# How do you find the inverse of [(3/4, -1/8), (5,1/2)]?

Dec 2, 2016

The inverse is $= \left(\begin{matrix}\frac{1}{2} & \frac{1}{8} \\ - 5 & \frac{3}{4}\end{matrix}\right)$

#### Explanation:

let the matrix be $A = \left(\begin{matrix}\frac{3}{4} & - \frac{1}{8} \\ 5 & \frac{1}{2}\end{matrix}\right)$

We calculate the determinant

$\det \left(A\right) = | \left(\frac{3}{4} , - \frac{1}{8}\right) , \left(5 , \frac{1}{2}\right) |$

$= \frac{3}{4} \cdot \frac{1}{2} - \left(- \frac{1}{8} \cdot 5\right) = \frac{3}{8} + \frac{5}{8} = 1 \ne 0$

For a matrix to be invertible, the determinant $\ne 0$

Now we calculate the matrix of the minor cofactor

${A}_{c} =$$\left(\begin{matrix}\frac{1}{2} & - 5 \\ \frac{1}{8} & \frac{3}{4}\end{matrix}\right)$

Now we calculate the transpose of ${A}_{c}^{t}$

$= \left(\begin{matrix}\frac{1}{2} & \frac{1}{8} \\ - 5 & \frac{3}{4}\end{matrix}\right)$

and the inverse is

${A}^{- 1} = \frac{1}{\det} A \left(\begin{matrix}\frac{1}{2} & \frac{1}{8} \\ - 5 & \frac{3}{4}\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{2} & \frac{1}{8} \\ - 5 & \frac{3}{4}\end{matrix}\right)$

Verification

$A \cdot {A}^{- 1} = \left(\begin{matrix}\frac{3}{4} & - \frac{1}{8} \\ 5 & \frac{1}{2}\end{matrix}\right) \cdot \left(\begin{matrix}\frac{1}{2} & \frac{1}{8} \\ - 5 & \frac{3}{4}\end{matrix}\right)$

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