Using elementary transformation, find the inverse of the matrix `A=[(a,b),(c,((1+bc)/a))]`.

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} a & b \\ c & \frac{1 + bc}{a} \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations to transform \( A \) into \( I \). The augmented matrix will look like this:

\[ \left( \begin{array}{cc|cc} a & b & 1 & 0 \\ c & \frac{1 + bc}{a} & 0 & 1 \end{array} \right) \]

### Step 1: Normalize the first row
We start by dividing the first row by \( a \):

\[ R_1 \rightarrow \frac{1}{a} R_1 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ c & \frac{1 + bc}{a} & 0 & 1 \end{array} \right) \]

### Step 2: Eliminate the first element of the second row
Next, we will eliminate the \( c \) in the second row by performing the operation:

\[ R_2 \rightarrow R_2 - c R_1 \]

This results in:

\[ \left( \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & \frac{1 + bc}{a} - c \cdot \frac{b}{a} & -c \cdot \frac{1}{a} & 1 \end{array} \right) \]

Simplifying the second row:

\[ \frac{1 + bc - cb}{a} = \frac{1}{a} \]

Thus, we have:

\[ \left( \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & \frac{1}{a} & -\frac{c}{a} & 1 \end{array} \right) \]

### Step 3: Normalize the second row
Now, we will normalize the second row by multiplying it by \( a \):

\[ R_2 \rightarrow a R_2 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & 1 & -c & a \end{array} \right) \]

### Step 4: Eliminate the second element of the first row
Next, we eliminate the \( \frac{b}{a} \) in the first row:

\[ R_1 \rightarrow R_1 - \frac{b}{a} R_2 \]

This results in:

\[ \left( \begin{array}{cc|cc} 1 & 0 & \frac{1}{a} + \frac{bc}{a} & -\frac{b}{a} \\ 0 & 1 & -c & a \end{array} \right) \]

### Final Result
The final augmented matrix is:

\[ \left( \begin{array}{cc|cc} 1 & 0 & \frac{1 + bc}{a} & -b \\ 0 & 1 & -c & a \end{array} \right) \]

Thus, the inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} \frac{1 + bc}{a} & -b \\ -c & a \end{pmatrix} \]