# How do you find the inverse of ((5, 2), (-1, a))?

Feb 14, 2016

Use the formula for the inverse of a $2 \times 2$ matrix to find:

${\left(\begin{matrix}5 & 2 \\ - 1 & a\end{matrix}\right)}^{- 1} = \left(\begin{matrix}\frac{a}{5 a + 2} & - \frac{2}{5 a + 2} \\ \frac{1}{5 a + 2} & \frac{5}{5 a + 2}\end{matrix}\right)$

#### Explanation:

In general the inverse of a $2 \times 2$ matrix is given by the formula:

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

In our example, let's calculate the determinant first:

$\left\mid \begin{matrix}5 & 2 \\ - 1 & a\end{matrix} \right\mid = 5 a + 2$

So provided $a \ne - \frac{2}{5}$ the determinant is non-zero and the inverse matrix can be written:

${\left(\begin{matrix}5 & 2 \\ - 1 & a\end{matrix}\right)}^{- 1} = \frac{1}{5 a + 2} \left(\begin{matrix}a & - 2 \\ 1 & 5\end{matrix}\right) = \left(\begin{matrix}\frac{a}{5 a + 2} & - \frac{2}{5 a + 2} \\ \frac{1}{5 a + 2} & \frac{5}{5 a + 2}\end{matrix}\right)$