Let `A=[(1//2, 3//4),(1, -1//2)]`, then the value of sum of all the elements of `A^(100)` is

Answer

To solve the problem, we need to find the sum of all the elements of the matrix \( A^{100} \), where \( A = \begin{pmatrix} \frac{1}{2} & \frac{3}{4} \\ 1 & -\frac{1}{2} \end{pmatrix} \).

### Step-by-Step Solution:

1. **Calculate \( A^2 \)**:
We start by calculating \( A^2 \) (which is \( A \times A \)):
\[ A^2 = A \cdot A = \begin{pmatrix} \frac{1}{2} & \frac{3}{4} \\ 1 & -\frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{2} & \frac{3}{4} \\ 1 & -\frac{1}{2} \end{pmatrix} \]

To perform the multiplication, we calculate each element of the resulting matrix:
- Element at (1,1):
\[ \frac{1}{2} \cdot \frac{1}{2} + \frac{3}{4} \cdot 1 = \frac{1}{4} + \frac{3}{4} = 1 \]
- Element at (1,2):
\[ \frac{1}{2} \cdot \frac{3}{4} + \frac{3}{4} \cdot -\frac{1}{2} = \frac{3}{8} - \frac{3}{8} = 0 \]
- Element at (2,1):
\[ 1 \cdot \frac{1}{2} + -\frac{1}{2} \cdot 1 = \frac{1}{2} - \frac{1}{2} = 0 \]
- Element at (2,2):
\[ 1 \cdot \frac{3}{4} + -\frac{1}{2} \cdot -\frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]
where \( I \) is the identity matrix.

2. **Calculate \( A^{100} \)**:
Since \( A^2 = I \), we can find \( A^{100} \) as follows:
\[ A^{100} = (A^2)^{50} = I^{50} = I \]

3. **Sum of all elements of \( A^{100} \)**:
The identity matrix \( I \) is given by:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
The sum of all elements of \( I \) is:
\[ 1 + 0 + 0 + 1 = 2 \]

### Final Answer:
The value of the sum of all the elements of \( A^{100} \) is \( 2 \).