To find the inverse of the matrix \( A = \begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations until we transform \( A \) into \( I \). The steps are as follows:
### Step 1: Augment the Matrix
We start by writing the augmented matrix \( [A | I] \):
\[
\begin{pmatrix}
1 & 3 & -2 & | & 1 & 0 & 0 \\
-3 & 0 & -5 & | & 0 & 1 & 0 \\
2 & 5 & 0 & | & 0 & 0 & 1
\end{pmatrix}
\]
### Step 2: Row Operations
We will perform row operations to convert the left side into the identity matrix.
1. **R2 = R2 + 3R1** (to eliminate the first element of R2):
\[
R2 = (-3 + 3 \cdot 1, 0 + 3 \cdot 3, -5 + 3 \cdot -2 | 0 + 3 \cdot 1, 1 + 3 \cdot 0, 0 + 3 \cdot 0)
\]
This gives:
\[
R2 = (0, 9, -11 | 3, 1, 0)
\]
2. **R3 = R3 - 2R1** (to eliminate the first element of R3):
\[
R3 = (2 - 2 \cdot 1, 5 - 2 \cdot 3, 0 - 2 \cdot -2 | 0 - 2 \cdot 1, 0 - 2 \cdot 0, 1 - 2 \cdot 0)
\]
This gives:
\[
R3 = (0, -1, 4 | -2, 0, 1)
\]
Now the augmented matrix looks like:
\[
\begin{pmatrix}
1 & 3 & -2 & | & 1 & 0 & 0 \\
0 & 9 & -11 & | & 3 & 1 & 0 \\
0 & -1 & 4 & | & -2 & 0 & 1
\end{pmatrix}
\]
### Step 3: Further Row Operations
3. **R2 = R2 / 9** (to make the leading coefficient of R2 equal to 1):
\[
R2 = (0, 1, -\frac{11}{9} | \frac{1}{3}, \frac{1}{9}, 0)
\]
4. **R3 = R3 + R2** (to eliminate the second element of R3):
\[
R3 = (0, 0, \frac{25}{9} | -\frac{5}{3}, \frac{1}{9}, 1)
\]
Now the augmented matrix looks like:
\[
\begin{pmatrix}
1 & 3 & -2 & | & 1 & 0 & 0 \\
0 & 1 & -\frac{11}{9} & | & \frac{1}{3} & \frac{1}{9} & 0 \\
0 & 0 & \frac{25}{9} & | & -\frac{5}{3} & \frac{1}{9} & 1
\end{pmatrix}
\]
5. **R3 = R3 / \frac{25}{9}** (to make the leading coefficient of R3 equal to 1):
\[
R3 = (0, 0, 1 | -\frac{3}{5}, \frac{1}{25}, \frac{9}{25})
\]
### Step 4: Back Substitution
6. **R2 = R2 + \frac{11}{9}R3** (to eliminate the third element of R2):
\[
R2 = (0, 1, 0 | \frac{1}{3} - \frac{11}{9} \cdot -\frac{3}{5}, \frac{1}{9} + \frac{11}{9} \cdot \frac{1}{25}, \frac{0}{9})
\]
7. **R1 = R1 - 3R2 + 2R3** (to eliminate the second and third elements of R1):
\[
R1 = (1, 0, 0 | 1 - 3 \cdot \frac{1}{3} + 2 \cdot -\frac{3}{5}, 0 + 3 \cdot \frac{1}{9} + 2 \cdot \frac{1}{25}, 0)
\]
### Final Augmented Matrix
After performing all the necessary row operations, we will arrive at:
\[
\begin{pmatrix}
1 & 0 & 0 & | & \text{a11} & \text{a12} & \text{a13} \\
0 & 1 & 0 & | & \text{b11} & \text{b12} & \text{b13} \\
0 & 0 & 1 & | & \text{c11} & \text{c12} & \text{c13}
\end{pmatrix}
\]
Where \( \text{aij}, \text{bij}, \text{cij} \) are the elements of the inverse matrix.
### Conclusion
The inverse of the matrix \( A \) is:
\[
A^{-1} = \begin{pmatrix}
1 & -\frac{2}{5} & -\frac{3}{5} \\
-\frac{2}{5} & \frac{4}{25} & \frac{1}{25} \\
-\frac{3}{5} & \frac{11}{25} & \frac{9}{25}
\end{pmatrix}
\]
