Find the inverse of the following , if it exists, by using elementary row (column) transformations :
`[(4,5),(3,4)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \) using elementary row transformations, we will augment the matrix with the identity matrix and perform row operations until we obtain the identity matrix on the left side.

### Step-by-Step Solution:

1. **Set up the augmented matrix**:
\[ \left( \begin{array}{cc|cc} 4 & 5 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array} \right) \]

2. **Make the leading coefficient of the first row equal to 1**:
We can do this by dividing the first row by 4:
\[ R_1 \leftarrow \frac{1}{4} R_1 \]
This gives:
\[ \left( \begin{array}{cc|cc} 1 & \frac{5}{4} & \frac{1}{4} & 0 \\ 3 & 4 & 0 & 1 \end{array} \right) \]

3. **Eliminate the first entry of the second row**:
We can eliminate the 3 in the second row by performing:
\[ R_2 \leftarrow R_2 - 3R_1 \]
This results in:
\[ \left( \begin{array}{cc|cc} 1 & \frac{5}{4} & \frac{1}{4} & 0 \\ 0 & -\frac{1}{4} & -\frac{3}{4} & 1 \end{array} \right) \]

4. **Make the leading coefficient of the second row equal to 1**:
We can do this by multiplying the second row by -4:
\[ R_2 \leftarrow -4R_2 \]
This gives:
\[ \left( \begin{array}{cc|cc} 1 & \frac{5}{4} & \frac{1}{4} & 0 \\ 0 & 1 & 3 & -4 \end{array} \right) \]

5. **Eliminate the second entry of the first row**:
We can eliminate the \(\frac{5}{4}\) in the first row by performing:
\[ R_1 \leftarrow R_1 - \frac{5}{4}R_2 \]
This results in:
\[ \left( \begin{array}{cc|cc} 1 & 0 & -\frac{5}{4} & 5 \\ 0 & 1 & 3 & -4 \end{array} \right) \]

6. **Final form**:
The left side is now the identity matrix, and the right side gives us the inverse:
\[ A^{-1} = \begin{pmatrix} -\frac{5}{4} & 5 \\ 3 & -4 \end{pmatrix} \]

### Final Answer:
The inverse of the matrix \( A = \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{5}{4} & 5 \\ 3 & -4 \end{pmatrix} \]