\( \left[\begin{array}{lll}1 & 1 & 2 \\ 3 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \) find the Rank of matrix

Find the inverse matrix of \( \begin{bmatrix}1 & 1 & 2 \\ 3 & 2 & 1 \\ 1 & 2 & 3\end{bmatrix} \). Matrices by following steps: - step0: Find the matrix inverse: \(\left[\begin{array}{rrr}{1}&{1}&{2}\\{3}&{2}&{1}\\{1}&{2}&{3}\end{array}\right]\) - step1: Begin by adjoining the identity matrix to form the matrix: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{3}&{2}&{1}&{0}&{1}&{0}\\{1}&{2}&{3}&{0}&{0}&{1}\end{array}\right]\) - step2: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{-1}&{-5}&{-3}&{1}&{0}\\{1}&{2}&{3}&{0}&{0}&{1}\end{array}\right]\) - step3: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{-1}&{-5}&{-3}&{1}&{0}\\{0}&{1}&{1}&{-1}&{0}&{1}\end{array}\right]\) - step4: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{1}&{5}&{3}&{-1}&{0}\\{0}&{1}&{1}&{-1}&{0}&{1}\end{array}\right]\) - step5: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{1}&{5}&{3}&{-1}&{0}\\{0}&{0}&{-4}&{-4}&{1}&{1}\end{array}\right]\) - step6: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{1}&{5}&{3}&{-1}&{0}\\{0}&{0}&{1}&{1}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right]\) - step7: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{2}&{1}&{0}&{0}\\{0}&{1}&{0}&{-2}&{\frac{1}{4}}&{\frac{5}{4}}\\{0}&{0}&{1}&{1}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right]\) - step8: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{1}&{0}&{-1}&{\frac{1}{2}}&{\frac{1}{2}}\\{0}&{1}&{0}&{-2}&{\frac{1}{4}}&{\frac{5}{4}}\\{0}&{0}&{1}&{1}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right]\) - step9: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{0}&{0}&{1}&{\frac{1}{4}}&{-\frac{3}{4}}\\{0}&{1}&{0}&{-2}&{\frac{1}{4}}&{\frac{5}{4}}\\{0}&{0}&{1}&{1}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right]\) - step10: Write the right half matrix: \(\left[\begin{array}{rrr}{1}&{\frac{1}{4}}&{-\frac{3}{4}}\\{-2}&{\frac{1}{4}}&{\frac{5}{4}}\\{1}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right]\) The inverse of the given matrix is: \[ \left[\begin{array}{rrr}1 & \frac{1}{4} & -\frac{3}{4} \\ -2 & \frac{1}{4} & \frac{5}{4} \\ 1 & -\frac{1}{4} & -\frac{1}{4}\end{array}\right] \] The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. In this case, the rank of the given matrix is 3, as all rows are linearly independent.