Using row elementary transformation find the inverse of the following... no links please...

$[\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}]$

A = [\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}] w e k n o w t h a t A = A I [\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{3} \to R_{1} + R_{3} [\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 1 & 0 & 4 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] R_{2} \to R_{3} - R_{2} [\begin{matrix} 1 & - 2 & 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 4 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 1 & - 1 & 1 \\ 1 & 0 & 1 \end{matrix}] R_{3} \to R_{3} - R_{1} [\begin{matrix} 1 & - 2 & 3 \\ - 1 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 1 & - 1 & 1 \\ 0 & 0 & 1 \end{matrix}] R_{2} \to \frac{- 1}{2} R_{3} + R_{2} [\begin{matrix} 1 & - 2 & 3 \\ 0 & 1 & - \frac{1}{2} \\ - 2 & 0 & 1 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 1 & - 1 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix}] R_{1} \to R_{1} + 2 R_{2} [\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & - \frac{1}{2} \\ - 2 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ 1 & - 1 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix}] R_{2} \to 2 R_{2} [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & - 1 \\ - 2 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ 2 & - 2 & 1 \\ 0 & 0 & 1 \end{matrix}] R_{3} \to \frac{1}{2} R_{3} [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & - 1 \\ - 1 & 0 & \frac{1}{2} \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ 2 & - 2 & 1 \\ 0 & 0 & \frac{1}{2} \end{matrix}] R_{3} \to R_{3} + R_{1} [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & - 1 \\ 0 & 0 & \frac{5}{2} \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ 2 & - 2 & 1 \\ 3 & - 2 & \frac{3}{2} \end{matrix}] R_{3} \to \frac{2}{5} R_{3} [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & - 1 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ 2 & - 2 & 1 \\ \frac{6}{5} & \frac{- 4}{5} & \frac{3}{5} \end{matrix}] R_{2} \to R_{2} + R_{3} [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ \frac{16}{5} & - \frac{14}{5} & \frac{8}{5} \\ \frac{6}{5} & \frac{- 4}{5} & \frac{3}{5} \end{matrix}] R_{2} \to \frac{1}{2} R_{2} [\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 2 & 1 \\ \frac{8}{5} & - \frac{7}{5} & \frac{4}{5} \\ \frac{6}{5} & \frac{- 4}{5} & \frac{3}{5} \end{matrix}] R_{1} \to R_{1} - 2 R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} \frac{3}{5} & \frac{- 2}{5} & \frac{- 1}{5} \\ \frac{8}{5} & - \frac{7}{5} & \frac{4}{5} \\ \frac{6}{5} & \frac{- 4}{5} & \frac{3}{5} \end{matrix}] S o A^{- 1} = [\begin{matrix} \frac{3}{5} & \frac{- 2}{5} & \frac{- 1}{5} \\ \frac{8}{5} & - \frac{7}{5} & \frac{4}{5} \\ \frac{6}{5} & \frac{- 4}{5} & \frac{3}{5} \end{matrix}] = \frac{1}{5} [\begin{matrix} 3 & - 2 & - 1 \\ 8 & - 7 & 4 \\ 6 & - 4 & 3 \end{matrix}]

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Using row elementary transformation find the inverse of the following... no links please... 1 - 2 3 0 - 1 4 - 2 2 1

Using row elementary transformation find the inverse of the following... no links please...

$[\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}]$