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

Answer

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

### Step 1: Write the augmented matrix
We start by writing the augmented matrix \([A | I]\), where \(I\) is the identity matrix of the same size as \(A\):
\[ \begin{pmatrix} -4 & 3 & | & 1 & 0 \\ -5 & 4 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Find the determinant of \(A\)
To check if the inverse exists, we calculate the determinant of \(A\):
\[ \text{det}(A) = (-4)(4) - (-5)(3) = -16 + 15 = -1 \]
Since the determinant is not zero, the inverse exists.

### Step 3: Make the leading coefficient of the first row equal to 1
To make the leading coefficient of the first row equal to 1, we can divide the first row by -4:
\[ R_1 \leftarrow \frac{1}{-4} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{3}{4} & | & -\frac{1}{4} & 0 \\ -5 & 4 & | & 0 & 1 \end{pmatrix} \]

### Step 4: Eliminate the first column of the second row
Next, we eliminate the first column of the second row by adding 5 times the first row to the second row:
\[ R_2 \leftarrow R_2 + 5R_1 \]
This results in:
\[ \begin{pmatrix} 1 & -\frac{3}{4} & | & -\frac{1}{4} & 0 \\ 0 & \frac{1}{4} & | & -\frac{5}{4} & 1 \end{pmatrix} \]

### Step 5: Make the leading coefficient of the second row equal to 1
Now, we need to make the leading coefficient of the second row equal to 1 by multiplying the second row by 4:
\[ R_2 \leftarrow 4R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{3}{4} & | & -\frac{1}{4} & 0 \\ 0 & 1 & | & -5 & 4 \end{pmatrix} \]

### Step 6: Eliminate the second column of the first row
Finally, we eliminate the second column of the first row by adding \(\frac{3}{4}\) times the second row to the first row:
\[ R_1 \leftarrow R_1 + \frac{3}{4}R_2 \]
This results in:
\[ \begin{pmatrix} 1 & 0 & | & 3 & -3 \\ 0 & 1 & | & -5 & 4 \end{pmatrix} \]

### Step 7: Read off the inverse
Now, we can read off the inverse matrix from the augmented matrix:
\[ A^{-1} = \begin{pmatrix} 3 & -3 \\ -5 & 4 \end{pmatrix} \]

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