# How do you solve 9a - 4b = -5 and 6a - 2b = -3 using matrices?

Sep 22, 2016

$x = - 0.33$
$y = 0.5$

#### Explanation:

Look at the steps -

Sep 22, 2016

$\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$

#### 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:

$9 a - 4 b = - 5 \text{ and } 6 a - 2 b = - 3$

First write them as matrices:

$\left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$

Now find the inverse matrix of $A = \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right)$

$\left\mid A \right\mid = \left(9 \times - 2\right) - \left(6 \times - 4\right) = - 18 + 24 = 6$

${A}^{-} 1 = \frac{1}{6} \left(\begin{matrix}- 2 & 4 \\ - 6 & 9\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right)$
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Multiply both sides of the matrix equation by the inverse matrix.

$\left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$

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

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

$\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$

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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.)