# How do you find the inverse of [(6,-10), (-4,-5)]?

Feb 19, 2017

The inverse matrix is $= \left(\begin{matrix}\frac{1}{14} & - \frac{1}{7} \\ - \frac{2}{35} & - \frac{3}{35}\end{matrix}\right)$

#### Explanation:

Let $A = \left(\begin{matrix}6 & - 10 \\ - 4 & - 5\end{matrix}\right)$

To determine if matrix $A$ is invertible, we calculate the determinant

$\det A = | \left(6 , - 10\right) , \left(- 4 , - 5\right) | = - 6 \cdot 5 - \left(- 10\right) \left(- 4\right) = - 30 - 40 = - 70$

As, $\det A \ne 0$, the matrix is invertible

We start, by calculating the matrix of co-factors

$C = \left(\begin{matrix}- 5 & 4 \\ 10 & 6\end{matrix}\right)$

Then, we calculate the transpose of matrix $C$

${C}^{T} = \left(\begin{matrix}- 5 & 10 \\ 4 & 6\end{matrix}\right)$

And the inverse is

${A}^{-} 1 = \frac{1}{\det} A \cdot {C}^{T}$

$= - \frac{1}{70} \cdot \left(\begin{matrix}- 5 & 10 \\ 4 & 6\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{14} & - \frac{1}{7} \\ - \frac{2}{35} & - \frac{3}{35}\end{matrix}\right)$

Verification,

$A \cdot {A}^{-} 1 = \left(\begin{matrix}6 & - 10 \\ - 4 & - 5\end{matrix}\right) \cdot \left(\begin{matrix}\frac{1}{14} & - \frac{1}{7} \\ - \frac{2}{35} & - \frac{3}{35}\end{matrix}\right)$

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

$= I$