If `A=[(1,2),(2,1)]` and `f(x)=(1+x)/(1-x)`, then f(A) is

Options

A. `[[1 ,1],[1,1]]`
B. `[[2 ,2],[2,2]]`
C. `[[-1 ,-1],[-1,-1]]`
D. None of these

Explanation

To find \( f(A) \) where \( A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \) and \( f(x) = \frac{1+x}{1-x} \), we can follow these steps:

### Step 1: Express \( f(A) \)
We can express \( f(A) \) using the formula:
\[ f(A) = I + A (I - A)^{-1} \]
where \( I \) is the identity matrix.

### Step 2: Calculate \( I - A \)
The identity matrix \( I \) for a 2x2 matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Now, calculate \( I - A \):
\[ I - A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]

### Step 3: Find the inverse of \( I - A \)
To find the inverse of \( I - A \), we first calculate its determinant:
\[ \text{det}(I - A) = (0)(0) - (-2)(-2) = 0 - 4 = -4 \]
Now, the adjoint of \( I - A \) is:
\[ \text{adj}(I - A) = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix} \]
Thus, the inverse is given by:
\[ (I - A)^{-1} = \frac{1}{\text{det}(I - A)} \cdot \text{adj}(I - A) = \frac{1}{-4} \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 0 \end{pmatrix} \]

### Step 4: Calculate \( A (I - A)^{-1} \)
Now compute \( A (I - A)^{-1} \):
\[ A (I - A)^{-1} = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 0 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} 1 \cdot 0 + 2 \cdot -\frac{1}{2} & 1 \cdot -\frac{1}{2} + 2 \cdot 0 \\ 2 \cdot 0 + 1 \cdot -\frac{1}{2} & 2 \cdot -\frac{1}{2} + 1 \cdot 0 \end{pmatrix} = \begin{pmatrix} -1 & -\frac{1}{2} \\ -\frac{1}{2} & -1 \end{pmatrix} \]

### Step 5: Calculate \( f(A) \)
Now, we can substitute back into the expression for \( f(A) \):
\[ f(A) = I + A (I - A)^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} -1 & -\frac{1}{2} \\ -\frac{1}{2} & -1 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 0 \end{pmatrix} \]

### Final Result
Thus, the final result is:
\[ f(A) = \begin{pmatrix} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 0 \end{pmatrix} \]