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

Options

A. `[{:(1,1),(1,1):}]`
B. `[{:(-1,-1),(-1,-1):}]`
C. `[{:(2,2),(2,2):}]`
D. none of these

Answer

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: Rewrite \( f(A) \)
We can express \( f(A) \) in terms of the identity matrix \( I \):
\[ f(A) = \frac{1 + A}{1 - A} \]
This means we need to compute \( 1 + A \) and \( 1 - A \).

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

### Step 3: Calculate \( 1 - A \)
Now, we subtract \( A \) from \( I \):
\[ 1 - A = I - A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1-1 & 0-2 \\ 0-2 & 1-1 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]

### Step 4: Compute \( f(A) = (1 + A)(1 - A)^{-1} \)
Next, we need to find the inverse of \( 1 - A \):
\[ 1 - A = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]
To find the inverse, we first calculate the determinant:
\[ \text{det}(1 - A) = (0)(0) - (-2)(-2) = -4 \]
Now, the adjoint of \( 1 - A \) is:
\[ \text{adj}(1 - A) = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]
Thus, the inverse is:
\[ (1 - A)^{-1} = \frac{1}{\text{det}(1 - A)} \cdot \text{adj}(1 - A) = \frac{1}{-4} \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix} \]

### Step 5: Multiply \( (1 + A) \) and \( (1 - A)^{-1} \)
Now we multiply \( 1 + A \) and \( (1 - A)^{-1} \):
\[ f(A) = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix} \]
Calculating the product:
\[ f(A) = \begin{pmatrix} 2 \cdot 0 + 2 \cdot \frac{1}{2} & 2 \cdot \frac{1}{2} + 2 \cdot 0 \\ 2 \cdot 0 + 2 \cdot \frac{1}{2} & 2 \cdot \frac{1}{2} + 2 \cdot 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

### Final Result
Thus, the final result is:
\[ f(A) = -\frac{1}{4} \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} = -\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]