Find the inverse of the following , if it exists, by using elementary row (column) transformations :
`[(-5,4),(-6,5)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} -5 & 4 \\ -6 & 5 \end{pmatrix} \) using elementary row transformations, we follow these steps:

### Step 1: Set up the augmented matrix
We start with the matrix \( A \) and augment it with the identity matrix \( I \):
\[ \begin{pmatrix} -5 & 4 & | & 1 & 0 \\ -6 & 5 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Make the leading coefficient of the first row 1
To make the leading coefficient of the first row equal to 1, we can multiply the first row by \(-\frac{1}{5}\):
\[ R_1 \leftarrow -\frac{1}{5} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{4}{5} & | & -\frac{1}{5} & 0 \\ -6 & 5 & | & 0 & 1 \end{pmatrix} \]

### Step 3: Eliminate the first column of the second row
Next, we want to eliminate the \(-6\) in the first column of the second row. We can do this by adding \(6\) times the first row to the second row:
\[ R_2 \leftarrow R_2 + 6R_1 \]
Calculating this gives:
\[ R_2 = (-6 + 6 \cdot 1, 5 + 6 \cdot -\frac{4}{5}, 0 + 6 \cdot -\frac{1}{5}, 1 + 6 \cdot 0) \]
\[ R_2 = (0, 5 - \frac{24}{5}, -\frac{6}{5}, 1) \]
This simplifies to:
\[ R_2 = (0, \frac{1}{5}, -\frac{6}{5}, 1) \]
So our augmented matrix now looks like:
\[ \begin{pmatrix} 1 & -\frac{4}{5} & | & -\frac{1}{5} & 0 \\ 0 & 1 & | & -\frac{6}{5} & 1 \end{pmatrix} \]

### Step 4: Make the leading coefficient of the second row 1
The second row already has a leading coefficient of 1, so we can proceed to eliminate the \(-\frac{4}{5}\) in the first row's second column. We do this by adding \(\frac{4}{5}\) times the second row to the first row:
\[ R_1 \leftarrow R_1 + \frac{4}{5}R_2 \]
Calculating this gives:
\[ R_1 = (1 + 0, -\frac{4}{5} + \frac{4}{5} \cdot 1, -\frac{1}{5} + \frac{4}{5} \cdot -\frac{6}{5}, 0 + \frac{4}{5} \cdot 1) \]
This simplifies to:
\[ R_1 = (1, 0, -\frac{1}{5} - \frac{24}{25}, \frac{4}{5}) \]
\[ R_1 = (1, 0, -\frac{25}{25}, \frac{4}{5}) = (1, 0, -1, \frac{4}{5}) \]
So our augmented matrix now looks like:
\[ \begin{pmatrix} 1 & 0 & | & -1 & \frac{4}{5} \\ 0 & 1 & | & -\frac{6}{5} & 1 \end{pmatrix} \]

### Step 5: Write the inverse
From the augmented matrix, we can read off the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} -1 & \frac{4}{5} \\ -\frac{6}{5} & 1 \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} -1 & \frac{4}{5} \\ -\frac{6}{5} & 1 \end{pmatrix} \]