# How do I find the inverse of a 2xx2 matrix?

Jun 9, 2018

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

#### Explanation:

Let's have a go a this without simply plugging in a remembered formula.

Given a matrix $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$, let's try multiplying it by $\left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$ and see what we get:

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

So if we multiply by $\frac{1}{a d - b c} = \frac{1}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid}$ then we find the inverse matrix:

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