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1) Using elementary transformations, find the inverse of the matrix

3 0 -1

2 3 0

0 4 1

$[\begin{matrix} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}] = A [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{1} \to \frac{1}{3} R_{1} [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1 \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ - \frac{\begin{matrix} 2 \end{matrix}}{3} & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{2} \to R_{2} - 2 R_{1} [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1 \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ - \frac{\begin{matrix} 2 \end{matrix}}{9} & \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 \\ 0 & 0 & 1 \end{matrix}] R_{2} \to \frac{1}{3} R_{2} [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{\begin{matrix} 1 \end{matrix}}{9} \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ - \frac{\begin{matrix} 2 \end{matrix}}{3} & \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 \\ \frac{8}{9} & - \frac{4}{3} & 1 \end{matrix}] R_{3} \to R_{3} - 4 R_{2} [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ - \frac{\begin{matrix} 2 \end{matrix}}{3} & \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 \\ 8 & - 12 & 9 \end{matrix}] R_{3} \to 9 R_{3} [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} \frac{\begin{matrix} 1 \end{matrix}}{3} & 0 & 0 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{matrix}] R_{2} \to R_{2} - \frac{2}{9} R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = A [\begin{matrix} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{matrix}] R_{1} \to R_{1} + \frac{1}{3} R_{3} T h u s, i n v e r s e o f A = [\begin{matrix} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{matrix}]$

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