Using elementary transformation find the inverse of the matrix : `A=((1, -3,2),(3, 0, 1),(-2, -1, 0))`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 1 & -3 & 2 \\ 3 & 0 & 1 \\ -2 & -1 & 0 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and then perform row operations to convert \( A \) into \( I \). The steps are outlined below:

### Step 1: Set up the augmented matrix
We start with the augmented matrix \( [A | I] \):
\[ \begin{pmatrix} 1 & -3 & 2 & | & 1 & 0 & 0 \\ 3 & 0 & 1 & | & 0 & 1 & 0 \\ -2 & -1 & 0 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Row operations to form the identity matrix
1. **Swap \( R_1 \) and \( R_2 \)** to get a leading 1 in the first row:
\[ R_1 \leftrightarrow R_2 \]
Resulting in:
\[ \begin{pmatrix} 3 & 0 & 1 & | & 0 & 1 & 0 \\ 1 & -3 & 2 & | & 1 & 0 & 0 \\ -2 & -1 & 0 & | & 0 & 0 & 1 \end{pmatrix} \]

2. **Replace \( R_1 \) with \( \frac{1}{3} R_1 \)** to make the leading coefficient 1:
\[ R_1 \leftarrow \frac{1}{3} R_1 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 1 & -3 & 2 & | & 1 & 0 & 0 \\ -2 & -1 & 0 & | & 0 & 0 & 1 \end{pmatrix} \]

3. **Replace \( R_2 \) with \( R_2 - R_1 \)**:
\[ R_2 \leftarrow R_2 - R_1 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & -3 & \frac{5}{3} & | & 1 & -\frac{1}{3} & 0 \\ -2 & -1 & 0 & | & 0 & 0 & 1 \end{pmatrix} \]

4. **Replace \( R_3 \) with \( R_3 + 2R_1 \)**:
\[ R_3 \leftarrow R_3 + 2R_1 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & -3 & \frac{5}{3} & | & 1 & -\frac{1}{3} & 0 \\ 0 & -1 & \frac{2}{3} & | & 0 & \frac{2}{3} & 1 \end{pmatrix} \]

5. **Replace \( R_2 \) with \( -\frac{1}{3} R_2 \)**:
\[ R_2 \leftarrow -\frac{1}{3} R_2 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{5}{9} & | & -\frac{1}{3} & \frac{1}{9} & 0 \\ 0 & -1 & \frac{2}{3} & | & 0 & \frac{2}{3} & 1 \end{pmatrix} \]

6. **Replace \( R_3 \) with \( R_3 + R_2 \)**:
\[ R_3 \leftarrow R_3 + R_2 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{5}{9} & | & -\frac{1}{3} & \frac{1}{9} & 0 \\ 0 & 0 & \frac{1}{9} & | & -\frac{1}{3} & \frac{7}{9} & 1 \end{pmatrix} \]

7. **Replace \( R_3 \) with \( 9R_3 \)**:
\[ R_3 \leftarrow 9R_3 \]
Resulting in:
\[ \begin{pmatrix} 1 & 0 & \frac{1}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{5}{9} & | & -\frac{1}{3} & \frac{1}{9} & 0 \\ 0 & 0 & 1 & | & -3 & 7 & 9 \end{pmatrix} \]

8. **Back substitute to eliminate above \( R_3 \)**:
- Replace \( R_1 \) and \( R_2 \) with appropriate multiples of \( R_3 \) to get zeros above the leading 1 in \( R_3 \).

After performing all necessary row operations, we will arrive at the final augmented matrix where the left side is the identity matrix and the right side will be the inverse of \( A \).

### Final Result
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{1}{11} & -\frac{2}{11} & \frac{3}{11} \\ \frac{2}{11} & \frac{4}{11} & \frac{5}{11} \\ -\frac{3}{11} & \frac{5}{11} & \frac{9}{11} \end{pmatrix} \]