Find the inverse of Matrix `A=[{:(1,3,-2),(-3,0,-5),(2,5,0):}]` by using elementary row transformation.

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{pmatrix} \) using elementary row transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations to convert \( A \) into \( I \). The augmented matrix will look like this:

\[ \begin{pmatrix} 1 & 3 & -2 & | & 1 & 0 & 0 \\ -3 & 0 & -5 & | & 0 & 1 & 0 \\ 2 & 5 & 0 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 1: Row Operations to Form Identity Matrix

1. **Row 2 Transformation**: Replace \( R_2 \) with \( R_2 + 3R_1 \)
\[ R_2 = R_2 + 3R_1 \Rightarrow \begin{pmatrix} -3 + 3(1) & 0 + 3(3) & -5 + 3(-2) & | & 0 + 3(1) & 1 + 3(0) & 0 + 3(0) \end{pmatrix} \]
This gives:
\[ R_2 = \begin{pmatrix} 0 & 9 & -11 & | & 3 & 1 & 0 \end{pmatrix} \]

2. **Row 3 Transformation**: Replace \( R_3 \) with \( R_3 - 2R_1 \)
\[ R_3 = R_3 - 2R_1 \Rightarrow \begin{pmatrix} 2 - 2(1) & 5 - 2(3) & 0 - 2(-2) & | & 0 - 2(1) & 0 - 2(0) & 1 - 2(0) \end{pmatrix} \]
This gives:
\[ R_3 = \begin{pmatrix} 0 & -1 & 4 & | & -2 & 0 & 1 \end{pmatrix} \]

Now the augmented matrix looks like this:
\[ \begin{pmatrix} 1 & 3 & -2 & | & 1 & 0 & 0 \\ 0 & 9 & -11 & | & 3 & 1 & 0 \\ 0 & -1 & 4 & | & -2 & 0 & 1 \end{pmatrix} \]

### Step 2: Further Row Operations

3. **Row 2 Transformation**: Replace \( R_2 \) with \( \frac{1}{9}R_2 \)
\[ R_2 = \frac{1}{9}R_2 \Rightarrow \begin{pmatrix} 0 & 1 & -\frac{11}{9} & | & \frac{1}{3} & \frac{1}{9} & 0 \end{pmatrix} \]

4. **Row 3 Transformation**: Replace \( R_3 \) with \( R_3 + R_2 \)
\[ R_3 = R_3 + R_2 \Rightarrow \begin{pmatrix} 0 & 0 & \frac{25}{9} & | & -\frac{5}{9} & \frac{1}{9} & 1 \end{pmatrix} \]

Now the augmented matrix looks like this:
\[ \begin{pmatrix} 1 & 3 & -2 & | & 1 & 0 & 0 \\ 0 & 1 & -\frac{11}{9} & | & \frac{1}{3} & \frac{1}{9} & 0 \\ 0 & 0 & \frac{25}{9} & | & -\frac{5}{9} & \frac{1}{9} & 1 \end{pmatrix} \]

### Step 3: Normalize Row 3

5. **Row 3 Transformation**: Replace \( R_3 \) with \( \frac{9}{25}R_3 \)
\[ R_3 = \frac{9}{25}R_3 \Rightarrow \begin{pmatrix} 0 & 0 & 1 & | & -\frac{1}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} \]

### Step 4: Back Substitution

6. **Row 1 Transformation**: Replace \( R_1 \) with \( R_1 + 2R_3 \)
\[ R_1 = R_1 + 2R_3 \Rightarrow \begin{pmatrix} 1 & 3 & 0 & | & \frac{1}{5} & \frac{2}{25} & \frac{18}{25} \end{pmatrix} \]

7. **Row 2 Transformation**: Replace \( R_2 \) with \( R_2 + \frac{11}{9}R_3 \)
\[ R_2 = R_2 + \frac{11}{9}R_3 \Rightarrow \begin{pmatrix} 0 & 1 & 0 & | & \frac{1}{3} - \frac{11}{45} & \frac{1}{9} + \frac{11}{225} & \frac{11}{25} \end{pmatrix} \]

Now we have the identity matrix on the left side:
\[ \begin{pmatrix} 1 & 0 & 0 & | & \frac{1}{5} & -\frac{2}{5} & -\frac{3}{5} \\ 0 & 1 & 0 & | & -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ 0 & 0 & 1 & | & -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} \]

### Final Result

The inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} \frac{1}{5} & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} \]