\( \left.\begin{array}{l}\text { Question } \\ \text { Matrix } A \text { is defined by }\left[\begin{array}{ccc}1 & 2 & -7 \\ -3 & -2 & 6 \\ -9 & -1 & 0\end{array}\right] \text {. Find the inverse of this matrix if it exists. } \\ \text { Provide your answer below: } \\ \text { Q } \\ \square \square \square \square \square \square \\ \square \\ \square\end{array}\right] \)

Matrices by following steps: - step0: Find the matrix inverse: \(\left[\begin{array}{lll}{1}&{2}&{-7}\\{-3}&{-2}&{6}\\{-9}&{-1}&{0}\end{array}\right]\) - step1: Begin by adjoining the identity matrix to form the matrix: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{-3}&{-2}&{6}&{0}&{1}&{0}\\{-9}&{-1}&{0}&{0}&{0}&{1}\end{array}\right]\) - step2: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{4}&{-15}&{3}&{1}&{0}\\{-9}&{-1}&{0}&{0}&{0}&{1}\end{array}\right]\) - step3: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{4}&{-15}&{3}&{1}&{0}\\{0}&{17}&{-63}&{9}&{0}&{1}\end{array}\right]\) - step4: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{1}&{-\frac{15}{4}}&{\frac{3}{4}}&{\frac{1}{4}}&{0}\\{0}&{17}&{-63}&{9}&{0}&{1}\end{array}\right]\) - step5: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{1}&{-\frac{15}{4}}&{\frac{3}{4}}&{\frac{1}{4}}&{0}\\{0}&{0}&{\frac{3}{4}}&{-\frac{15}{4}}&{-\frac{17}{4}}&{1}\end{array}\right]\) - step6: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{1}&{-\frac{15}{4}}&{\frac{3}{4}}&{\frac{1}{4}}&{0}\\{0}&{0}&{1}&{-5}&{-\frac{17}{3}}&{\frac{4}{3}}\end{array}\right]\) - step7: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{-7}&{1}&{0}&{0}\\{0}&{1}&{0}&{-18}&{-21}&{5}\\{0}&{0}&{1}&{-5}&{-\frac{17}{3}}&{\frac{4}{3}}\end{array}\right]\) - step8: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{0}&{-34}&{-\frac{119}{3}}&{\frac{28}{3}}\\{0}&{1}&{0}&{-18}&{-21}&{5}\\{0}&{0}&{1}&{-5}&{-\frac{17}{3}}&{\frac{4}{3}}\end{array}\right]\) - step9: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{0}&{0}&{2}&{\frac{7}{3}}&{-\frac{2}{3}}\\{0}&{1}&{0}&{-18}&{-21}&{5}\\{0}&{0}&{1}&{-5}&{-\frac{17}{3}}&{\frac{4}{3}}\end{array}\right]\) - step10: Write the right half matrix: \(\left[\begin{array}{lll}{2}&{\frac{7}{3}}&{-\frac{2}{3}}\\{-18}&{-21}&{5}\\{-5}&{-\frac{17}{3}}&{\frac{4}{3}}\end{array}\right]\) The inverse of matrix \( A \) is given by: \[ A^{-1} = \left[\begin{array}{ccc} 2 & \frac{7}{3} & -\frac{2}{3} \\ -18 & -21 & 5 \\ -5 & -\frac{17}{3} & \frac{4}{3} \end{array}\right] \] This matrix is the inverse of the original matrix \( A \).