# How do you find the inverse of A=((-7, 5), (5, -4))?

Oct 27, 2016

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

#### Explanation:

DetA=Det $\left(\begin{matrix}- 7 & 5 \\ 5 & - 4\end{matrix}\right) = \left(- 7 \cdot - 4\right) - \left(5 \cdot 5\right) = 28 - 25 = 3$

The determinant is $\ne 0$, so the inverse exists

Then we calculate the cofactors

${A}_{11} = - 4$ ${A}_{12} = - 5$
${A}_{21} = - 5$ ${A}_{22} = - 7$

So the cofactor matrix is $\left(\begin{matrix}- 4 & - 5 \\ - 5 & - 7\end{matrix}\right)$

And the transpose is $\left(\begin{matrix}- 4 & - 5 \\ - 5 & - 7\end{matrix}\right)$

And finally we divide all the components by the determinant
${A}^{- 1} = \left(\begin{matrix}- \frac{4}{3} & - \frac{5}{3} \\ - \frac{5}{3} & - \frac{7}{3}\end{matrix}\right)$

We have to check by multiplying $A \cdot {A}^{- 1}$

$\left(\begin{matrix}- 7 & 5 \\ 5 & - 4\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{4}{3} & - \frac{5}{3} \\ - \frac{5}{3} & - \frac{7}{3}\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$