Using elementary transformations, find the inverse of the matrix. `({:(0,0,-1),(3,4,5),(-2,-4,-7):})`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{pmatrix} \) using elementary transformations, we will follow these steps:

### Step 1: Form the Augmented Matrix
We start by forming the augmented matrix \( [A | I] \), where \( I \) is the identity matrix of the same size as \( A \).

\[ [A | I] = \begin{pmatrix} 0 & 0 & -1 & | & 1 & 0 & 0 \\ 3 & 4 & 5 & | & 0 & 1 & 0 \\ -2 & -4 & -7 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Apply Elementary Row Operations
We will use elementary row operations to transform the left side of the augmented matrix into the identity matrix.

1. **Swap Row 1 and Row 2** to get a leading 1 in the first row:
\[ R_1 \leftrightarrow R_2 \]
Resulting matrix:
\[ \begin{pmatrix} 3 & 4 & 5 & | & 0 & 1 & 0 \\ 0 & 0 & -1 & | & 1 & 0 & 0 \\ -2 & -4 & -7 & | & 0 & 0 & 1 \end{pmatrix} \]

2. **Make the first column below the leading 1 zero**:
- For Row 3: \( R_3 + \frac{2}{3} R_1 \)
\[ R_3 \rightarrow R_3 + \frac{2}{3} R_1 \]
Resulting matrix:
\[ \begin{pmatrix} 3 & 4 & 5 & | & 0 & 1 & 0 \\ 0 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 0 & -\frac{1}{3} & | & 0 & \frac{2}{3} & 1 \end{pmatrix} \]

3. **Scale Row 1 to make the leading coefficient 1**:
\[ R_1 \rightarrow \frac{1}{3} R_1 \]
Resulting matrix:
\[ \begin{pmatrix} 1 & \frac{4}{3} & \frac{5}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 0 & -\frac{1}{3} & | & 0 & \frac{2}{3} & 1 \end{pmatrix} \]

4. **Make the third row leading coefficient positive**:
\[ R_2 \rightarrow -R_2 \]
Resulting matrix:
\[ \begin{pmatrix} 1 & \frac{4}{3} & \frac{5}{3} & | & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 0 \\ 0 & 0 & -\frac{1}{3} & | & 0 & \frac{2}{3} & 1 \end{pmatrix} \]

5. **Eliminate the third column in Row 1 and Row 3**:
- For Row 1: \( R_1 - \frac{5}{3} R_2 \)
- For Row 3: \( R_3 + \frac{1}{3} R_2 \)
Resulting matrix:
\[ \begin{pmatrix} 1 & \frac{4}{3} & 0 & | & \frac{5}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 0 \\ 0 & 0 & 0 & | & -\frac{1}{3} & \frac{2}{3} & 1 \end{pmatrix} \]

6. **Make Row 3 leading coefficient 1**:
\[ R_3 \rightarrow -3 R_3 \]
Resulting matrix:
\[ \begin{pmatrix} 1 & \frac{4}{3} & 0 & | & \frac{5}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 & | & -1 & 0 & 0 \\ 0 & 0 & 0 & | & 1 & -2 & -3 \end{pmatrix} \]

7. **Make the second column leading coefficient zero**:
- For Row 1: \( R_1 - \frac{4}{3} R_2 \)
Resulting matrix:
\[ \begin{pmatrix} 1 & 0 & 0 & | & 5 & 3 & 4 \\ 0 & 0 & 1 & | & -1 & 0 & 0 \\ 0 & 0 & 0 & | & 1 & -2 & -3 \end{pmatrix} \]

### Step 3: Extract the Inverse
The right side of the augmented matrix now represents the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} 5 & 3 & 4 \\ -1 & 0 & 0 \\ 1 & -2 & -3 \end{pmatrix} \]

### Summary
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 5 & 3 & 4 \\ -1 & 0 & 0 \\ 1 & -2 & -3 \end{pmatrix} \]