# How do you find the inverse of [(2,-3), (-2,-2)]?

Jan 11, 2017

The answer is $= \left(\begin{matrix}\frac{1}{5} & - \frac{3}{10} \\ - \frac{1}{5} & - \frac{1}{5}\end{matrix}\right)$

#### Explanation:

The inverse of the matrix $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is

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

Let $A = \left(\begin{matrix}2 & - 3 \\ - 2 & - 2\end{matrix}\right)$

$D e t A = | \left(a , b\right) , \left(c , d\right) | = a d - b c = - 4 - 6 = - 10$

As $D e t A \ne 0$, the matrix is invertible

${A}^{-} 1 = - \frac{1}{10} \left(\begin{matrix}- 2 & 3 \\ 2 & 2\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{5} & - \frac{3}{10} \\ - \frac{1}{5} & - \frac{1}{5}\end{matrix}\right)$

Verification

$A \cdot {A}^{-} 1 = \left(\begin{matrix}2 & - 3 \\ - 2 & - 2\end{matrix}\right) \cdot \left(\begin{matrix}\frac{1}{5} & - \frac{3}{10} \\ - \frac{1}{5} & - \frac{1}{5}\end{matrix}\right)$

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