# How to solve X ? ((-5,5),(-2,5))*X = ((4,5),(5,-1))

Sep 16, 2015

To solve $A X = B$, where $A$ and $B$ are matrices, we instead solve $X = B \cdot {A}^{-} 1$.

#### Explanation:

You might be tempted to try to divide the matrices, e.g. $X = \frac{B}{A}$ where $B = \left(\begin{matrix}4 & 5 \\ 5 & - 1\end{matrix}\right)$ and $A = \left(\begin{matrix}- 5 & 5 \\ - 2 & 5\end{matrix}\right)$. But actually, whenever you want to divide matrices, you instead turn the problem into matrix multiplication. In particular, to solve $A X = B$, we instead solve $X = B \cdot {A}^{-} 1$.

So first we compute ${A}^{-} 1$. The formula for the inverse of a $2 x 2$ matrix is:

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

That gives us:

${\left(\begin{matrix}- 5 & 5 \\ - 2 & 5\end{matrix}\right)}^{-} 1 = \frac{1}{\left(- 5 \cdot 5\right) - \left(5 \cdot - 2\right)} \left(\begin{matrix}5 & - 5 \\ 2 & - 5\end{matrix}\right)$
$= - \frac{1}{15} \left(\begin{matrix}5 & - 5 \\ 2 & - 5\end{matrix}\right)$
$= \left(\begin{matrix}- \frac{1}{3} & \frac{1}{3} \\ - \frac{2}{15} & \frac{1}{3}\end{matrix}\right)$

Now we multiply $B \cdot {A}^{-} 1$:

$\left(\begin{matrix}4 & 5 \\ 5 & - 1\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{1}{3} & \frac{1}{3} \\ - \frac{2}{15} & \frac{1}{3}\end{matrix}\right)$
$= \left(\begin{matrix}- 2 & 3 \\ - \frac{23}{15} & \frac{4}{3}\end{matrix}\right)$