5. Caleule a matria inversa através do método de Gauss Jordan: (a) \( \left(\begin{array}{lll}6 & 2 & 1 \\ 0 & 4 & 1 \\ 2 & 1 & 0\end{array}\right) \)

Find the inverse matrix of \( \begin{pmatrix}6 & 2 & 1 \\ 0 & 4 & 1 \\ 2 & 1 & 0\end{pmatrix} \). Matrices by following steps: - step0: Find the matrix inverse: \(\left[\begin{array}{rrr}{6}&{2}&{1}\\{0}&{4}&{1}\\{2}&{1}&{0}\end{array}\right]\) - step1: Begin by adjoining the identity matrix to form the matrix: \(\left[\begin{array}{rrr|rrr}{6}&{2}&{1}&{1}&{0}&{0}\\{0}&{4}&{1}&{0}&{1}&{0}\\{2}&{1}&{0}&{0}&{0}&{1}\end{array}\right]\) - step2: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{4}&{1}&{0}&{1}&{0}\\{2}&{1}&{0}&{0}&{0}&{1}\end{array}\right]\) - step3: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{4}&{1}&{0}&{1}&{0}\\{0}&{\frac{1}{3}}&{-\frac{1}{3}}&{-\frac{1}{3}}&{0}&{1}\end{array}\right]\) - step4: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{1}&{\frac{1}{4}}&{0}&{\frac{1}{4}}&{0}\\{0}&{\frac{1}{3}}&{-\frac{1}{3}}&{-\frac{1}{3}}&{0}&{1}\end{array}\right]\) - step5: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{1}&{\frac{1}{4}}&{0}&{\frac{1}{4}}&{0}\\{0}&{0}&{-\frac{5}{12}}&{-\frac{1}{3}}&{-\frac{1}{12}}&{1}\end{array}\right]\) - step6: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{1}&{\frac{1}{4}}&{0}&{\frac{1}{4}}&{0}\\{0}&{0}&{1}&{\frac{4}{5}}&{\frac{1}{5}}&{-\frac{12}{5}}\end{array}\right]\) - step7: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{\frac{1}{6}}&{\frac{1}{6}}&{0}&{0}\\{0}&{1}&{0}&{-\frac{1}{5}}&{\frac{1}{5}}&{\frac{3}{5}}\\{0}&{0}&{1}&{\frac{4}{5}}&{\frac{1}{5}}&{-\frac{12}{5}}\end{array}\right]\) - step8: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{1}{3}}&{0}&{\frac{1}{30}}&{-\frac{1}{30}}&{\frac{2}{5}}\\{0}&{1}&{0}&{-\frac{1}{5}}&{\frac{1}{5}}&{\frac{3}{5}}\\{0}&{0}&{1}&{\frac{4}{5}}&{\frac{1}{5}}&{-\frac{12}{5}}\end{array}\right]\) - step9: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{0}&{0}&{\frac{1}{10}}&{-\frac{1}{10}}&{\frac{1}{5}}\\{0}&{1}&{0}&{-\frac{1}{5}}&{\frac{1}{5}}&{\frac{3}{5}}\\{0}&{0}&{1}&{\frac{4}{5}}&{\frac{1}{5}}&{-\frac{12}{5}}\end{array}\right]\) - step10: Write the right half matrix: \(\left[\begin{array}{rrr}{\frac{1}{10}}&{-\frac{1}{10}}&{\frac{1}{5}}\\{-\frac{1}{5}}&{\frac{1}{5}}&{\frac{3}{5}}\\{\frac{4}{5}}&{\frac{1}{5}}&{-\frac{12}{5}}\end{array}\right]\) A matriz inversa da matriz dada através do método de Gauss Jordan é: \[ \left[\begin{array}{rrr}\frac{1}{10} & -\frac{1}{10} & \frac{1}{5} \\ -\frac{1}{5} & \frac{1}{5} & \frac{3}{5} \\ \frac{4}{5} & \frac{1}{5} & -\frac{12}{5}\end{array}\right] \]