# How do you find the inverse of A=((3, 2), (4, 5))?

Nov 25, 2016

The inverse is ${A}^{- 1} = \left(\begin{matrix}\frac{5}{7} & - \frac{2}{7} \\ - \frac{4}{7} & \frac{3}{7}\end{matrix}\right)$

#### Explanation:

To calculate the inverse of a matrix, verify that the determinant is $\ne 0$

Det A=$| \left(3 , 2\right) , \left(4 , 5\right) | = 3 \cdot 5 - 2 \cdot 4 = 15 - 8 = 7$

As $D e t A \ne 0$, the inverse exists.

Let's calculate the minors cofactor

Minor cofactor is $\left(\begin{matrix}5 & - 4 \\ - 2 & 3\end{matrix}\right)$

Then we take the tranpose, $\left(\begin{matrix}5 & - 2 \\ - 4 & 3\end{matrix}\right)$

The inverse is ${A}^{- 1} = \frac{1}{7} \left(\begin{matrix}5 & - 2 \\ - 4 & 3\end{matrix}\right)$

${A}^{- 1} = \left(\begin{matrix}\frac{5}{7} & - \frac{2}{7} \\ - \frac{4}{7} & \frac{3}{7}\end{matrix}\right)$

Verification

$A \cdot {A}^{- 1} = \left(\begin{matrix}3 & 2 \\ 4 & 5\end{matrix}\right) \cdot \left(\begin{matrix}\frac{5}{7} & - \frac{2}{7} \\ - \frac{4}{7} & \frac{3}{7}\end{matrix}\right)$

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