}^{-} 1 = \frac{1}{\det} A \cdot \left(\left(3 , - 2\right) \left(- 4 , 2\right)\right)$

$= - \frac{1}{2} \left(\begin{matrix}3 & - 2 \\ - 4 & 2\end{matrix}\right) = \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right)$

Verification

${\forall}^{-} 1 = \left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = I$

Therefore,

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right) \left(\begin{matrix}2 \\ 7\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$