Using elementary row operations, find the inverse of matrix [3 -2 3; 2 1 -1; 4 -3 2].

Steps to Solve

1. Set Up the Augmented Matrix
   We start by creating an augmented matrix consisting of the given matrix and the identity matrix. $$ \begin{bmatrix} 3 & -2 & 3 & | & 1 & 0 & 0 \ 2 & 1 & -1 & | & 0 & 1 & 0 \ 4 & -3 & 2 & | & 0 & 0 & 1 \end{bmatrix} $$

2. Row Operations to Achieve Row Echelon Form
   We want to make the left-hand side of the augmented matrix into an upper triangular form.

• First, let's swap Row 1 and Row 2: $$ \begin{bmatrix} 2 & 1 & -1 & | & 0 & 1 & 0 \ 3 & -2 & 3 & | & 1 & 0 & 0 \ 4 & -3 & 2 & | & 0 & 0 & 1 \end{bmatrix} $$

• Next, we can replace Row 2 with Row 2 minus ( \frac{3}{2} ) times Row 1: $$ R_2 \rightarrow R_2 - \frac{3}{2}R_1 $$ This gives: $$ \begin{bmatrix} 2 & 1 & -1 & | & 0 & 1 & 0 \ 0 & -\frac{7}{2} & \frac{9}{2} & | & 1 & -\frac{3}{2} & 0 \ 4 & -3 & 2 & | & 0 & 0 & 1 \end{bmatrix} $$

• Then, we replace Row 3 with Row 3 minus ( 2R_1 ): $$ R_3 \rightarrow R_3 - 2R_1 $$ Resulting in: $$ \begin{bmatrix} 2 & 1 & -1 & | & 0 & 1 & 0 \ 0 & -\frac{7}{2} & \frac{9}{2} & | & 1 & -\frac{3}{2} & 0 \ 0 & -5 & 4 & | & 0 & -2 & 1 \end{bmatrix} $$

3. Further Row Operations for Simplification
   Continue simplifying Row 2 by multiplying it by ( -\frac{2}{7} ): $$ R_2 \rightarrow -\frac{2}{7}R_2 $$ Resulting in: $$ \begin{bmatrix} 2 & 1 & -1 & | & 0 & 1 & 0 \ 0 & 1 & -\frac{9}{7} & | & -\frac{2}{7} & \frac{3}{7} & 0 \ 0 & -5 & 4 & | & 0 & -2 & 1 \end{bmatrix} $$

4. Eliminate below the pivot in Column 2
   Replace Row 3 with Row 3 plus ( 5R_2 ): $$ R_3 \rightarrow R_3 + 5R_2 $$ We will arrive at: $$ \begin{bmatrix} 2 & 1 & -1 & | & 0 & 1 & 0 \ 0 & 1 & -\frac{9}{7} & | & -\frac{2}{7} & \frac{3}{7} & 0 \ 0 & 0 & \frac{3}{7} & | & -\frac{10}{7} & \frac{1}{7} & 1 \end{bmatrix} $$

5. Back Substitution to Achieve Reduced Row Echelon Form
   Now we will work upwards to eliminate all elements above our pivots:

• Replace Row 1 with Row 1 minus Row 2: $$ R_1 \rightarrow R_1 - R_2 $$ This results in: $$ \begin{bmatrix} 2 & 0 & 6 & | & \frac{2}{7} & \frac{4}{7} & 0 \ 0 & 1 & -\frac{9}{7} & | & -\frac{2}{7} & \frac{3}{7} & 0 \ 0 & 0 & \frac{3}{7} & | & -\frac{10}{7} & \frac{1}{7} & 1 \end{bmatrix} $$ Then simplify Row 1 and Row 3 as needed.

6. Scale Rows
   Continue scaling to achieve final identity form and find the inverse. The resulting matrix will yield the inverse.