# How do you find the inverse of A=((6, 7, 8), (1, 0, 1), (0, 1, 0))?

Dec 17, 2016

The answer is $= \left(\begin{matrix}- \frac{1}{2} & 4 & \frac{7}{2} \\ 0 & 0 & 1 \\ \frac{1}{2} & - 3 & - \frac{7}{2}\end{matrix}\right)$

#### Explanation:

First we calculate the determinant of matrix A

$\det A = | \left(6 , 7 , 8\right) , \left(1 , 0 , 1\right) , \left(0 , 1 , 0\right) |$

$= 6 \left(0 - 1\right) - 7 \left(0\right) + 8 \left(1\right) = 2$

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

Then, we calculate the matrix of minor cofactors

${A}^{c} = \left(\left(\begin{matrix}\begin{matrix}0 & 1 \\ 1 & 0\end{matrix} \\ - \begin{matrix}1 & 1 \\ 0 & 0\end{matrix} \\ \begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\end{matrix}\right) , \left(- \left(\begin{matrix}7 & 8 \\ 1 & 0\end{matrix}\right) , \left(\begin{matrix}6 & 8 \\ 0 & 0\end{matrix}\right) , - \left(\begin{matrix}6 & 7 \\ 0 & 1\end{matrix}\right)\right) , \left(\begin{matrix}\begin{matrix}7 & 8 \\ 0 & 1\end{matrix} \\ - \begin{matrix}6 & 8 \\ 1 & 1\end{matrix} \\ \begin{matrix}6 & 7 \\ 1 & 0\end{matrix}\end{matrix}\right)\right)$

$= \left(\begin{matrix}- 1 & 0 & 1 \\ 8 & 0 & - 6 \\ 7 & 2 & - 7\end{matrix}\right)$

Then we calculate the transpose of ${A}^{c}$

${\overline{A}}^{c} = \left(\begin{matrix}- 1 & 8 & 7 \\ 0 & 0 & 2 \\ 1 & - 6 & - 7\end{matrix}\right)$

Then, the inverse is

${A}^{- 1} = {\overline{A}}^{c} / \det A$

$= \frac{1}{2} \left(\begin{matrix}- 1 & 8 & 7 \\ 0 & 0 & 2 \\ 1 & - 6 & - 7\end{matrix}\right)$

$= \left(\begin{matrix}- \frac{1}{2} & 4 & \frac{7}{2} \\ 0 & 0 & 1 \\ \frac{1}{2} & - 3 & - \frac{7}{2}\end{matrix}\right)$

Verification, by doing $A {A}^{- 1}$

$= \left(\begin{matrix}6 & 7 & 8 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{1}{2} & 4 & \frac{7}{2} \\ 0 & 0 & 1 \\ \frac{1}{2} & - 3 & - \frac{7}{2}\end{matrix}\right)$

$= \left(\left(1 , 00\right) , \left(0 , 10\right) , \left(0 , 0 , 1\right)\right)$

$= I$