# How do you solve ((8, 7), (1, 1))x=((3, -6), (-2, 9))?

Aug 14, 2018

$x = \left(\begin{matrix}17 & - 69 \\ - 19 & 78\end{matrix}\right)$

#### Explanation:

Let's try multiplying both sides by $\left(\begin{matrix}1 & - 7 \\ - 1 & 8\end{matrix}\right)$ and see what happens:

$\left(\begin{matrix}1 & - 7 \\ - 1 & 8\end{matrix}\right) \left(\begin{matrix}8 & 7 \\ 1 & 1\end{matrix}\right) x = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) x = x$

whereas:

$\left(\begin{matrix}1 & - 7 \\ - 1 & 8\end{matrix}\right) \left(\begin{matrix}3 & - 6 \\ - 2 & 9\end{matrix}\right) = \left(\begin{matrix}17 & - 69 \\ - 19 & 78\end{matrix}\right)$

So:

$x = \left(\begin{matrix}1 & - 7 \\ - 1 & 8\end{matrix}\right) \left(\begin{matrix}8 & 7 \\ 1 & 1\end{matrix}\right) x = \left(\begin{matrix}1 & - 7 \\ - 1 & 8\end{matrix}\right) \left(\begin{matrix}3 & - 6 \\ - 2 & 9\end{matrix}\right) = \left(\begin{matrix}17 & - 69 \\ - 19 & 78\end{matrix}\right)$

In general, the multiplicative inverse of $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is:

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

and in our example:

$\left\mid \begin{matrix}8 & 7 \\ 1 & 1\end{matrix} \right\mid = 8 \cdot 1 - 7 \cdot 1 = 8 - 7 = 1$