Using elementary transformations, find the inverse of the matrices `A=[(2,0,-1),(5,1,0),(0,1,3)]A^-1=?`

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) of the same order and perform row operations to transform \( A \) into \( I \). The resulting transformations on \( I \) will give us \( A^{-1} \).

### Step-by-Step Solution:

1. **Set Up the Augmented Matrix**:
We start with the augmented matrix \( [A | I] \):
\[ \begin{pmatrix} 2 & 0 & -1 & | & 1 & 0 & 0 \\ 5 & 1 & 0 & | & 0 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

2. **Swap Rows**:
We can swap \( R_1 \) and \( R_2 \) to make the leading coefficient of the first row larger:
\[ \begin{pmatrix} 5 & 1 & 0 & | & 0 & 1 & 0 \\ 2 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

3. **Row Operation**:
Perform \( R_1 \leftarrow R_1 - 2R_2 \):
\[ R_1 = (5 - 2 \cdot 2, 1 - 2 \cdot 0, 0 - 2 \cdot -1 | 0 - 2 \cdot 1, 1 - 2 \cdot 0, 0 - 2 \cdot 0) \]
This gives:
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 2 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

4. **Row Operation**:
Perform \( R_2 \leftarrow R_2 - 2R_1 \):
\[ R_2 = (2 - 2 \cdot 1, 0 - 2 \cdot 1, -1 - 2 \cdot 2 | 1 - 2 \cdot -2, 0 - 2 \cdot 1, 0 - 2 \cdot 0) \]
This gives:
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & -2 & -5 & | & 5 & -2 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

5. **Row Operation**:
Perform \( R_2 \leftarrow -\frac{1}{2}R_2 \):
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

6. **Row Operation**:
Perform \( R_3 \leftarrow R_3 - R_2 \):
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 0 & \frac{1}{2} & | & \frac{5}{2} & -1 & 1 \end{pmatrix} \]

7. **Row Operation**:
Perform \( R_3 \leftarrow 2R_3 \):
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

8. **Row Operation**:
Perform \( R_2 \leftarrow R_2 - \frac{5}{2}R_3 \):
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & 1 & 0 & | & -\frac{15}{2} & 6 & -5 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

9. **Row Operation**:
Perform \( R_1 \leftarrow R_1 - 2R_3 \):
\[ \begin{pmatrix} 1 & 1 & 0 & | & -12 & 5 & -4 \\ 0 & 1 & 0 & | & -\frac{15}{2} & 6 & -5 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

10. **Final Row Operation**:
Perform \( R_1 \leftarrow R_1 - R_2 \):
\[ \begin{pmatrix} 1 & 0 & 0 & | & 3 & -1 & 1 \\ 0 & 1 & 0 & | & -\frac{15}{2} & 6 & -5 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

The resulting matrix on the right side is \( A^{-1} \):
\[ A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \]