# How do you solve 2x+3y=-1 and 4x-5y=7 using matrices?

Sep 22, 2016

$x = \frac{8}{11} \mathmr{and} y = - \frac{9}{11}$

#### Explanation:

Although the method might seem quite daunting, once the preparation process is mastered, the method itself is surprisingly quick and easy, involving a few simple calculations.

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We have the following equations:

$2 x + 3 y = - 1 \text{ and } 4 x - 5 y = 7$

First write them as matrices:

$\left(\begin{matrix}2 & \text{ } 3 \\ 4 & - 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 7\end{matrix}\right)$

Now find the inverse matrix of $A = \left(\begin{matrix}2 & \text{ } 3 \\ 4 & - 5\end{matrix}\right)$

$\left\mid A \right\mid = \left(2 \times - 5\right) - \left(4 \times 3\right) = - 10 - 12 = - 22$

${A}^{-} 1 = \frac{1}{-} 22 \left(\begin{matrix}- 5 & - 3 \\ - 4 & 2\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}\frac{5}{22} & \frac{3}{22} \\ \frac{2}{11} & - \frac{1}{11}\end{matrix}}\right)$
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Multiply both sides of the matrix equation by the inverse matrix.

$\left(\textcolor{red}{\begin{matrix}\frac{5}{22} & \frac{3}{22} \\ \frac{2}{11} & - \frac{1}{11}\end{matrix}}\right) \left(\begin{matrix}2 & \text{ } 3 \\ 4 & - 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}\frac{5}{22} & \frac{3}{22} \\ \frac{2}{11} & - \frac{1}{11}\end{matrix}}\right) \left(\begin{matrix}- 1 \\ 7\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times \times x} \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{3} \\ \frac{1}{2}\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times} \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{8}{11} \\ - \frac{9}{11}\end{matrix}\right)$

$\therefore x = \frac{8}{11} \mathmr{and} y = - \frac{9}{11}$

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Background knowledge... to help with the method above..
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A 2 x 2 matrix multiplied by the unit matrix remains unchanged

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

A matrix multiplied by its inverse gives the unit matrix -
also known as the Identity Matrix.

$A \times {A}^{-} 1 = I = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

To find the inverse matrix (${M}^{-} 1$) of matrix M

$M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

1. Find the determinant $\left(\left\mid M \right\mid\right) = a d - b c$

2. ${M}^{-} 1 = \frac{1}{\left(\left\mid M \right\mid\right)} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

(swop a and d and change the signs of b and c, then divide by the determinant.)