# How do you find the inverse of A=((0, 2, 4), (1, 3, 3), (1, 5, 8))?

Nov 1, 2017

The inverse is ${A}^{-} 1 = \left(\begin{matrix}- \frac{9}{2} & - 2 & 3 \\ \frac{5}{2} & 2 & - 2 \\ - 1 & - 1 & 1\end{matrix}\right)$

#### Explanation:

Proceed as follows

Our matrix is $A = \left(\begin{matrix}0 & 2 & 4 \\ 1 & 3 & 3 \\ 1 & 5 & 8\end{matrix}\right)$

Write side by side $A$ and $I$

$\left(\begin{matrix}0 & 2 & 4 \\ 1 & 3 & 3 \\ 1 & 5 & 8\end{matrix}\right) \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

Perform the following row operations

$R 1 \leftrightarrow R 3$

$\left(\begin{matrix}1 & 5 & 8 \\ 1 & 3 & 3 \\ 0 & 2 & 4\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{matrix}\right)$

$R 2 \leftarrow R 2 - R 1$

$\left(\begin{matrix}1 & 5 & 8 \\ 0 & - 2 & - 5 \\ 0 & 2 & 4\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & - 1 \\ 1 & 0 & 0\end{matrix}\right)$

$R 3 \leftarrow R 3 + R 2$

$\left(\begin{matrix}1 & 5 & 8 \\ 0 & - 2 & - 5 \\ 0 & 0 & - 1\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & - 1 \\ 1 & 1 & - 1\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) \times \left(- 1\right)$

$\left(\begin{matrix}1 & 5 & 8 \\ 0 & - 2 & - 5 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 0 & 1 & - 1 \\ - 1 & - 1 & 1\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) \times \left(- 1\right)$

$\left(\begin{matrix}1 & 5 & 8 \\ 0 & 2 & 5 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 0 & - 1 & 1 \\ - 1 & - 1 & 1\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) - 5 \left(R 3\right)$

$\left(\begin{matrix}1 & 5 & 8 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}0 & 0 & 1 \\ 5 & 4 & - 4 \\ - 1 & - 1 & 1\end{matrix}\right)$

$R 1 \leftarrow \left(R 1\right) - 8 \left(R 3\right)$

$\left(\begin{matrix}1 & 5 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}8 & 8 & - 7 \\ 5 & 4 & - 4 \\ - 1 & - 1 & 1\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{2}$

$\left(\begin{matrix}1 & 5 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}8 & 8 & - 7 \\ \frac{5}{2} & 2 & - 2 \\ - 1 & - 1 & 1\end{matrix}\right)$

$R 1 \leftarrow \left(R 1\right) - 5 \left(R 2\right)$

$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right) \left(\begin{matrix}- \frac{9}{2} & - 2 & 3 \\ \frac{5}{2} & 2 & - 2 \\ - 1 & - 1 & 1\end{matrix}\right)$

Therefore,

${A}^{-} 1 = \left(\begin{matrix}- \frac{9}{2} & - 2 & 3 \\ \frac{5}{2} & 2 & - 2 \\ - 1 & - 1 & 1\end{matrix}\right)$