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

Options

A. `-4[(1,1),(1,1)]`
B. `-8[(1,1),(1,1)]`
C. `4[(1,1),(1,1)]`
D. None of these

Answer

To find \( f(A) \) where \( A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \) and \( f(x) = (1+x)(1-x) \), we first need to evaluate \( f(x) \) in terms of a matrix.

### Step 1: Define the function \( f(x) \)
The function is given as:
\[ f(x) = (1+x)(1-x) = 1 - x^2 \]

### Step 2: Apply the function to the matrix \( A \)
To find \( f(A) \), we will replace \( x \) in \( f(x) \) with the matrix \( A \):
\[ f(A) = 1 - A^2 \]

### Step 3: Calculate \( A^2 \)
First, we need to compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4 \)
- Second row, first column: \( 2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4 \)
- Second row, second column: \( 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} \]

### Step 4: Compute \( f(A) \)
Now, substituting \( A^2 \) back into the function:
\[ f(A) = 1 - A^2 = I - A^2 \]
Where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Now, we compute \( I - A^2 \):
\[ f(A) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 1-5 & 0-4 \\ 0-4 & 1-5 \end{pmatrix} = \begin{pmatrix} -4 & -4 \\ -4 & -4 \end{pmatrix} \]

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