# How do you find the inverse of [(4,2) (1,2)]?

Dec 8, 2016

The inverse is $= \left(\begin{matrix}\frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{6} & \frac{2}{3}\end{matrix}\right)$

#### Explanation:

If the matrix is

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

the inverse is

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

First we calculate the determinant of the matrix

$| \left(4 , 2\right) , \left(1 , 2\right) | = 8 - 2 = 6 \ne 0$

The determinant is $\ne 0$, so the matrix is invertible

The inverse is $= \frac{1}{6} \left(\begin{matrix}2 & - 2 \\ - 1 & 4\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{6} & \frac{2}{3}\end{matrix}\right)$

Verification,

We calculate $A \cdot {A}^{- 1} = \left(\begin{matrix}4 & 2 \\ 1 & 2\end{matrix}\right) \cdot \left(\begin{matrix}\frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{6} & \frac{2}{3}\end{matrix}\right)$

$= \left(\begin{matrix}\frac{4}{3} - \frac{2}{6} & - \frac{4}{3} + \frac{4}{3} \\ \frac{1}{3} - \frac{2}{6} & - \frac{1}{3} + \frac{4}{3}\end{matrix}\right)$

$= \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = I$